Chemistry MCQ

$\begin{aligned}& \text {(d): We know, } \Delta T_f=K_f \times \frac{w_B \times 1000}{M_B \times w_A} \\& M_B=\frac{4.9}{2.5} \times \frac{5 \times 1000}{70}=\frac{24500}{175}=140 \\ & M_B=140 \mathrm{~g} \mathrm{~mol}^{-1}\end{aligned}$

(c) : $\Delta T_f=K_f \times m=K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}$

Given, $K_f=1.17, w_1=98 \mathrm{~g}, M_2=60, \Delta T_f=0.2 \mathrm{~K}$

$\therefore 0.2=1.17 \times \frac{w_2 \times 1000}{60 \times 98}$

or, $w_2=\frac{0.2 \times 60 \times 98}{1.17 \times 1000} \mathrm{~g}=1.005 \mathrm{~g}$

(a): $\frac{P_1^{\circ}-P_s}{P_1^{\circ}}=\frac{n_2}{n_1+n_2}$

$P_1^{\circ}=$ vapour pressure of pure water

$P_s=$ vapour pressure of the solution

$n_2=$ number of moles of the solute

$n_1=$ number of moles of the solvent

$\begin{aligned}& \frac{P_1^{\circ}-P_s}{P_1^{\circ}}=\frac{0.1}{0.1+9.9}=\frac{0.1}{10}=0.01\left[\because n_2=\frac{9}{90}=0.1\right] \\& P_1^{\circ}-P_s=0.01 P_1^{\circ} \Rightarrow P_1^{\circ}-0.01 P_1^{\circ}=P_s \Rightarrow P_s=0.99 P_1^{\circ}\end{aligned}$