Question

$1.5\, g$ of a non-volatile, non-electrolyte is dissolved in $50 \,g$ benzene $(K_b\, =\, 2.5\, K kg \,mol^{-1}$ ). The elevation of the boiling point of the solution is $0.75\, K$. The molecular weight of the solute in $g$ $mol^{-1}$ is

• A. 200
• B. 50
• C. 75
• D. 100

Answer 1

Answer: (D)

100

Answer 2

$\Delta Tb = Kb \cdot \,m$
where,
$\Delta Tb$ is the boiling point elevation,
$Tb$ (solution) $-$ Tb (pure solvent),
$Kb$ is the ebullioscopic constant,
$m$ is the molality of the solution
$\Delta Tb =0.75 K$
weight of solute $=1.5 g$
weight of solvent $=50 g$
molality $=$
weight of solute molar weight of solute $1000$ weight of solvent in $g =1.5 \, M \times 100050 \,g$
Now,$0.75 \, K =2.5 \, K \, kg \, mol ^{-1} \times 1.5 \, M \times 100050$
$M =2.5 \,K \,kg \,mol^ {-1} \times 1.5 \times 10000.70 \,K \times 50=100.8 \,g \, mol ^{-1}$
Molecular weight of solute is $100.8 \, g \,mol^{-1}.$