# Some Basic Concepts of Chemistry

## Chemistry Class 11

### NCERT

1   Calculate the molecular mass of the following:$\\$ (i)$H_2O$$\\ (ii)CO_2$$\\$ (iii)$CH_4$

##### Solution :

(i)$H_ 2 O$$\\ The molecular mass of water, H _2 O$$\\$ $= (2 *$ Atomic mass of hydrogen$) + (1 *$Atomic mass of oxygen)$\\$ $= [2(1.0084) + 1(16.00 u)]$$\\ = 2.016 u + 16.00 u$$\\$ $= 18.016$$\\ = 18.02 u (ii) CO _2$$\\$ The molecular mass of carbon dioxide, $CO _2$$\\ = (1 *Atomic mass of carbon) + (2 * Atomic mass of oxygen)\\ = [1(12.011 u) + 2 (16.00 u)]$$\\$ $= 12.011 u + 32.00 u$$\\ = 44.01 u$$\\$ (iii) $CH _4$$\\ The molecular mass of methane, CH_ 4$$\\$ $= (1 *$ Atomic mass of carbon$) + (4 *$ Atomic mass of hydrogen)$\\$ $= [1(12.011 u) + 4 (1.008 u)]$$\\ = 12.011 u + 4.032 u$$\\$ $= 16.043 u$

2   Calculate the mass percent of different elements present in sodium sulphate $(Na_ 2 SO_ 4 )$ .

##### Solution :

The molecular formula of sodium sulphate is $Na_2 SO_ 4$$\\ Molar mass of Na _2 SO _4 =[(2 *23.0) + (32.066) + 4 (16.00)]$$\\$ $142.066 g$$\\ Mass percent of an element =\dfrac{\text{Mass ofthat element in thecompound}}{\text{Molar mass of thecompound}}*100$$\\$ $\therefore$ Mass percent of sodium:$\\$ $\dfrac{46.0g}{142.066g}*100$$\\ =32.379$$\\$ $=32.4\%$$\\ Mass percent of sulphur:\\ =\dfrac{32.066g}{142.066g}*100$$\\$ $=22.57$$\\ =22.6\%$$\\$ Mass percent of oxygen:$\\$ $=\dfrac{64.0g}{142.066g}*100$$\\ =45.049$$\\$ $=45.05\%$

3   Determine the empirical formula of an oxide of iron which has $69.9$% iron and $30.1$% dioxygen by mass.

% of iron by mass $69.9$ % [Given]$\\$ % of oxygen by mass $30.1$ % [Given]$\\$ Relative moles of iron in iron oxide:$\\$ $\dfrac{\%\text{of iron by mass}}{\text{Atomic mass of iron}}$$\\ =\dfrac{69.9}{55.85}=1.25 Relative moles of oxygen in iron oxide:\\ =\dfrac{\%\text{of oxygen by mass}}{\text{Atomic mass or oxygen}}$$\\$ $=\dfrac{30.1}{16.00}$$\\ =1.88$$\\$ Simplest molar ratio of iron to oxygen:$\\$ $= 1.25 : 1.88$$\\ =1:1.5$$\\$ $2 : 3$$\\ \therefore The empirical formula of the iron oxide is Fe _2 O _3 . 4 Calculate the amount of carbon dioxide that could be produced when\\ (i) 1 mole of carbon is burnt in air.\\ (ii) 1 mole of carbon is burnt in 16 g of dioxygen.\\ (iii) 2 moles of carbon are burnt in 16 g of dioxygen. ##### Solution : The balanced reaction of combustion of carbon can be written as:\\ C+ O _2 \to C O _2$$\\$ (i) As per the balanced equation, $1$ mole of carbon burns in $1$ mole of dioxygen (air) to produce $1$ mole of carbon dioxide.$\\$ (ii) According to the question, only $16$ g dioxygen is available. Hence, it will react with $0.5$ mole of carbon to give $22$ g of carbon dioxide. Hence, it is a limiting reactant.$\\$ (iii) According to the question, only $16$ g dioxygen is available. It is a limiting reactant. Thus, $16$ g of dioxygen can combine with only $0.5$ mole of carbon to give $22$ g of carbon dioxide.

5   Calculate the mass of sodium acetate $(CH _3 COONa)$ required to make $500$ mL of $0.375$ molar aqueous solution. Molar mass of sodium acetate is $82.0245$ g mol $^{-1}$

$0.375$ M aqueous solution of sodium acetate$\\$ $\equiv 1000$ mL of solution containing $0.375$ moles of sodium acetate $\therefore$ Number of moles of sodium acetate in $500$ mL$\\$ $=\dfrac{0.375}{1000}*500$$\\ =0.1875 mole\\ Molar mass of sodium acetate=\\ 82.0.245 g mole^{-1}(Given)\\ \therefore Required mass of sodium acetate = (82.0245 g mol^{ -1} ) (0.1875 mole)\\ = 15.38 g 6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41g mL ^-1 and the mass percent of nitric acid in it being 69\%. ##### Solution : Mass percent of nitric acid in the sample = 69\% [Given]\\ Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.\\ Molar mass of nitric acid (HNO _3 ) \\ =\{1+14+3(16)\}g mol^{-1} \\ = 1 + 14 + 18 \\ = 63 g mol ^{-1} \\ \therefore Number of moles in 69 g of HNO _3 \\ =\dfrac{69 g}{63 g mol^{-1}} \\ = 1.095 mol \\ Volume of 100g of nitric acid solution\\ =\dfrac{\text{Mass of solution}}{\text{density of solution}} \\ =\dfrac{100g}{1.41 gmL^{-1}} \\ = 70.92 mL = 70.92 * 10 ^{- 3} L \\ Concentration of nitric acid \\ \dfrac{1.095 mole}{70.92*10^{-3}L} \\ = 15.44 mol / L \\ \therefore Concentration of nitric acid = 15.44 mol/L 7 How much copper can be obtained from 100 g of copper sulphate (CuSO_ 4 )? ##### Solution : 1 mole of CuSO_ 4 contains 1 mole of copper.\\ Molar mass of CuSO _4 = (63.5) + (32.00) + 4(16.00) \\ = 63.5 + 32.00 + 64.00 \\ = 159.5 g \\ 159.5 g of CuSO _4 contains 63.5 g of copper.\\ \Rightarrow 100g of CuSO_4 will contain \dfrac{63.5*100g}{159.5} of copper \\ \therefore Amount of copper that can be obtained from 100g CuSO_ 4 =\dfrac{63.5*100}{159.5} \\ =39.81 g 8 Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69g \ mol ^{- 1} ##### Solution : Mass percent of iron (Fe) = 69.9\% (Given)\\ Mass percent of oxygen (O) = 30.1\% (Given)\\ Number of moles of iron present in the oxide =\dfrac{69.90}{55.85} \\ = 1.25 \\ Number of moles of oxygen present in the oxide =\dfrac{30.1}{16.0} \\ =1.88 \\ Ratio of iron to oxygen in the oxide,\\ = 1.25 : 1.88 \\ =\dfrac{1.25}{1.25}:\dfrac{1.88}{1.25} \\ = 1 : 1.5 \\ = 2 : 3 \\ \therefore The empirical formula of the oxide is Fe _2 O_ 3 . \\ Empirical formula mass of Fe _2 O_ 3 = [2(55.85) + 3(16.00)] g \\ Molar mass of Fe _2 O_ 3 = 159.69 g \\ \therefore n=\dfrac{\text{Molar mass}}{\text{Empiricical formula mass}} \\ =\dfrac{159.69 g}{159.7 g} \\ =0.999 \\ =1(approx)\\ Molecular formula of a compound is obtained by multiplying the empirical formula with n. \\ Thus, the empirical formula of the given oxide is Fe_ 2 O_ 3 and n is 1.$$\\$ Hence, the molecular formula of the oxide is $Fe _2 O_ 3 .$

9   Calculate the atomic mass (average) of chlorine using the following data:$\\$ $\begin{array}{|c|c|}\hline & \text{% Natural Abundance} & \text{Molar mass}\\ \hline ^{35}Cl & 75.77 & 34.9689\\ \hline ^{37}Cl & 24.23 & 36.9659 \\ \hline \end{array}$

##### Solution :

The average atomic mass of chlorine$\\$ =[(Fractional abundance of $^{35}Cl$)(Molar mass of $^{35}Cl$)+(Fractional abundance of $^{37}Cl$)(Molar mass of $^{37}Cl$)]$\\$ $=[\{(\dfrac{75.77}{100})(34.9689 u) \}+\{(\dfrac{24.23}{100})(36.9659 u)\}]$ $\\$ $=26.4959 + 8.9568$$\\ = 35.4527 u$$\\$ $\therefore$ The average atomic mass of chlorine =$35.4527 u$

10   In three moles of ethane $(C_ 2 H_ 6 )$, calculate the following:$\\$ (i) Number of moles of carbon atoms.$\\$ (ii) Number of moles of hydrogen atoms.$\\$ (iii)Number of molecules of ethane.

##### Solution :

(i) $1$ mole of $C _2 H _6$ contains $2$ moles of carbon atoms.$\\$ $\therefore$ Number of moles of carbon atoms in $3$ moles of $C _2 H_ 6 .$ $\\$ $= 2 * 3 = 6$ $\\$ (ii)$1$ mole of $C_ 2 H_ 6$ contains $6$ moles of hydrogen atoms.$\\$ $\therefore$ Number of moles of carbon atoms in $3$ moles of $C_ 2 H_ 6 .$ $\\$ $= 3 * 6 = 18$ $\\$ (iii) $1$ mole of $C _2 H _6$ contains $6.023 * 10^{ 23}$ molecules of ethane.$\\$ $\therefore$ Number of molecules in $3$ moles of $C_ 2 H_ 6 .$ $\\$ $= 3 * 6.023 * 10^{ 23} = 18.069 *10 ^{23}$

11   What is the concentration of sugar ($C _{12} H_{ 22} O_{ 11}$ ) in mol $L^{ -1}$ if its 20 g are dissolved in enough water to make a final volume up to 2 L?

##### Solution :

Pressure is defined as force acting per unit area of the surface.$\\$ $P=\dfrac{F}{A}\\ =\dfrac{1034g*9.8ms^{-2}}{cm^2}*\dfrac{1kg}{1000g}*\dfrac{(100)^2cm^2}{1m^2}\\ =1.01332*10^5kg m^{-1}s^{-2}$ $\\$ We know,$\\$ $1N=1kg ms^{-2}$ $\\$ Then,$\\$ $1Pa=1Nm^{ -2} = 1 kg m^{ -2} s^{ -2}\\ 1 Pa = 1 kg m ^{-1} s ^{-2}\\ \therefore$ Pressure = $1.01332* 10^ 5 Pa$

14   What is the SI unit of mass? How is it defined?

##### Solution :

The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of the international prototype of kilogram.

15   Match the following prefixes with their multiples:$\\$ $\begin{array}{|c|c|}\hline & \text{Prefixes}& \text{Multiples}\\ \hline (i) & \text{micro}& 10^6 \\ \hline (ii) &\text{deca} & 10^9 \\ \hline (iii)& \text{mega} & 10^{-6}\\ \hline (iv)& \text{giga} & 10^{-15} \\ \hline (v)&\text{femto} & 10\\ \hline \end{array}$

##### Solution :

$\begin{array}{|c|c|}\hline & \text{Prefixes}& \text{Multiples}\\ \hline (i) & \text{micro}& 10^{-6} \\ \hline (ii) &\text{deca} & 10 \\ \hline (iii)& \text{mega} & 10^6\\ \hline (iv)& \text{giga} & 10^9 \\ \hline (v)&\text{femto} & 10^{-15}\\ \hline \end{array}$

16   What do you mean by significant figures?

##### Solution :

Significant figures are those meaningful digits that are known with certainty. They indicate uncertainty in an experiment or calculated value. For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures are 3. Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result.

17   A sample of drinking water was found to be severely contaminated with chloroform, $CHCl _3$, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).$\\$ (i) Express this in percent by mass.$\\$ (ii) Determine the molality of chloroform in the water sample.$\\$

##### Solution :

(i) 1 ppm is equivalent to 1 part out of 1 million ($10^ 6$ ) parts.$\\$ $\therefore$ Mass percent of 15 ppm chloroform in water$\\$ $=\dfrac{15}{10^6}*100\\= 1.5* 10^{-3}\%$ $\\$ (ii) $100 g$ of the sample contains $1.5 * 10 ^{-3} g$of $CHCI_ 3$ .$\\$ $\implies 1000 g$ of the sample contains $1.5 * 10^{ -2} g of CHCI_ 3$$\\ \therefore Molality of chloroform in water\\ =\dfrac{1.5*10^{-2}g}{\text{Molar massof CHCI_ 3}} \\ Molar mass of CHCI _3 = 12.00 + 1.00 + 3(35.5)\\ = 119.5 g mol^{ -1}$$\\$ $\therefore$ Molality of chloroform in water =$0.0125 * 10^{ -2} m\\ = 1.25 * 10^{ -4} m$

18   Express the following in the scientific notation:$\\$ (i) 0.0048$\\$ (ii) 234,000$\\$ (iii)8008$\\$ (iv) 500.0$\\$ (v) 6.0012$\\$

##### Solution :

$(i) 0.0048 = 4.8 * 10^{ -3}\\ (ii) 234,000 = 2.34 * 10^ 5\\ (iii)8008 = 8.008 * 10^ 3\\ (iv) 500.0 = 5.000 * 10 ^2\\ (v) 6.0012 = 6.0012 * 10 ^0$

19   How many significant figures are present in the following?$\\$ (i) 0.0025$\\$ (ii) 208$\\$ (iii)5005$\\$ (iv) 126,000$\\$ (v) 500.0$\\$ (vi) 2.0034$\\$

##### Solution :

(i) 0.0025$\\$ There are 2 significant figures.$\\$ (ii) 208$\\$ There are 3 significant figures.$\\$ (iii)5005$\\$ There are 4 significant figures.$\\$ (iv) 126,000$\\$ There are 3 significant figures.$\\$ (v) 500.0$\\$ There are 4 significant figures.$\\$ (vi) 2.0034$\\$ There are 5 significant figures.$\\$

20   Round up the following upto three significant figures.$\\$ (i) 34.216$\\$ (ii) 10.4107$\\$ (iii)0.04597$\\$ (iv) 2808$\\$

##### Solution :

(i) 34.2$\\$ (ii) 10.4$\\$ (iii)0.0460$\\$ (iv) 2810$\\$

##### Solution :

(i) Balancing the given chemical equation,$\\$ $N_{ 2( g )} + 3 H_ {2( g )} \to 2 NH {3( g )}$$\\ From the equation, 1 mole (28g) of dinitrogen reacts with 3 mole (6g) of dihydrogen to give 2 mole (34g) of ammonia.\\ = 2.00 * 10 ^3 g of dinitrogen will react with \dfrac{6g}{28g}*2.00*10^3g dihydrogen i.e.,\\ 2.00*10^3g of dinitrogen will react with 428.6 g of dihydrogen. \\ Given,\\ Amount of dihydrogen = 1.00 * 10^ 3 g$$\\$ Hence, $N _2$ is the limiting reagent.$\\$ $\therefore 28 g$ of$N _2$ produces $34 g$ of $NH_ 3$.$\\$ Hence, mass of ammonia produced by $2000 g$ of $N _2=\dfrac{34g}{28g}*2000g\\ =2428.57g$ $\\$ (ii) $N_ 2$ is the limiting reagent and $H_ 2$ is the excess reagent. Hence, $H_ 2$ will remain unreacted.$\\$ (iii) Mass of dihydrogen left unreacted =$1.00 * 10^ 3 g – 428.6 g$ = 571.4 g$\\$

25   How are $0.50 mol Na _2 CO _3$ and $0.50 M Na _2 CO _3$ different?

##### Solution :

Molar mass of $Na _2 CO_ 3 = (2 * 23) + 12.00 + (3 * 6)\\ = 106 g mol ^{-1}$$\\ Now, 1 mole of Na_ 2 CO_ 3 means 106 g of Na_ 2 CO _3$$\\$. $\therefore 0.5$ mol of $Na_2CO_3=\dfrac{106g}{1 mole}*0.5 mol \ \ Na_2CO_3\\ =53gNa_2CO_3\\ =0.50 M\ \ of \ \ Na_2CO_3$ $\\$ Hence, $0.50 M\ \ of \ \ Na_2CO_3$is present in 1 L of water or $53 g Na _2 CO _3$ is present in 1 L of water.

26   If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

##### Solution :

27   Convert the following into basic units:$\\$ (i) 28.7 pm$\\$ (ii) 15.15 pm$\\$ (iii)25365 mg$\\$

##### Solution :

$(i) 28.7 pm:\\ 1 pm = 10^{ -12} m\\ \therefore 28.7 pm = 28.7 * 10 ^{- 12} m\\ = 2.87 * 10 ^{- 11} m\\ (ii) 15.15 pm:\\ 1 pm = 10^{ -12} m\\ \therefore 15.15 pm = 15.15 * 10 ^{- 12} m\\ = 1.515 * 10 ^{- 12} m\\ (iii) 25365 mg:\\ 1 mg = 10^{ -3} g\\ 25365 mg = 2.5365 * 10^ 4 * 10^{ -3} g\\ Since,\\ 1 g = 10 ^{-3} kg\\ 2.5365 * 10 ^1 g = 2.5365 * 10^{ -1} * 10 ^{-3} kg\\ \therefore 25365 mg = 2.5365 * 10^{ -2} kg$

28   Which one of the following will have largest number of atoms?$\\$ (i) 1 g Au (s)$\\$ (ii) 1 g Na (s)$\\$ (iii)1 g Li (s)$\\$ (iv) 1 g of CI 2 (g)

##### Solution :

$(i) 1 g$of Au $(s) =\dfrac{1}{197}$ mol of Au (s)$\\$ $=\dfrac{6.022*10^{23}}{197}$ atoms of Au (s)$\\$ $3.06*10^{21}$atoms of Au (s)$\\$ $(ii) 1 g$ of Na (s) $=\dfrac{1}{23}$mol of Na (s)$\\$ $\dfrac{6.022*10^{23}}{23}$ atoms of Na (s)$\\$ =$0.262 * 10 ^{23 }$atoms of Na (s)$\\$ = $26.2 * 10 ^{21}$ atoms of Na (s)$\\$ $(iii) 1 g$of Li (s) $=\dfrac{1}{7}$mol of Li (s)$\\$ $=\dfrac{6.022*10^{23}}{7}$atoms of Li (s) $\\$ $= 0.86 * 10 ^{23}$ atoms of Li (s)$\\$ $= 86.0 * 10 ^{21}$ atoms of Li (s)$\\$ $(iv) 1 g$ of $CI_ 2 (g) =\dfrac{1}{71}$ mol of $CI_ 2 (g)$$\\ \dfrac{6.022*10^{23}}{71} atoms of CI_ 2 (g)$$\\$ $= 0.0848 * 10 ^{23}$ atoms of $CI_ 2 (g)$$\\ = 8.48 *10 ^{21} atoms of CI _2 (g)$$\\$ Hence,$1 _g$of Li (s) will have the largest number of atoms.

29   Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

##### Solution :

Mole fraction of $C_ 2 H_ 5 OH =\dfrac{\text{Number of moles}C _2 H _5 OH}{\text{Number of molesof solu tion}}\\ 0.040=\dfrac{n_{C_2H_5OH}}{n_{C_2H_5OH}+n_{H_2o}}....(1)$ $\\$ Number of moles present in 1 L water:$\\$ $n_{H_2O}=\dfrac{1000g}{18g mol^{-1}}\\ n_{H_2O}=55.55 mol$ $\\$ Substituting the value of $n_{ H _2 O}$ in equation (1),$\\$ $\dfrac{n_{C_2H_5OH}}{n_{C_2H_5OH}+55.55}=0.040\\ n_{C_2H_5OH}=0.040n_{C_2H_5OH}+(0.040)(55.55)\\ 0.96n_{C_2H_5OH}=2.222 mole\\ n_{C_2H_5OH}=\dfrac{2.222}{0.96} mole\\ n_{C_2H_5OH}=2.314$mole$\\$ Molarity of solution$=\dfrac{2.314 mol}{1 L}\\ =2.314M$ $\\$

30   What will be the mass of one $^{12 }C$ atom in g?

##### Solution :

1 mole of carbon atoms = $6.023 * 10^{ 23}$ atoms of carbon$\\$ = 12 g of carbon$\\$ $\therefore$Mass of one $^{12} C$ atom =$\dfrac{12g}{6.022*10^{23}}\\ =1.993*10^{-23}g$

31   How many significant figures should be present in the answer of the following calculations?$\\$ $(i)=\dfrac{0.2856 * 298.15 * 0.112}{0.5785}\\ (ii)=5 * 5.364\\ (iii)0.0125 + 0.7864 + 0.0215$

$(i)=\dfrac{0.2856 * 298.15 * 0.112}{0.5785}$$\\ Least precise number of calculation = 0.112 \therefore Number of significant figures in the answer\\ = Number of significant figures in the least precise number\\ = 3\\ (ii) 5 * 5.364\\ Least precise number of calculation = 5.364\\ \therefore Number of significant figures in the answer = Number of significant figures in 5.364\\ = 4\\ (iii) 0.0125 + 0.7864 + 0.0215\\ Since the least number of decimal places in each term in four, the number of significant figures in the answer is also 4.\\ 32 Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:\\ \begin{array}{|c|c|}\hline \text{Isotope}&\text{Isotopic molar mass}&\text{Abundance}\\ \hline ^{36}Ar&35.96755 g \ mol ^{-1}&0.337\%\\ \hline ^{36}Ar&37.96272 g \ mol^{ -1}&0.063\%\\ \hline ^{40}Ar&39.9624 g \ mol^{-1}&99.600\%\\ \hline \end{array} ##### Solution : Molar mass of argon\\ =[(35.96755* \dfrac{0.337}{100})+(37.96272 * \dfrac{0.063}{100})+(39.9624* \dfrac{90.60}{100})]g mol^{-1} \\ =[ 0.121 + 0.024 + 39.802 ] g \ \ mol ^{- 1}\\ =39.947 g \ \ mol ^{-1}$$\\$

33   Calculate the number of atoms in each of the following$\\$ (i) 52 moles of Ar$\\$ (ii) 52 u of He$\\$ (iii) 52 g of He.$\\$

##### Solution :

(i) 1 mole of Ar =$6.022 * 10^{ 23}$ atoms of Ar$\\$ $\therefore$ 52 mole of Ar =$52 * 6.022 * 10 ^{23 }$ atoms of Ar$\\$ =$3.131 * 10 ^{25}$ atoms of Ar$\\$ (ii) 1 atom of He = 4 u of the$\\$ Or,$\\$ 4 u of He = 1 atom of He$\\$ 1 u of He=$\dfrac{1}{4}$ atom of He $\\$ 52u of He=$\dfrac{52}{4}$ atom of He$\\$ =13 atoms of He$\\$ (iii) 4 g of He =$6.022 * 10^{ 23}$ atoms of He$\\$ $\therefore$ 52 g of He =$\dfrac{6.022*10^{23}*52}{4}$ atoms of He $\\$ $=7.8286*10^{24}$ atoms of He

34   A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen$\\$ gives 3.38 g carbon dioxide, 0.690 g water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate$\\$ (i) empirical formula,$\\$ (ii) molar mass of the gas, and$\\$ (iii) molecular formula.

(i) 1 mole (44 g) of $CO_ 2$ contains 12 g carbon. $\\$ $\therefore$ 3.38 g of $CO_ 2$ will contain carbon $=\dfrac{12g}{44 g} =3.38 g\\ = 0.9217 g$$\\ 18 g of water contains 2 g of hydrogen.\\ \therefore 0.690 g of water will contain hydrogen =\dfrac{ 2 g}{18 g}=0.690\\ = 0.0767 g$$\\$ Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:$\\$ $= 0.9217 g + 0.0767 g\\ = 0.9984 g$$\\ \therefore Percent of C in the compound =\dfrac{0.9217 g}{0.9984 g}* 100\\ = 92.32\%$$\\$ Percent of H in the compound =$\dfrac{ 0.0767 g}{0.9984 g}* 100\\ = 7.68\%$$\\ Moles of carbon in the compound =\dfrac{ 92.32} {12.00}\\ = 7.69$$\\$ Moles of hydrogen in the compound = $\dfrac{ 7.68}{ 1}\\ = 7.68$$\\ \therefore Ration of carbon to hydrogen in the compound = 7.69 : 7.68\\ = 1.1$$\\$ Hence, the empirical formula of the gas is CH.$\\$ (ii) Given,$\\$ Weight of $10.0L$ of the gas (at S.T.P) = $11.6 g$ $\therefore$ Weight of 22.4 L of gas at STP = $\dfrac{11.6 g}{10.0 L}* 22.4 L\\ = 25.984 g\\ \approx 26 g$$\\ Hence, the molar mass of the gas is 26 g.\\ (iii) Empirical formula mass of CH = 12 + 1 = 13 g n=\dfrac{\text{Molar mass of gas}} {\text{Empirical Formula mass of gas}}\\ =\dfrac{26 g}{ 13 g}\\ n = 2$$\\$ $\therefore$ Molecular formula of gas = $(CH)_ n$$\\ = C _2 H_ 2 35 Calcium carbonate reacts with aqueous HCl to given CaCl _2 and CO _2 according to the reaction, CaCO _{3( s )} + 2 HCl_{ ( aq ) }+ CaCl _{2( aq ) }+ CO_ {2( g )}+ H 2 O_{ ( l )} What mass of CaCO _{3} is required to react completely with 25 mL of 0.75 M HCl? ##### Solution : 0.75 M of HCI = 0.75 mol of HCl are present in 1 L of water\\ =[( 0.75\ mol )*( 36.5 g \ mol ^{- 1 })] HCl is present in 1 L of water\\ = 27.375 g of HCI is present in 1 L of water\\ Thus, 1000 mL of solution contains 27.375 g of HCl.\\ \therefore Amount of HCl present in 25 mL of solution \dfrac{27.375 g}{1000 mL}* 25 mL\\ = 0.6844 g$$\\$ From the given chemical equation,$\\$ $CaCO_ 3( s ) + 2 HCl ( aq ) \to CaCl _2( aq ) + CO_ 2( g ) + H _2 O_{ ( l )}$$\\ 2 mol of HCl (2 * 36.5 = 71 g) react with 1 mol of CaCO _3 (100 g).$$\\$ $\therefore$ Amount of CaCO 3 that will react with $0.6844 g =\dfrac{100}{71} * 0.6844 g\\ = 0.9639 g$

36   Chlorine is prepared in the laboratory by treating manganese dioxide (MnO 2 ) with aqueous hydrochloric acid according to the reaction$\\$ $4 HCl ( aq ) + MnO_ 2( l ) + MnCl_ 2( aq ) + Cl_ 2( g )$$\\$ How many grams of HCl react with 5.0 g of manganese dioxide?

##### Solution :

1 mol$[55 + 2 * 16 = 87 g] MnO _2$ reacts completely with 4 mol $[4 * 36.5 = 146 g]$ of HCl.$\\$ $\therefore 5.0 g \ of\ MnO_ 2$will react with$\\$ $\dfrac{146 g}{87 g} * 5.0 g\\ =8.4 g$ of HCl$\\$ Hence, $8.4 g$ of HCl will react completely with$5.0 g$ of manganese dioxide.