Question

A solution prepared by dissolving $0.8gm$ of naphthalene in $100g$ of $CC{l}_4$ has boiling point elevation of ${0.4}^{o}C$. A $1.24g$ of an unknown solute in same amount of $CC{l}_4$ produced boiling point elevation of ${0.62}^{o}C$, then molar mass of unknown solute is:

• A. $25g$
• B. $50g$
• C. $75g$
• D. $128g$

Answer 1

Answer: (A)

$25g$

In 1st case

$w_2=0.8$ g

$M_2$ of naphthalne $C_{10}{H}_8=128$

$w_1=100$ g

$\Delta T_b={0.4}^0$ C

$\Delta T_b=K_{b}m$

Where m is the molality

$K_b$=Molal elevation constant

$\Delta T_b=K_b\frac{w_2\times 1000}{M_2{w}_1}$

$K_b=\frac{128\times 0.4}{8}=6.4$ K Kg ${mol}^{-}$

In second case

$w_2=1.24$ g

$w_1=100$ g

$K_b=6.4$ K Kg ${mol}^{-}$

$\Delta T_b={0.62}^0$ C

$M_2=\frac{1.24\times 6.4\times 10}{0.62}=128g$