Question

Boiling point of chloroform was raised by $0.323 \,K$, when $0.5143\, g$ of anthracene was dissolved in its $35 \,g$. Molecular mass of anthracene is $(K_b$ for $CHCl_3 = 3.9 \,K \,kg\,, mol^{-1})$

• A. $132.32 \,g/mol$
• B. $242.32 \,g/mol$
• C. $177.42 \,g/mol$
• D. $79.42 \,g/mol$

Answer 1

Answer: (C)

$\Delta T_{b}=0.323 \, K$, $w=0.5143\,g, W=35\,g$
$K_{b} \left(CHCl_{3}\right)=3.9\,Kg\,mol^{-1}, M=$?
As, $\Delta T_{b}=\frac{1000\times K_{b} \times w}{M \times W}$
$\Rightarrow $ Molecular mass of anthracene $(M)$
$=\frac{1000\times3.9\times0.5143}{35\times0.323}$
$=177.42\, g / mol$