Question

Vapour pressure of ${CCI}_4$ at ${24}^{o}C$ is $143mmHg0.05$g of the non-volatile solute (mol.wt.=65) is dissolved in $100ml{CCI}_4$. Find the vapour pressure of the solution (Density of ${CCI}_4$=$1.58g/{cm}^2$)

• A. $143.99mm$
• B. $94.39mm$
• C. $199.34mm$
• D. $14.197mm$

Answer 1

Answer: (A)

$143.99mm$

The equation of relative lowering vapour pressure is given as-

$\frac{p^0-p}{p^0}=X_2$

Whereas,

$p^0=$ Vapour prssure of solvent

$p=$ Vapour pressure of solution

$X_2=$ Mole fraction of solute

Given that density of $C{Cl}_4$ is $1.58g/{cm}^3$

As we know that,

$\text{density}=\frac{\text{mass}}{\text{volume}}$

$\Rightarrow 1.58=\frac{\text{mass}}{100}$

$\Rightarrow \text{mass of}C{Cl}_4=158g$

Molecular weight of $C{Cl}_4=154g$

$\therefore$ No. of moles of $C{Cl}_4=\frac{158}{154}=1.026\text{mol}$

Given weight of solute $=0.05g$

Molecular weight of solute $=65g$

$\therefore$ No. of moles of solute $=\frac{0.05}{65}=0.00077\text{mol}$

$\therefore$ Total no. of moles $=1.026+0.00077=1.02677\text{mol}$

$\therefore$ Mole fraction of solute $=\frac{0.00077}{1.02677}=0.00075$

Vapour pressure of solvent $=143\text{mm of Hg}$

Noe from equation of relative lowering vapour pressure, we have

$\frac{143-p}{143}=0.00075$

$\Rightarrow 143-p=0.107$

$\Rightarrow p=142.893\text{mm of Hg}$

Hence the vapour pressure of solution is $142.893\text{mm of Hg}$.