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.

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 $