Answer

Mass percentage of $\mathrm{C_6} \mathrm{H_6}$

$$ =\frac{\text { Mass of } \mathrm{C_6} \mathrm{H_6}}{\text { Total mass of the solution }} \times 100 \% $$

$$ \begin{aligned} & =\frac{\text { Mass of } \mathrm{C_6} \mathrm{H_6}}{\text { Mass of } \mathrm{C_6} \mathrm{H_6}+\text { Mass of } \mathrm{CCl_4}} \times 100 \% \\ & =\frac{22}{22+122} \times 100 \% \\ & =15.28 \% \end{aligned} $$

Mass percentage of $\mathrm{CCl_4} \quad$ Total mass of the solution

$$ =\frac{\text { Mass of } \mathrm{CCl_4}}{\text { Total mass of the solution }} \times 100 \% $$

$$ =\frac{\text { Mass of } \mathrm{CCl_4}}{\text { Mass of } \mathrm{C_6} \mathrm{H_6}+\text { Mass of } \mathrm{CCl_4}} \times 100 \% $$

$=\frac{122}{22+122} \times 100 \%$

$=84.72 \%$

Alternatively,

Mass percentage of $\mathrm{CCl_4}=(100-15.28) \%$

$=84.72 \%$

Answer

Let the total mass of the solution be $100 \mathrm{~g}$ and the mass of benzene be $30 \mathrm{~g}$.

$\therefore$ Mass of carbon tetrachloride $=(100-30) \mathrm{g}$

$=70 \mathrm{~g}$

Molar mass of benzene $\left(\mathrm{C_6} \mathrm{H_6}\right)=(6 \times 12+6 \times 1) \mathrm{g} \mathrm{mol}^{-1}$

$=78 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Number of moles of

$$ \mathrm{C_6} \mathrm{H_6}=\frac{30}{78} \mathrm{~mol} $$

$=0.3846 \mathrm{~mol}$

Molar mass of carbon tetrachloride $\left(\mathrm{CCl_4}\right)=1 \times 12+4 \times 355$

$=154 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Number of moles of $\mathrm{CCl_4}=\frac{70}{154} \mathrm{~mol}$

$=0.4545 \mathrm{~mol}$

Thus, the mole fraction of $\mathrm{C_6} \mathrm{H_6}$ is given as:

Number of moles of $\mathrm{C_6} \mathrm{H_6}$

Number of moles of $\mathrm{C_6} \mathrm{H_6}+$ Number of moles of $\mathrm{CCl_4}$

$=\frac{0.3846}{0.3846+0.4545}$

$=0.458$

Answer

Molarity is given by:

$$ \text { Molarity }=\frac{\text { Moles of solute }}{\text { Volume of solution in litre }} $$

(a) Molar mass of $\mathrm{Co}\left(\mathrm{NO_3}\right)_{2} \cdot 6 \mathrm{H_2} \mathrm{O}=59+2(14+3 \times 16)+6 \times 18$

$=291 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Moles of $\mathrm{Co}\left(\mathrm{NO_3}\right)_{2} \cdot 6 \mathrm{H_2} \mathrm{O}=\frac{30}{291} \mathrm{~mol}$

$=0.103 \mathrm{~mol}$

Therefore, molarity $=\frac{0.103 \mathrm{~mol}}{4.3 \mathrm{~L}}$

$=0.023 \mathrm{M}$

(b) Number of moles present in $1000 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H_2} \mathrm{SO_4}=0.5 \mathrm{~mol}$ $\therefore$ Number of moles present in $30 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H_2} \mathrm{SO_4}=\frac{0.5 \times 30}{1000} \mathrm{~mol}$

$=0.015 \mathrm{~mol}$

Therefore, molarity

$$ =\frac{0.015}{0.5 \mathrm{~L}} \mathrm{~mol} $$

$=0.03 \mathrm{M}$

Answer

Molar mass of urea $\left(\mathrm{NH_2} \mathrm{CONH_2}\right)=2(1 \times 14+2 \times 1)+1 \times 12+1 \times 16$

$=60 \mathrm{~g} \mathrm{~mol}^{-1}$

0.25 molar aqueous solution of urea means:

$1000 \mathrm{~g}$ of water contains $0.25 \mathrm{~mol}=(0.25 \times 60) \mathrm{g}$ of urea

$=15 \mathrm{~g}$ of urea

That is,

$(1000+15) \mathrm{g}$ of solution contains $15 \mathrm{~g}$ of urea

Therefore, $2.5 \mathrm{~kg}(2500 \mathrm{~g})$ of solution contains

$$ =\frac{15 \times 2500}{1000+15} \mathrm{~g} $$

$=36.95 \mathrm{~g}$

$=37 \mathrm{~g}$ of urea (approximately)

Hence, mass of urea required $=37 \mathrm{~g}$

Note:There is a slight variation in this answer and the one given in the NCERT textbook.

Answer

(a) Molar mass of $\mathrm{KI}=39+127=166 \mathrm{~g} \mathrm{~mol}^{-1}$

$20 \%$ (mass/mass) aqueous solution of $\mathrm{KI}$ means $20 \mathrm{~g}$ of $\mathrm{KI}$ is present in $100 \mathrm{~g}$ of solution.

That is, $20 \mathrm{~g}$ of KI is present in $(100-20) \mathrm{g}$ of water $=80 \mathrm{~g}$ of water

Therefore, molality of the solution

$$ =\frac{\text { Moles of KI }}{\text { Mass of water in } \mathrm{kg}} $$

$=\frac{\frac{20}{166}}{0.08} \mathrm{~m}$

$=1.506 \mathrm{~m}$

$=1.51 \mathrm{~m}$ (approximately)

(b) It is given that the density of the solution $=1.202 \mathrm{~g} \mathrm{~mL}^{-1}$

$=\frac{100 \mathrm{~g}}{1.202 \mathrm{~g} \mathrm{~mL}^{-1}}$

$$ =\frac{\text { Mass }}{\text { Density }} $$

$=83.19 \mathrm{~mL}$

$=83.19 \times 10^{-3} \mathrm{~L}$

Therefore, molarity of the solution

$$ =\frac{\frac{20}{166} \mathrm{~mol}}{83.19 \times 10^{-3} \mathrm{~L}} $$

$=1.45 \mathrm{M}$

(c) Moles of KI

$$ =\frac{20}{166}=0.12 \mathrm{~mol} $$

Moles of water

$$ =\frac{80}{18}=4.44 \mathrm{~mol} $$

Therefore, mole fraction of $\mathrm{KI}$

$$ =\frac{\text { Moles of KI }}{\text { Moles of KI }+ \text { Moles of water }} $$

$$ =\frac{0.12}{0.12+4.44} $$

$=0.0263$