Question

$0.15g$ of a substance dissolved in $15g$ of a solvent boiled at a temp. higher by ${0.216}^{0}C$ then that of the pure solvent. Find out the molecular weight of the substance.

[$k_b$ for solvent is ${2.16}^{o}C$]

• A. $1.01$
• B. $10.1$
• C. $100$
• D. $10$

Answer 1

The correct option is C $100$

Weight of substance = 0.15g

Weight of solvent = 15g = 0.015 kg

m = molality

As we know that,

molality $\left(m\right)$ = $\frac{\text{moles of solute}}{\text{weight of solvent}\left(\text{in Kg}\right)}$

Let the molecular weight of substance dissolved be $x$.

$\therefore m=\frac{\left(\frac{\text{weight of substance}}{\text{mol. weight of substance}}\right)}{\text{weight of solvent}\left(\text{in Kg}\right)}=\frac{\left(\frac{0.15}{x}\right)}{0.015}=\frac{10}{x}$

As we know that,

$\Delta T_b=k_b.m$

where,

$\Delta T_b$ = Elevation in boiling point = $0.216℃$

$k_b=2.16℃$

m = molality = $\frac{10}{x}$

$\therefore 0.216=2.16\times \frac{10}{x}$

$\Rightarrow x=100$

Hence, the molecular weight of the substance is 100 g.

Hence, the correct option is $C$