 # Solutions

## Chemistry

### NCERT

1   Calculate the mass percentage of benzene $(C_6 H_6 )$ and carbon tetrachloride $(CCl_4 )$ if $22 g$ of benzene is dissolved in $122 g$ of carbon tetrachloride.

##### Solution :

Mass percentage of $C_6 H_6 =\dfrac{\text{Mass of$C_6H_6$}}{\text{Total mass of the solution}}*100$%$\\$ $=\dfrac{\text{Mass of$C_6H_6$}}{\text{Mas of$C_6H_6$+ Mass of$CCl_4$}}*100$%$\\$ $=\dfrac{22}{22+122}*100$%$\\$ $=15.28$%$\\$ Mass percentage of $CCl_4$=$\\$ $\dfrac{\text{Mass of$CCl_4$}}{\text{Mass of$C_6H_6$+Mass of$CCl_4$}}*100$%$\\$ $=\dfrac{122}{22+122}*100$%$\\$ $=84.72$%$\\$ Alternatively,$\\$ Mass percentage of $CCl_4 = (100-15.28) = 84.72$%$\\$

2   Calculate the mole fraction of benzene in solution containing $30$% by mass in carbon tetrachloride.

##### Solution :

$30$% by mass of $C_6 H_6$ in $CCl_4 \Rightarrow 30 g C_6 H_6$ in $100 g$ solution$\\$ $\therefore$ Number of moles of $C_6 H_6 ( n^ C_6 H_6 ) = 30/78 = 0.385$ mol$\\$ (molar mass of $C_6 H_6 = 78g$)$\\$ No. of moles of $\\$ $CCl_4(n^CCl_4)=\dfrac{70}{154}=0.455\\ x_{C_{6}H_{6}}=\dfrac{n^C_6H_6}{n^C_6H_6+n^CCl_4}\\ =\dfrac{0.385}{0.385+0.455}\\ =\dfrac{0.385}{0.84}\\ =0.458\\ X_{CCl_{14}}=1-0.458=0.542$

3   Calculate the molarity of each of the following solution: $\\$ (a) $30 g$ of $Co(NO _3 ) _2 . 6H _2 O$ in $4.3 L$ of solution$\\$ (b) $30 mL$ of $0.5 M H_2 SO_4$ diluted to $500 mL.$

##### Solution :

(a) Molar mass of $Co(NO _3 )_ 2 .6H _2 O = 310.7 g mol ^{-1}$$\\ no. of moles = 30/310.7 = 0.0966$$\\$ Vol. of solution $= 4.3 L$$\\ Molarity = 0.0966/4.3 = 0.022 M$$\\$ (b) $1000 mL$ of $0.5M H_2 SO_4$ contain $H_2 SO_4 = 0.5$ mole$\\$ $30 mL$ of $0.5 M H_2 SO_4$ contain $H_2 SO_4$$\\ = 0.5/1000 × 30 = 0.015 mole \\ Volume of solution = 500mL = 0.5L$$\\$ Molarity =$0.015/0.5 = 0.03M$

4   Calculate the mass of urea $(NH_2 CONH_2 )$ required in making $2.5 kg$ of $0.25$ molal aqueous solution.

$0.25$ molal aqueous solution to urea means that Moles of urea$= 0.25$ mole$\\$ Mass of solvent $(NH_2 CONH_2 ) = 60 g$ mol $^{-1}$$\\ \therefore 0.25 mole of urea = 0.25 × 60 = 15g$$\\$ Mass of solution $= 1000 + 15g = 1.015 kg$$\\ 1.015 kg of urea solution contains 15g of urea\\ \therefore 2.5 kg of solution contains urea = 15/1.015 × 2.5 = 37 g. 5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL ^{-1} ##### Solution : 20% aq. KOH solution \Rightarrow 20g of KI in 100g solution \\ \therefore Mass of solvent = 100 - 20= 80 g$$\\$ (i) Molality=$\dfrac{\text{no of moles of KI}}{\text{mass of solvent(kg)}}\\ =\dfrac{0.120}{0.080}=1.5 mol kg^{-1}$$\\ (ii)Density of solution = 1.202 g mL ^{-1}$$\\$ Volume of solution $=\dfrac{100}{1.202}=83.2 mL$$\\ =0.0832L$$\\$ $\therefore$ Molarity $= \dfrac{0.120}{0.0832}=1.44M$$\\ (iii) No of moles of Kl = 0.120$$\\$ $n_{H_2O}=\dfrac{80}{18}=4.44$$\\ x_{Kl}=\dfrac{0.120}{0.120+4.44}$$\\$ $=\dfrac{0.120}{4.560}=0.0263$

6   $H_ 2 S$, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of $H _2 S$ in water at $STP$ is $0.195 m$, calculate Henry’s law constant.

##### Solution :

Solubility of $H_ 2 S \ gas = 0.195 m = 0.195$ mole in $1 kg$ of solvent$\\$ $1 kg$ of solvent =$1000g$ $\\$ $=\dfrac{1000}{18}=55.55$ moles $\\$ $\therefore X_{H_2S}=\dfrac{0.195}{0.195+55.55}$ $\\$ $=\dfrac{0.195}{55.745}0.0035$ $\\$ Pressure at $STP = 0.987$ bar$\\$ Applying Henry’s law$\\$ $P_{H_2S}=K_H*X_{H_2S}$ $\\$ $\Rightarrow K_H=\dfrac{P_{H_2S}}{X_{H_2S}}=\dfrac{0.987}{0.0035}=282$ bar $\\$

7   Heny’s law constant for $CO^ 2$ in water is $1.67 × 10^ 8$ Pa at $298 K$. Calculate the quantity of $CO_ 2$ in $500 mL$ of soda water when packed under $2.5$ atm $CO _2$ pressure at $298 K.$

##### Solution :

$KH = 1.67 × 10 ^8$Pa$\\$ $P_{ CO_ 2} = 2.5$ atm $= 2.5 × 101325$ Pa$\\$ $\therefore X_{ CO_ 2} \dfrac{p_{ CO _2}} {k_ H} = \dfrac{2.5 * 101325}{1.67 * 10^8}= 1.517 *10 ^{- 3}$ $\\$ For $500 mL$ of soda water, water $\cong 500 mL$ $\\$ $=500 g =\dfrac{500}{18} = 27.78$ moles$\\$ $\therefore n_{ H _2 O} =27.78$ moles$\\$ $\therefore \dfrac{n_{ CO_ 2}}{27.78}= 27.78$ moles $\\$ $n_{ CO _2} = 42.14 * 10 ^{- 3}$ mole$\\$ $= 42.14m$ mole $\\$ $= 42.14 × 10^{ -3} × 44g$ $\\$ $= 1.854 g$

8   The vapour pressure of pure liquids A and B are $450$ and $700 mm$ Hg respectively, at $350 K$. Find out the composition of the liquid mixture if total vapour pressure is $600 mm$ Hg. Also find the composition of the vapour phase.

##### Solution :

$P_A=450 mm, P_B=700mm, P_{total }=600 mm$ $\\$ As P total =$P_A+P_B$ $\\$ $=X_AP_A +(1-X_A)P_B$ $\\$ $=P_B+(P_A-P_B)X_A$ $\\$ $\Rightarrow 600=700+(450-700)X_A$ $\\$ Or $X_A=0.40$ $\\$ $\therefore X_B=1-X_A =1-1.40=0.60$ $\\$ $\therefore P_A=X_AP_A=0.40*450=180 mm$ $\\$ $\therefore P_B=X_BP_A=0.60*700 =420 mm$ $\\$ $\therefore$ Mole fraction of $A$ in vapour phase $\\$ $=\dfrac{P_A}{P_A+P_B}=\dfrac{180}{180+420}=0.30$ $\\$ And, mole fraction of $B$ in vapour phase $\\$ $=1-0.30=0.70$

9   Vapour pressure of pure water at $298 K$ is $23.8 mm Hg. 50 g$ of urea $(NH_ 2 CONH _2 )$ is dissolved in $850 g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

##### Solution :

$P^o = 23.8 mm Hg$ $\\$ $W2 = 50g, M _2 ($urea$) = 60g$ mol$^ {-1}$ $\\$ $w1 = 850g, M _1 ($water$) = 18g$mol $^{ -1}$ $\\$ To find: Ps and $(P^o- P _s )/ P^o$ $\\$ Solution: Applying Raoult’s law, $\\$ $\dfrac{P^o-P_s}{P^o}=\dfrac{n_2}{n_1+n_2}\dfrac{w_2/M_2}{w_1/M_1+w_2/M_2}$ $\\$ $\therefore \dfrac{P^o-P_s}{P^o}=\dfrac{50/60}{850/18+50/60}$ $\\$ $=\dfrac{0.83}{47.22+0.83}=0.017$ $\\$ Putting $P^o=23.8 mm,$ we have$\\$ $\dfrac{23.8-P_s}{P_s}=0.017$ $\\$ $\Rightarrow 23.8-P_s=0.017 P_s$ $\\$ or $1.017 P_s=23.8$ $\\$ or $P_s=23.4 mm$

10   Boiling point of water at $750 mm Hg$ is $99.63^oC$. How much sucrose is to be added to $500 g$ of water such that it boils at $100^oC.$

##### Solution :

Given $\Delta T_ b = 100 - 99.63 = 3.37^o$ $\\$ Mass of water $w_ 1 = 500g$ $\\$ Molar mass of water, $M_ 1 = 18g$ mol$^{ -1}$ $\\$ Molar mass of sucrose, $M _2 = 342g$ mol$^{ -1}$ $\\$ To find Mass of sucrose, $w _2 = ?$ $\\$ We know, $\Delta T _b = K_ b × m$ $\\$ $=K_b*\dfrac{w_2}{M_2}*\dfrac{100}{w_1}$ $\\$ $\Rightarrow w_2=\dfrac{M_2*w_1*\Delta T_b}{1000* K_b}\\ =\dfrac{342*500*3.37}{1000*0.52}$ $\\$ $w_2=1108.2g$ $\\$ $\therefore$ Mass of solute, $w_2=1.11 kg$

11   Calculate the mass of ascorbic acid (Vitamin $C, C _6 H_ 8 O_ 6$ ) to be dissolved in $75 g$ of acetic acid to lower its melting point by $1.5*C. K f = 3.9\ kg \ \ mol ^{-1}$

##### Solution :

Given: $\Delta T _f = 1.5^o$ $\\$ Mass of $CH _3 COOH, w_ 1 =75g\\ M_ 1 = 60g mol ^{-1}\\ M _2 (C _6 H_ 8 O) = 176g mol^{ -1}\\ K_ f = 3.9\ kg \ mol ^{-1}$ $\\$ To find: $w _2 = ?$ $\\$ Solution: Applying $M_ 2=\dfrac{1000K_fw_2}{w_1\Delta T_f }$ $\\$ Or $w _2 =\dfrac{M_2*w_1*\Delta T_f}{1000*K_f}\\ W_2=\dfrac{176*75*1.5}{1000*3.9}=5.077g$

12   Calculate the osmotic pressure in pascals exerted by solution prepared by dissolving $1.0 g$ of polymer of molar mass $185,000$ in $450 mL$ of water at $37^oC$

Given $V = 450mL = 0.45L\\ T = 37^oC = 310K\\ R = 8.314 kPaL\ \ k ^{-1} mol ^{ -1}$$\\ To find:\\ Solution: Applying the formula,\\ \pi=CRT=\dfrac{n}{V}RT\\ n=\dfrac{1.0g}{185,000g \ mol^{-1}}\\ \therefore P=\dfrac{1}{185,000}*\dfrac{1}{0.45}*8.314*10^3 PaLK^{-1}*310K\\ =30.96 Pa \\ 13 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example. ##### Solution : A Solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solution: There are nine types of solution.\\ Types of Solution Examples\\ Gaseous solutions:\\ (a) Gas in gas: Air, mixture of O _2 and N _2 , etc.\\ (b) Liquid in gas: Water vapour\\ (c) Solid in gas: Camphor vapours in N _2 gas, smoke etc.\\ Liquid solutions\\ (a) Gas in liquid: CO _2 dissolved in water (aerated water), and O _2 dissolved in water, etc.\\ (b) Liquid in liquid: Ethanol dissolved in water, etc.\\ (c) Solid in liquid: Sugar dissolved in water, saline water, etc.\\ Solid Solutions\\ (a) Gas in solid: Solution of hydrogen in palladium\\ (b) Liquid in solid:Amalgams,eg.Na-Hg.\\ (c) Solid in solid: Gold ornaments (Cu/Ag with Au).\\ 14 Give an example of a solid solution in which the solute is a gas. ##### Solution : Solution of hydrogen in palladium and dissolved gases in minerals 15 Define the following terms:\\ (i) Mole fraction\\ (ii) Molality\\ (iii) Molarity\\ (iv) Mass percentage\\ ##### Solution : (i) \text{Mole Fraction:}$$\\$ It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute.$\\$ $(X_A=\dfrac{n_A}{n_A+n_B})....(1)$ $\\$ Mole fraction of solvent $X_ B =\dfrac{n_b}{n_A+n_B}....(2)$ $\\$ Adding the above two equations, we get$\\$ $X_A+X_B=\dfrac{n_A}{n_A+n_B}+\dfrac{n_B}{n_A+n_B}+\dfrac{n_A+n_B}{n_A+n_B}=1\\ i.e.,X_A+X_B=1\\ \therefore X_A=1-X_B$ or $X_B=1-X_A$ $\\$

(ii) Molality: It is defined as the number of moles of a solute present in 1000g (1kg) of a solvent.$\\$ Molality (m) =$\dfrac{\text{Number of moles of solute}}{\text{Weight of solvent int in kg}}=\dfrac{n}{W}$ $\\$Note: Molality is considered better way of expressing concentration of solutions, as compared to molarity because molality does not change with change in temperature since the mass of solvent does not vary with temperature.$\\$ (iii) Molarity: It is defined as the number of moles of solute present in one litre of solution.$\\$ Moloarity(M)=$\dfrac{\text{Number of moles of solute}}{\text{Volume of Solution in litre}}=\dfrac{n}{V}\\ n=\dfrac{\text{Weight in grams}}{\text{Molecular weight of solute}}\\ \therefore M=\dfrac{\text{Weight in grams}}{\text{Volumeof solutioninlitres}}*\dfrac{1}{\text{Molecular weight of solute}}$ $\\$ Strength: This is weight (in gms) of solute per litre of solution$\\$ $\therefore Molarity=\dfrac{\text{Strength}}{\text{Molecular weight of solute}}$ $\\$ or Strength = Molarity × Molecular weight$\\$ Note: Molarity is the most common way of expressing concentration of a solution in laboratory. However, it has one disadvantage. It changes with temperature because volume of a solution alters due to expansion and contraction of the liquid with temperature.$\\$ (iv) Mass Percentage: It is the amount of solute in grams present in 100g of solution.$\\$ $=\dfrac{\text{Mass of solute}}{\text{Massof solution}}*100$

16   Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is $1.504\ g m \ L^{ -1 }?$

##### Solution :

68% nitric acid by mass means that 68g mass of nitric acid is dissolved in 100g mass the solution. Molar mass of $HNO_3 = 63 g mol ^{-1}\\ \therefore 68 g \ \ of \ \ HNO_3=\dfrac{68}{63}=1.079 \ mole$ $\\$ Density of solution$=1.504 g \ mL^{-1}$ given $\\$ $\therefore$ Volume of solution $\\$ $-\dfrac{\text{mass}}{\text{density}}=\dfrac{100}{1.504}=66.5 mL$ $\\$ $\therefore$ Molarity of solution $\\$ $=\dfrac{\text{moles of slolute}*1000}{\text{Volume of solution in mL}}\\ =\dfrac{1.079*1000}{65}=16.23 M$

17   A solution of glucose in water is labelled as 10% w/w, that would be the molality and mole fraction of each component in the solution? If the density of solution is $1.2g mL^{ -1}$ then what shall be the molarity of the solution?

##### Solution :

10 percent w/w solution of glucose in water means 10g glucose and 90g of water.$\\$ $\therefore$ 10g of glucose $=\dfrac{10}{180}=0.0555$ moles $\\$ and $90g \ of \ \ H2O=\dfrac{90}{18}=5$ moles $\\$ $\therefore$ Molality of solution $\\$ $=\dfrac{\text{Moles of solute}*1000}{\text{Mass of solution in grams}}\\ =\dfrac{0.0555}{90}*1000=0.617 m$ $\\$ moles fraction of glucose$\\$ $=X_g=\dfrac{\text{no.of moles of glucose}}{\text{No.of molesof glucose + No.of moles of water}}\\ =\dfrac{5}{5+0.0555}=0.99$ $\\$ volume of 100g solution$\\$ $=\dfrac{\text{mass of solution}}{\text{Density}}=\dfrac{100}{1.2}=83.33 ml$ $\\$ $\therefore$ Molarity of solution $=\dfrac{0.0555}{83.33}*1000\\ =0.67 M$

18   How many $mL$ of $0.1 M HCl$ are required to react completely with $1 g$ mixture of $Na _2 CO _3$and $NaHCO _3$ containing equimolar amounts of both?

Calculation of no. of moles of components in the mixture.$\\$ Let $x$ g of $Na _2 CO _3$ is present in the mixture.$\\$ $\therefore (1 - x)g$ of $NaHCO _3$ is present in the mixture.$\\$ Molar mass of $Na _2 CO _3$$\\ = 2 × 23 + 12 + 3 × 16 = 84g mol ^{-1}$$\\$ and molar mass of $NaHCO _3$$\\ = 23 × 1 + 1 + 12 + 3 × 16 = 84g mol ^{-1}$$\\$ No. of moles of $Na _2 CO _3$ in x g =$\dfrac{x}{106}$ $\\$ No. of moles of $NaHCO _3$ in $(1 - x) g = (1 - x)/84$$\\ As given that the mixture contains equimolar amounts of Na_ 2 CO_ 3 and NaHCO _3 , therefore\\ \dfrac{x}{106}=\dfrac{1-x}{84}\\ 106-106x=84x\\ 106=190x\\ \therefore x=\dfrac{106}{190}=0.558g \\ \therefore No. of moles of Na_2CO_3 present\\ =\dfrac{0.558}{106}=0.00526 \\ and no. of moles of NaHCO _3 present\\ =\dfrac{1-0.558}{84}=0.00526 \\ Calculation of no. of moles of HCl required\\ Na _2 CO _3 +2HCl \to 2NaCl + H _2 O + CO _2$$\\$ $NaHCO _3 + HCl \to NaCl + H _2 O + CO _2$$\\ As can be seen, each mole of Na _2 CO_ 3 needs 2 moles of HCl,\\ \therefore 0.00526 mole of Na _2 CO _3 needs = 0.00526 × 2 = 0.01052 mole\\ Each mole of NaHCO_3 needs 1 mole of HCl$$\\$ $\therefore 0.00526$ mole of $NaCHO_ 3$ needs = $1 × 0.00526 = 0.00526$ mole$\\$ Total amount of $HCl$ needed will be$\\$ $= 0.01052 + 0.00526 = 0.01578$mole.$\\$ $0.1$ mole of $0.1 M HCl$ are present in $1000 mL$ of $HCl$$\\ \therefore 0.01578 mole of 0.1 M HCl will be present in\\ =\dfrac{1000}{0.1}*0.01578=157.8 mL 19 A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution. ##### Solution : 300 g of 25% solution will contain =\dfrac{25*300}{100}=75g of solute.\\ 400g of 40% solution will contain=\dfrac{40*400}{100}=160g of solute \\ \therefore Total mass of solute =160+75=235g \\ Total mass of solution =300+400=700g \\ Now,the percentage of solute in solution =\dfrac{235}{700}*100=33.5%\\ And the percentage of water in solution = 100 - 33.5 = 66.5%\\ 20 An antifreeze solution is prepared from 222.6 g of ethylene glycol (C_ 2 H_ 6 O_ 2 ) and 200g of water. Calculate the molality of the solution. If the density of the solution is 1.072 mL ^{-1} , then what shall be the molarity of the solution? ##### Solution : Mass of solute = 222.6g$$\\$ Molar mass of solute $C _2 H_ 6 O_ 2$$\\ = 12 × 2 + 4 + 2 (12 + 1) = 62 g\ mol ^{-1} \\ \therefore Moles of solute = \dfrac{222.6}{62}=3.59 \\ Mass of solvent = 200 g$$\\$ $\therefore$ Molality $=\dfrac{3.59}{200}*1000=17.95 \ mol\ kg^{-1}$ $\\$ Total mass of solution = $422.6g$$\\ Volume of solution =\dfrac{422.6}{1.072}=394.21 mL \\ \therefore Molarity=\dfrac{3.59}{394.2}*1000=9.1 \ mol\ \ L^{-1} 21 A sample of drinking water was found to be severely contaminated with chloroform (CHCl _3 ), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).\\ (i) express this in percent by mass.\\ (ii) determine the molality of chloroform in water sample\\ ##### Solution : 15 ppm means 15 parts in million (10^ 6 ) by mass in the solution\\ \therefore percentage by mass=\dfrac{15}{10^6}*100=15*10^{-4}\% \\ As only 15g of chloroform is present in 10^ 6 g of the solution, mass of the solvent = 10 ^6 g\\ Molar mass of CHCl _3 = 12 + 1 + 3 × 35.5 = 119.5 mol ^{-1 }$$\\$ Moles of $CHCl_ 3 =\dfrac{15}{119.5}$$\\ \therefore Molality =\dfrac{15/119.5*1000}{10^6}=1.25*10^{-4}m 22 What role does the molecular interaction play in a solution of alcohol and water? ##### Solution : Alcohol and water both have strong tendency to form intermolecular hydrogen bonding. On mixing the two, a solution is formed as a result of formation of H-bonds between alcohol and H _2 O molecules but these interactions are weaker and less extensive than those in pure H_ 2 O. Thus they show a positive deviation from ideal behavior. As a result of this, the solution of alcohol and water will have higher vapour pressure and lower boiling point than that of water and alcohol. 23 Why do gases always tend to be less soluble in liquids as the temperature is raised? ##### Solution : When gases are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Lechatlier’s Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids. 24 State Henry’s law and mention some important applications. ##### Solution : The effect of pressure on the solubility of a gas in a liquid is governed by Henry’s Law. It states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas. Mathematically, P = KHX where P is the partial pressure of the gas; and X is the mole fraction of the gas in the solution and KH is Henry’s Law constant.\\ Applications of Henry’s law:\\ (i) In the production of carbonated beverages (as solubility of CO _2 increases at high pressure)\\ (ii) In the deep sea diving.\\ (iii) For climbers or people living at high altitudes, where low blood O _2 causes climbers to become weak and make them unable to think clearly\\ 25 The partial pressure of ethane over a solution containing 6.56 × 10 ^{-3} g of ethane is 1 bar. If the solution contains 5.00 × 10^{ -2} g of ethane, then what shall be the partial pressure of the gas? ##### Solution : We know that, m = KH × P$$\\$ $\therefore 6.56 × 10 ^{-2} g = KH × 1 bar .........(i)\\ \therefore 5.00 × 10^{ -2} g = KH × P .........(ii)\\ KH = 6.56 × 10 ^{-2} /1\ \ bar \ (from \ \ i)\\ KH = 5.00 × 10^{ -2} /P\ \ bar\ (from\ \ ii),\\ \therefore \dfrac{6.56*10^{-2}}{1}=\dfrac{5.00*10^{-2}}{P}\\ \therefore P=\dfrac{5.00}{6.56}=0.762$ bar $\\$

26   What is meant by positive and negative deviations from Raoult’s law and how is the sign of $\Delta _{ mix}\ H$ related to positive and negative deviations from Raoult’s law?

##### Solution :

Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and $\Delta _{ mix}\ H$ is positive because stronger A -A or B -B interactions are replaced by weaker A - B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall $\Delta _{ mix}\ H$ is positive. Similarly $\Delta _{ mix}\ V$ is positive i.e., the volume of solution is some what more than sum of volumes of solvent and solute. So there is expansion in volume on solution formation. Similarly in case of solutions exhibiting negative deviations, A-B interaction are stronger than A-A and B-B. So weaker interactions are replaced by stronger interactions, so there is release of enenergy i.e. $\Delta _{ mix}\ H$ is negative.

27   An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

##### Solution :

Vapour pressure of pure water at the boiling point $(P^o) = 1.013$ bar$\\$ Vapour pressure of solution $(Ps) = 1.004$bar$\\$ Mass of solute$(w _2 ) = 2g$$\\ Molar mass of solvent, water (M_ 1 ) = 18g$$\\$ Mass of solvent $(w_ 1 ) = 98g$$\\ Mass of solution = 100g$$\\$ Applying Raoult’s Law for dilute solutions,$\\$ $\dfrac{P^o-P_s}{P^o}=\dfrac{n_2}{n_1+n_2} \cong \dfrac{n_2}{n_1}$ [Dilute solution being 2%]$\\$ $\dfrac{P^o -P_s}{P^o}=\dfrac{n_2}{n_1} =\dfrac{W_2/W_1}{W_1/W_1}\\ \dfrac{(1.013-1.004)}{(1.013)}=\dfrac{2*18}{M_2*98}\\ \therefore M_2=\dfrac{2*18}{98*0.009}*1.013=41.35g \ mol^{-1}$ $\\$

28   Heptane and octane form an ideal solution. At $373 K$, the vapour pressure of the two liquid components $105.2 kP_a$ and $46.8 kP_a$ respectively. What will be the vapour pressure of a mixture of $26.0 g$ of heptane and $35.0 g$ of octane?

Molar mass heptane ($C _7 H _{16} ) = 7 × 12 + 16 = 100g mol ^{-1}$$\\ Molar mass of octane (C _8 H _{18} ) = 8 × 12 + 18 = 114 g mol ^{-1}$$\\$ Moles of heptane present in mixture =$\dfrac{26.0}{100}=0.26 \ mol$ $\\$ Moles of octane present in mixture =$\dfrac{35.0}{114}=0.307 \ mol$ $\\$ Mole fraction of heptane $x_ H = \dfrac{0.26}{0.26+0.307}=0.458$ $\\$ Mole fraction of octane, $x O – (1 - 0.458) = 0.542$$\\ Vapour pressure of heptane = x H × P^o= 0.458 × 105.2 kPa = 48.18 kPa$$\\$ Vapour pressure of octane = $x O × P^o= 0.542 × 46.8 kPa = 25.36 kPa$$\\ Vapour pressure of mixture = 48.18 + 25.36 = 73.54 kPa 29 The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it ##### Solution : 1 molal solution of solute means 1 mole of solute in 1000g of the solvent.\\ Molar mass of water (solvent) = 18 g mol^{ -1} \\ therefore Moles of water = \dfrac{1000}{8}=55.5 moles \\ \therefore Mole fraction of solute =\dfrac{1}{1+55.5}=0.0177 \\ Now,\dfrac{P^o-P_s}{P^o}=x_2\\ \dfrac{12.3-P_s}{12.3}=0.0177\\ =P_s=12.08kPa \\ 30 Calculate the mass of a non-volatile solute (molar mass 40 g\ mol ^{-1} ) which should be dissolved in 114 g octane to reduce its vapour pressure to 80% ##### Solution : P_ x = 80\% of P^o\\ =\dfrac{80}{100}P^o=0.8p^o \\ Let Wg of solute is present in mixture.\\ Moles of solute present =\dfrac{W}{40} moles\\ Molar mass of octane, C8H18 = 8 * 12 + 8 = 114 mol ^{-1}$$\\$ $\therefore$Moles of octance =$\dfrac{114}{114}=1 mol$ $\\$ Now$\dfrac{p^o-P_s}{P^o}=x_2=\dfrac{W/40}{\frac{W}{40}+1}\\ \dfrac{P^o-0.80 P^o}{P^o}=\dfrac{W/40}{W/40+1}\\ 1-0.80=\dfrac{W*40}{40(W+40)}=\dfrac{W}{W+40}\\ 0.20=\dfrac{W}{W+40}\\ 0.20W+8=W\\ 8=W(1-0.2)\\ 8=0.8W\\ \therefore W=\dfrac{8}{0.8}=10g$

31   A solution containing $30g$ of non-volatile solute exactly in $90 g$ of water has a vapour pressure of $2.8 kPa$ at $298 K$. Further, $18g$ of water is then added to the solution and the new of vapour pressure becomes $2.9 kPa$ at $298 K$. Calculate$\\$ (i) molar mass of the solute$\\$ (ii) vapour pressure of water at $298 K$.$\\$

##### Solution :

Let the molar mass of solute = $Mg mol^{ -1}$ $\\$ $\therefore$ Moles of solute present =$\dfrac{30g}{Mg\ mol^{-1}}=\dfrac{30}{M} mol$ $\\$ Moles of solvent present,$(n_1)=\dfrac{90}{18}=5 \ moles.$ $\\$ $\dfrac{P^o-P_s}{P^o}=\dfrac{n_2}{n_1+n_2}\\ \dfrac{P^o-2.8}{P^o}=\dfrac{30/W}{5+30/W}\\ 1-\dfrac{2.8}{P^o}=\dfrac{30}{(5M+30)}\\ 1-\dfrac{30}{5M+30}=\dfrac{2.8}{P^o}\\ 1-\dfrac{6}{M+6}=\dfrac{2.8}{P^o}\\ \dfrac{M+6-6}{M+6}=\dfrac{2.8}{P^o}\\ \dfrac{M}{M+6}=\dfrac{2.8}{P^o}\\ \dfrac{P^o}{2.8}=1+\dfrac{6}{M}.........(i)$ $\\$ After adding $18 g$ of water, moles of water becomes$\\$ $=\dfrac{90+18}{18}=\dfrac{108}{18}=6 moles \\ \therefore \dfrac{P^o-P_s}{P^o}=\dfrac{30M}{M(6M+30)}=\dfrac{5}{M+5}\\ 1-\dfrac{2.9}{P}=\dfrac{5}{M+5}\\ 1-\dfrac{5}{M+5}=\dfrac{2.9}{P^o}\\ \dfrac{M+5-5}{M+5}=\dfrac{2.9}{p^o}\\ \dfrac{p^o}{2.9}=\dfrac{M+5}{M}=1+\dfrac{5}{M}\\ \dfrac{P^o}{2.9}=1+\dfrac{5}{M}.........(ii)$ $\\$ Dividing equation (i) by (ii) we get$\\$ $\dfrac{2.9}{2.8}=\dfrac{1+6/M}{1+5/M}\\ 2.9(1+\dfrac{5}{M})=2.8(1+\dfrac{6}{M})\\ 2.9+\dfrac{2.9*5}{M}=2.8+\dfrac{2.8*6}{M}\\ 2.9+\dfrac{14.5}{M}=2.8+\dfrac{16.8}{M}\\ 0.1=\dfrac{16.8}{M}-\dfrac{14.5}{M}=\dfrac{2.3}{M}\\ M=\dfrac{2.3}{0.1}\\ M=23\ g \ mol^{-1}$ $\\$ Putting $M = 23$, in equation (i) we get$\\$ $\dfrac{P^o}{.8}=1+\dfrac{6}{23}=\dfrac{29}{23}\\ P^o=\dfrac{29}{23}*2.8=3.53 kPa$ $\\$

32   A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K.

Molar mass of sugar = $342g mol^{ -1}$ $\\$ Molality of sugar solution=$\dfrac{5*1000}{342*100}=0.146$$\\ \therefore \Delta T_f foor sugar solution=273.15-271=2.15^o \\ \therefore \Delta T_f=K*m\\ \Delta T_f=K_f*0.146=K_f=2.15/0.146 \\ Molality of glucose solution =\dfrac{5}{180}*\dfrac{1000}{100}=0.278 \\ (Molar mass of glucose =180\ g \ mol^{-1})\\ \Delta T_f=K_f*m=\dfrac{2.15}{0.146}*0.278=4.09^o \\ \therefore Freezing point of glucose solution = 273.15 - 4.09 = 269.06K. 33 Two elements A and B form compounds having formula AB_ 2 and AB _4 . When dissolved in 20g of benzene (C_ 6 H _6 ). 1 g of AB_2 lowers the freezing point by 2.3 K whereas 1.0 g of AB _4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg\ mol^{ -1} . Calculate atomic masses of A and B. ##### Solution : Using the relation, M _2 =\dfrac{1000*k_f*w_2}{w_1*\Delta T_f}\\ \therefore M_{AB_2}=\dfrac{1000*5.1*1}{20*2.3}=110.8\ g \ mol^{-1}\\ M_{AB_4}=\dfrac{1000*5.1*1}{20*1.3}=196.15\ g \ mol^{-1} \\ Let the atomic masses of and B are ‘p’ and ‘q’ respectively.\\ Then molar mass of AB _2 = p + 2q = 110.8\ g\ mol^{ -1} ....(i)$$\\$ And molar mass of $AB _4 = p + 4q = 196.15\ g\ mol ^{-1} ......(ii)$$\\ Subtracting equation (ii) from equation (i),\\ we get 2q = 85.28 = q = 42.64$$\\$ Putting $q= 42.64$ in eqn. (i) wed get$\\$ $P = 110.87 - 85.28\\ P = 25.59$$\\ Thus, atomic mass of A = 25.59 u and\\ atomic mass of B = 42.64 u 34 At 300 K, 36g of glucose present in a liter of its solution has an osmotic pressure of 4.08 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration? ##### Solution : \pi= CRT\\ 4.98=\dfrac{w_1}{M_1}*R*300\\ 4.98=\dfrac{36}{180}*R*300\\ 4.98=60R.......(i) \\ In second case 1.52 = C * R * 300 .....(ii)$$\\$ Diving equation (ii) by equation (i), we get$\\$ $C=\dfrac{60*1.52}{300*4.98}=0.06M$ $\\$

35   Suggest the most important type of intermolecular attractive interaction in the following pairs:$\\$ (i) n-hexane and n-octane$\\$ (ii) $I _2$ and $CCl _4$.$\\$ (iii) $NaClO _4$ and water$\\$ (iv) methanol and acetone$\\$ (v) acetonitrile ($CH_ 3 CN$) and acetone ($C _3 H _6 O$)$\\$

##### Solution :

(i) Both n-hexane and n-octane are non-polar. Thus, the intermolecular interactions will be London dispersion forces.$\\$ (ii) Both $I _2$ and $CCl _4$ are non-polar. Thus the intermolecular interactions will be London dispersion forces.$\\$ (iii)$NaClO _4$ is an ionic compound and gives $Na ^+$ and $ClO ^-_ 4$ ions in the Solution. Water is a polar molecule. Thus, the intermolecular interactions will be ion-dipole interactions.$\\$ (iv) Both methanol and acetone are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.$\\$ (v) Both $CH _3 CN$ and $C _3 H_ 6 O$ are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.$\\$

36   Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain.$\\$ Cyclohexane, $KCl, CH_ 3 OH, CH _3 CN.$$\\ ##### Solution : (a) Cyclohexane and n-octane both are non-polar. They mix completely in all proportions.\\ (b) KCl is an ionic compound,KCl will not dissolve in n-octane.\\ (c) CH _3 OH is polar. CH_ 3 OH will dissolve in n-octane.\\ (d) CH _3 CN is polar but lesser than CH _3 OH. Therefore, it will dissolve in n-octane but to a greater extent as compared to CH 3 OH Hence, the order is KCl < CH_ 3 OH < CH _3 CN < Cyclohexane.\\ 37 Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water\\ (i) phenol\\ (ii) toluene\\ (iii) formic acid\\ (iv) ethylene glycol\\ (v) chloroform\\ (vi) Pentanol ##### Solution : (i) Phenol (having polar – OH group) – Partially soluble.\\ (ii) Toluene (non-polar) – insoluble.\\ (iii) Formic acid (form hydrogen bonds with water molecules) – Highly soluble\\ (iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble\\ (v) Chloroform (non-polar) insoluble.\\ (vi) Pentanol (having polar – OH) – Partially soluble.\\ 38 If the density of some lake water is 1.25 g mL ^{-1} and contains 92g\ f\ Na ^+ ions per kg of water, calculate the molality of Na ^+ ions in the lake. ##### Solution : Molar mass of Na = 23g\ mol ^{-1} \\ \therefore No. of moles of Na + ions present =\dfrac{92}{23}=4 moles \\ \therefore Molality= \dfrac{4*1000}{1000}=4m \\ 39 If the solubility product of CuS is 6 * 10^{ -16} , calculate the maximum molarity of CuS aqueous solution. ##### Solution : CuS \leftrightarrow Cu-S^{2-},K_{sp}=6*10^{-16}$$\\$ Maximum molarity of $CuS$ in aqueous solution means solubility of $CuS$.$\\$ Let the solubility of $CuS$ be $S \ \ mol\ \ L^{ -1}$ $\\$ $\therefore K_{ sp} = [ Cu^{2+} ][ S ]^{2-} 6* 10^{- 16}= S * S = S^ 2\\ \therefore S=\sqrt{6*10^{-16}}=2.45*10^{-8}\ mol \ \ L^{-1}$

40   Calculate the mass percentage of aspirin ($C_ 9 H _8 O _4$ ) in acetonitrile ($CH _3 CN$) when $6.5 g$ of $C _9 H _8 O _4$ is dissolved in $450 g$ of $CH_ 3 CN$.

$6.5 g$ of $C _9 H_ 8 O_ 4$ is dissolved in $450 g$ of $CH _3 CN.$ Then, total mass of the solution =$(6.5 + 450) g = 456.5 g$$\\ Therefore, mass percentage of C_ 9 H _8 O _4 \\ =\dfrac{6.5}{456.5}*100\%\\ =1.424\% \\ 41 Calculate the amount of benzoic acid (C _6 H _5 COOH) required for preparing 250 mL of 0.15M solution in methanol. ##### Solution : 0.15 M solution of benzoic acid in methanol means,1000 mL of solution contains 0.15 mol of benzoic acid.\\ \dfrac{0.15*250}{1000} \\ Therefore, 250 mL of solution contains = mol of benzoic acid = 0.0375 mol of benzoic acid.\\ Molar mass of benzoic acid (C _6 H_ 5 COOH) = 7 × 12 + 6 × 1 + 2 × 16 = 122\ g\ mol ^{-1}$$\\$ Hence, required benzoic acid = $0.0375\ \ mol × 122 \ \ g \ \ mol ^{-1}$$\\ = 4.575 g$$\\$

The molar mass of nalorphene ($C _{19} H _{21} NO _3$ ), $19 * 12 + 21 * 1 + 1 * 14 + 3 * 16= 311g\ mol ^{- 1}$$\\ In 1.5 × 10^{ -3}m aqueous solution of nalorphene, 1.5 * 10 ^{- 3 }* 311g$$\\$ 1 kg (1000 g) of water contains $1.5 × 10 ^{-3}$ mol = $0.4665 g$$\\ Therefore, total mass of the solution is given as\\ = (1000 + 0.4665) g$$\\$ = $1000.4665 g$$\\ This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665g.$$\\$ Therefore, mass of the solution containing $1.5 mg$ of nalorphene is:$\\$ $\dfrac{1000.4665*1.5*10^{-3}}{0.4665}g\\ =3.22g$ $\\$ Hence, the mass of aqueous solution required is $3.22 g.$

42   The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

##### Solution :

Among $H, Cl,$ and $F, H$ is least electronegative while $F$ is most electronegative. Then,$F$ can withdraw electrons towards itself more than $Cl$and $H$. $\\$Thus, trifluoroacetic acid can easily lose $H ^+$ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:$\\$ Acetic acid < trichloroacetic acid < trifluoroacetic acid.

43   Calculate the depression in the freezing point of water when $10 g$ of $CH_ 3 CH _2 CHClCOOH$ is added to $250 g$ of water. $K_a = 1.4 × 10^{-3}, K_f = 1.86 K\ \ kg\ \ mol^{-1}.$

Molar mass of $CH _3 CH_ 2 CHClCOOH$ is= $15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1 = 122.5g\ \ mol ^{- 1}$ $\\$ sent in $10 g$ of $CH_ 3 CH _2 CHClCOOH$ is =$\dfrac{10g}{122.5 \ g \ mol^{-1}}=0.0816 \ mol$ $\\$ It is given that $10 g$ of $CH _3 CH _2 CHClCOOH$ is added to $250 g$ of water.$\\$ Molality of the solution, $=\dfrac{0.0186}{250}*1000=0.3264 \ mol kg^{-1}$ $\\$ Let $\alpha$ be the degree of dissociation of $CH _3 CH _2 CHClCOOH$ $\\$ $CH _3 CH _2 CHClCOOH$ undergoes dissociation according to the following equation:$\\$ $\qquad \qquad CH _3 CH_ 2 CHClCOOH \leftrightarrow CH _3 CH _2 CHClCOO ^-+ H ^+$$\\ Initial conc. \qquad \qquad C\ mol \ L^{-1} \qquad 0 \qquad 0 \\ At equilibrium \qquad C(1-\alpha) \qquad C\alpha \qquad C \alpha \\ \therefore K_a = \dfrac{C\alpha^2}{1}\\ =K_a=C\alpha^2\\ =\alpha =\sqrt{\dfrac{K_a}{C}}\\ =\sqrt{\dfrac{1.4*10^{-3}}{0.3264}} (K_a =1.4*10^{-3})\\ =0.0655 \\ Again \\ CH _3 CH_ 2 CHClCOOH \leftrightarrow CH_ 3 CH _2 CHClCOO ^-+ H ^+ \\ Initial moles \qquad 1 \qquad 0 \qquad 0 \\ At equilibrium \quad 1-\alpha \qquad \alpha \qquad \alpha \\ Total moles of equilibrium =1-\alpha+ \alpha+ \alpha \\ =1+\alpha \\ \therefore i=\dfrac{1+\alpha }{1}\\ =1+\alpha \\ 1+0.0655\\ 1.0655 \\ Hence, the depression in the freezing point of water is given as:\\ \Delta T_f=i.K_fm\\ =1.0655*1.86K kg \ mol^{-1}*0.3264 \ mol \ kg^{-1}\\ =0.65 K \\ 44 19.5 g of CH _2 FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid. ##### Solution : It is given that:\\ w _1 = 500 g\\ w _2 = 19.5 g\\ K _f = 1.86 K\ kg\ mol ^{- 1}\\ \Delta T_ f = 1 K \\ We know that:\\ M_2=\dfrac{K_f*w_2*1000}{\Delta T_f*w_1}\\ =\dfrac{1.86K \ kg \ mol^{-1}*19.5g * 1000 g \ kg^{-1}}{500g*1 K}\\ =72.54 \ g \ mol^{-1} \\ Therefore, observed molar mass of CH _2 FCOOH , ( M _2 )_{ obs} = 72.54 g\ mol$$\\$ The calculated molar mass of $CH _2 FCOOH:$$\\ (M_2)_{cal}=14+19+12+16+16+1\\ =78\ g\ mol^{-1} \\ Therefore, van’t Hoff factor \\ i=\dfrac{(M_2)_{cal}}{(M_2)_{obs}} is \\ =\dfrac{78\ g \ mol^{-1}}{72.54 \ g \ mol^{-1}}\\ =1.0753 \\ Let \alpha be the degree of dissociation of CH _2 FCOOH: CH_ 2 FCOOH \leftrightarrow CH_ 2 FCOO ^-+ H ^+ \\ Initial conc. \qquad C\ mol \ L^{-1} \qquad 0 \qquad 0 \\ At equilibrium \qquad C(1-\alpha ) \qquad C\alpha \qquad C\alpha \\ Total =C(1+\alpha )\\ \therefore i=\dfrac{C(1+\alpha)}{C}\\ =i=1+\alpha \\ =\alpha =i-1\\ =1.0753-1\\ = 0.0753 \\ Now, the value of K _a is given as:\\ K_a=\dfrac{[CH_2FCOO^-][H^+]}{[CH^2FCOOH]}\\ =\dfrac{C\alpha.C\alpha}{C(1-\alpha )}\\ =\dfrac{C \alpha^2}{1-\alpha} \\ Taking the volume of the solution as 500 mL, we have the concentration:\\ C=\dfrac{\frac{19.5}{78}}{500}*1000 M\\ =0.5 M\\ K_a=\dfrac{C \alpha^2}{1-\alpha } \\ Therefore,\\ =\dfrac{0.5*(0.0753)^2}{1-0.0753}\\ =\dfrac{0.5*0.00567}{0.9247}\\ =0.00307 (approximately)\\ =3.07 * 10^{-3} \\ 45 Vapour pressure of water at 293 K is 17.535 mm\ Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water. ##### Solution : Vapour pressure of water, P _1^ 0 = 17.535 mm of Hg$$\\$ Mass of glucose, $w_ 2 = 25 g$$\\ Mass of water, w_ 1 = 450 g$$\\$ We know that,$\\$ Molar mass of glucose $(C_6H_{12}O_6), M_ 2 = 6 * 12 + 12 * 1 + 6 * 16 =180 g\ mol^{ -1}$$\\ Molar mass of water, M _1 = 18 g\ mol ^{-1}$$\\$ Then, number of moles of glucose,$\\$ $n_2=\dfrac{25}{180 g \ mol^{-1}}\\ =0.139 \ mol$ $\\$ And, number of moles of water,$\\$ $n_1=\dfrac{450g}{18g\ mol^{-1}}\\ =25 \ mol$ $\\$ We know that,$\\$ $\dfrac{P^0_1-P_1}{P^0_1}=\dfrac{n_1}{n_2+n_1}\\ =\dfrac{17.535-P_1}{17.535}=\dfrac{0.139}{0.139+25}\\ =17.535-P_1=\dfrac{0.139*17.535}{25.139}\\ =17.535-P_1=0.097$ $\\$ $p_ 1 = 17.44 mm$ of Hg$\\$ Hence, the vapour pressure of water is $17.44 mm\ of \ Hg.$

46   Henry’s law constant for the molality of methane in benzene at $298 K$ is $4.27 × 10 ^5\ mm\ Hg.$ Calculate the solubility of methane in benzene at $298 K$ under $760\ mm\ Hg.$

##### Solution :

Here,$p = 760$$\\ mm Hg\ kH = 4.27 × 10 5$$\\$ mm Hg According to Henry’s law,$\\$ $p = kHx\\ =x=\dfrac{p}{k_H}\\ =\dfrac{760 mm \ Hg}{4.27*10^{5} mm \ hg}\\ =177.99*10^{-5}\\ =178*10^{-5}$ (approximately)$\\$ Hence, the mole fraction of methane in benzene is$178 × 10^{ -5}$

47   $100 g$ of liquid A (molar mass $140 g mol ^{-1}$ ) was dissolved in $1000 g$ of liquid B (molar mass $180 \ g\ mol^{ -1}$ ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

##### Solution :

Number of moles of liquid A,$\\$ $n_A=\dfrac{100}{140}\ mol =0.714 \ mol$ $\\$ Number of moles of liquid B,$\\$ $n_B=\dfrac{1000}{180} \ mol =5.556 \ mol$ $\\$ Then, mole fraction of A,$\\$ $x_A=\dfrac{n_A}{n_A+n_B}\\ =\dfrac{0.714}{0.714+5.556}\\ =0.114$ $\\$ And, mole fraction of $B, x_ B = 1 - 0.114 = 0.886$ $\\$ Vapour pressure of pure liquid $B, p_ B^ 0 = 500 torr$ Therefore, vapour pressure of liquid B in the solution,$\\$ $p _B = p _B^ 0 x _B\\ = 500 × 0.886\\ = 443 torr$$\\ Total vapour pressure of the solution, p total = 475 torr$$\\$ Vapour pressure of liquid A in the solution,$p _A = p total - p_ B = 475 – 443\\ = 32 torr$ $\\$ Now,$\\$ $p _A= p _A^ 0 x _A\\ =p^0_A=\dfrac{p_A}{x_A}\\ =\dfrac{32}{0.114}\\ =280.7 torr$ $\\$ Hence, the vapour pressure of pure liquid A is $280.7$torr.

48   Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at $300 K$ are $50.71\ mm\ Hg$and $32.06 mm\ Hg$ respectively. Calculate the mole fraction of benzene in vapour phase if $80 g$ of benzene is mixed with $100 g$ of toluene.

##### Solution :

It can be observed from the graph that the plot for the ptotal of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

Molar mass of benzene$\\$ $( C_ 6 H_ 6 )= 6 * 12 +6*1=78 g mol^{ -1}$ $\\$ Molar mass of toluene$\\$ $( C_ 6 H_ 6 )= 7 * 12 +8*1=92 g mol^{ -1}$ $\\$ Now, no. of moles present in 80 g of benzene$\\$ $=\dfrac{80}{78} \ mol =1.026 \ mol$ $\\$ And, no. of moles present in $100 g$ of toluene$\\$ $=\dfrac{100}{92} \ mol =1.087 \ mol$ $\\$ Mole fraction of benzene, $x _b$$\\ =\dfrac{1.026}{1.026+1.087}=0.486 \\ And, mole fraction of toluene,\\ x_ t = 1 - 0.486 =0.514$$\\$ It is given that vapour pressure of pure benzene,$\\$ $p_ b ^0 = 50.71 mm Hg$$\\ And, vapour pressure of pure toluene,\\ p_ t ^0 = 32.06 mm Hg$$\\$ Therefore, partial vapour pressure of benzene,$\\$ $P _b = x_ b * p_ b\\ = 0.514 * 32.06\\ = 16.479 mm\ Hg$$\\ Hence, mole fraction of benzene in vapour phase is given by:\\ \dfrac{p_b}{p_b+p_t}\\ =\dfrac{24.645}{24.645 + 16.479}\\ \dfrac{24.645}{41.124}\\ = 0.599\\ = 0.6 49 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen are 3.30 × 10 ^7 mm and 6.51 × 10^ 7 mm respectively, calculate the composition of these gases in water. ##### Solution : Percentage of oxygen (O2) in air = 20 %\\ Percentage of nitrogen (N2) in air = 79%\\ Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is,\\ (10 × 760) mm\ Hg = 7600 mm\ Hg$$\\$ Therefore, Partial pressure of oxygen, $p_{O_2}=\dfrac{20}{100}*7600\ mm\ Hg = 1520\ mm\ Hg$$\\ Partial pressure of nitrogen,\\ P_{N_2}=\dfrac{79}{100}*7600 \ mm \ hg =6004 \ mm \ Hg \\ Now, according to Henry’s law:\\ p = KH.x$$\\$ For oxygen:$\\$ $p_{ O _2} = k _H . x _{O_ 2}\\ =x_{O_2}=\dfrac{p_{O_2}}{k_{H}}\\ \dfrac{1520 \ mm \ Hg}{3.30* 10^7\ mm \ Hg}(given k_H=3.30*10^7\ mm Hg)\\ =4.61*10^{-5}$ $\\$ For nitrogen:$\\$ $P_{N_2}=K_H.x_{N_2}\\ =x_{N_2}=\dfrac{P_{N_2}}{K_H}\\ =\dfrac{6004 \ mm \ Hg}{6.51*10^7 \ mm \ Hg}\\ =9.22*10^{-5}$ $\\$ Hence, the mole fractions of oxygen and nitrogen in water are $4.61 ×10^{ -5}$and $9.22 × 10^{ -5}$ respectively.

50   Determine the amount of $CaCl 2 (i = 2.47)$ dissolved in $2.5$ litre of water such that its osmotic pressure is $0.75$ atm at $27°C$.

##### Solution :

We know that,$\\$ $\pi = i\dfrac{n}{V}RT\\ =\pi =i\dfrac{w}{MV}RT\\ =w=\dfrac{\pi MV}{iRT}\\ \pi=0.75 \ atm \\ V=2.5 L\\ i=2.47\\ T=(27+273)K=300K$ $\\$ Here, $\\$ $R=0.0821 L \ atm K^{-1} mol^{-1}\\ M=1*40+2*35.5\\ 111 g \ mol^{-1}$ $\\$ Therefore , w $\\$ $=\dfrac{0.75 * 111 * 2.5}{2.47 * 0.0821 * 300}\\ =3.42 g$ $\\$ Hence, the required amount of $CaCl _2$ is $3.42 g$.

51   Determine the osmotic pressure of a solution prepared by dissolving $25 mg$of $K _2 SO _4$ in 2 litre of water at 25° C, assuming that it is completely dissociated.

##### Solution :

When $K _2 SO _4$is dissolved in water,$\\$ $K ^+$ and $SO_ 4^{ 2 -}$ ions are produced.$\\$ $K _2 SO _4 \to 2 K ^++ SO_ 4 ^{2 -}$$\\$ Total number of ions produced = 3$\\$ i =3 Given, w = 25 mg = 0.025g$\\$ V = 2 L$\\$ T = 250C = (25 + 273) K = 298 K Also,$\\$ we know that:$\\$ $R = 0.0821\ L\ atm\ K ^{-1}\ mol ^{-1}\\ M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol ^{-1}$ Appling the following relation,$\\$ $\pi=i \dfrac{n}{v}RT\\ =i\dfrac{w}{M}\dfrac{1}{v}RT\\ =3*\dfrac{0.025}{174}*\dfrac{1}{2}*0.0821*298\\ =5.27*10^{-3} \ atm$ $\\$