Question

When a non-volatile solute of $10g$ is added to $150g$ of carbon disulphide, the boiling point of the solution raised to ${0.55}^{\circ }C$ above that of pure solvent. The boiling point of pure carbon disulphide is ${46.30}^{\circ }C$ and its heat of vapourisation is $84cal/g$. Determine the molar mass of the solute.

• A. $294.5g/mol$
• B. $320g/mol$
• C. $284.8g/mol$
• D. $193.5g/mol$

Answer 1

The correct option is A $294.5g/mol$

We know,
Boiling point elevation constant,
$K_b=\frac{R{T}_b^2}{1000L_v}$
where,
$T_b$ is boiling point of pure solvent in $K$
$R\text{in}cal/mol/K$
$L_v$ is latent heat of vaourisation of solvent in $cal/g$
Thus,
$K_b=\frac{2\times (46.30+273{)}^2}{1000\times 84}=2.43$
Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
$\therefore m=\frac{△T_b}{K_b}=\frac{0.55}{2.43}$
And also, $m=\frac{\text{moles of solute}}{\text{wt. of solvent in grams}}\times 1000$
$=\frac{10}{M}\times \frac{1000}{150}=\frac{200}{3M}$
where, M is the molar weight of the non-volatile solute.
Thus, $\frac{200}{3M}=\frac{0.55}{2.43};M=294.5g/mol$
Thus, the molar mass of non-volatile solute is $294.5g/mol$