Question

The boiling point of benzene is $353.23\, K$. When $1.80 \,g$ of a nonvolatile solute was dissolved in $90 \,g$ of benzene, the boiling point is raised to $354.11\, K$. The molar mass of the solute is :
$\left[K_{b}\right.$ for benzene $=2.53 \,K\, kg\, mol ^{-1}$ ]

• A. $5.8 \,g \,mol ^{-1}$
• B. $0.58\, g\, mol^{-1} $
• C. $58 \,g \, mol^{-1} $
• D. $0.88\, g \,mol^{-1} $

Answer 1

Answer: (C)

Use the following formula to find the molecular mass of solute.
$M=\frac{K_{b} \times 1000 \times w}{\Delta T_{b} \times W}$
where, $w =$ weight of solute $=1.80 g$
$W =$ weight of solvent $=90 g$
$\Delta T_{b}=$ elevation in boiling point
$=354.11 K -353.23 K =0.88 K$
$K_{b}=2.53 \,K \,kg \,mol ^{-1}$
$\therefore M=\frac{2.53 \times 1000 \times 1.80}{0.88 \times 90}$
$=58 \,g \,mol ^{-1}$