Question

The addition of $3g$ of substance to $100gCC{l}_4(M=154gmol6-1)$ raises the boiling point of
$CC{l}_4$ by ${0.60}^{\circ }C$ of $K_b\left(CC{l}_4\right)$ is $5.03kgmo{l}^{-1}K$. Calculate:
(a)the freezing point depression
(b)the relative lowering of vapour pressure
(c)the osmotic pressure at $298K$
(d)the molar mass of the substance
Given $K_f\left(CC{l}_4\right)=31.8kgmo{l}^{-1}K$ and $\rho \left(density\right)ofsolution=1.64g/c{m}^3$.

• A. (a)${3.793}^{\circ }C$,(b)$0.018$, (c)$4.65atm$,(d)$251.5$
• B. (a)${379.3}^{\circ }C$,(b)$0.18$, (c)$4.65atm$,(d)$251.5$
• C. (a)${3.793}^{\circ }C$,(b)$0.018$, (c)$46.5atm$,(d)$25$
• D. none of the above

Answer 1

The correct option is A (a)${3.793}^{\circ }C$,(b)$0.018$, (c)$4.65atm$,(d)$251.5$

(a)$\frac{\Delta T_f}{\Delta T_b}=\frac{k_f}{k_b}=\Delta T_f=\frac{0.6\times 31.8}{5.03}={3.793}^{\circ }C$
(b)Relative lowering of vapour pressure$=\frac{n}{n+N}=\frac{\frac{3}{251.5}}{\frac{3}{251.5}+\frac{100}{154}}=0.018$
(c)$\pi =CRT$
$n=\frac{3}{251.5}=0.012$

$v=\frac{103}{1.64}=62.8mL$

$\pi =\frac{0.012}{0.0628}\times 0.0821\times 298=4.65atm$

(d)$0.6=\frac{5.03\times 3\times 1000}{M{}_w\times 100}\Rightarrow M_w=251.5$