Question

When $2\, g$ of a non-volatile solute was dissolved in $90 \,g$ of benzene the boiling point of benzene is raised by $0.88 \,K$. Which of the following may be the solute? $\left(K_{b}\right.$ for benzene $\left.=2.53 \,K \,kg\, mol ^{-1}\right)$

• A. $CO \left( NH _{2}\right)_{2}$
• B. $C _{6} H _{12} O _{6}$
• C. $NaCl$
• D. None of these.

Answer 1

Answer: (D)

$w_{A}=2 \,g , w_{B}=90 \,g , M_{B}=78$

$\Delta T_{b}=0.88, K_{b}=2.53 \,K \,kg\, mol ^{-1}$

$\Delta T_{b}=K_{b} \times m$

$0.88=2.53 \times \frac{2 / m_{A}}{90} \times 1000 $

$\Rightarrow 0.88=\frac{2.53 \times 2}{90 \times m_{A}} \times 1000$

$m_{A}=\frac{2.53 \times 2 \times 1000}{90 \times 0.88}=63.8 \approx 64$

Molecular Mass of $CO \left( NH _{2}\right)_{2} \Rightarrow 60$

Molecular Mass of $C _{6} H _{12} O _{6} \Rightarrow 180$

Molecular Mass of $NaCl \Rightarrow 58.5$