Question

If $0.15\, g$ of a solute dissolved in $15 \,g$ of solvent is boiled at a temperature higher by $0.216^{\circ} C$, than that of the pure solvent, the molecular weight of the substance is: (Molal elevation constant for the solvent is $216^{\circ} C$ )

• A. 10
• B. 10.1
• C. 100
• D. 1.01

Answer 1

Answer: (C)

Key Idea: Use the following formula to find molecular formula of solute.
$M=\frac{K_{b} \times w \times 1000}{\Delta T_{b} \times W}$
where $M=$ molecular mass of solute
$w=$ weight of solute $=0.15\, g$
$W=$ weight of solvent $=15\, g $
$\Delta T_{b}=$ elevation in boiling point
$=0.216^{\circ} C M$
$=\frac{2.16 \times 0.15 \times 100}{0.216 \times 15}=100$