Question

The boiling point of benzene is $353.23K$. When $1.80g$ of a nonvolatile solute was dissolved in $90g$ of benzene, the boiling point is raised to $354.11K$. The molar mass of the solute is :
$[K_b$ for benzene $=2.53Kkgmo{l}^{-1}$ ]

• A. $5.8gmo{l}^{-1}$
• B. $0.58gmo{l}^{-1}$
• C. $58gmo{l}^{-1}$
• D. $0.88gmo{l}^{-1}$

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 $=1.80g$
$W=$ weight of solvent $=90g$
$\Delta T_b=$ elevation in boiling point
$=354.11K-353.23K=0.88K$
$K_b=2.53Kkgmo{l}^{-1}$
$\therefore M=\frac{2.53\times 1000\times 1.80}{0.88\times 90}$
$=58gmo{l}^{-1}$