Question

If $0.15\, g$ of a solute dissolved in $15$ got solvent is boiled at a temperature higher by $ 0.216^{\circ}C $ , than that of the pure solvent, the molecular weight of the substance, (molal elevation constant for the solvent is $ 2.16^{\circ }C$ ):

• A. 100
• B. 10.1
• C. 10
• D. 1.01

Answer 1

Answer: (A)

Molecular weight of solute is calculated by using following formula.
$M_{B}=\frac{K_{b} \times W_{B} \times 1000}{\Delta T_{b} \times W_{a}}$
where, $M_{B}=$ molecular v/eight of solute
$W_{B}=$ weight of solute $=0.15\, g$
$\Delta T_{b}=$ elevation in boiling point
$=0.216^{\circ} C W_{a}=$ weight of solvent $=15.0\, g$
$K_{b}=2.16$
$\therefore M_{B}=\frac{2.16 \times 0.15 \times 1000}{0.216 \times 15}$
$=100$