Question

What is the boiling point of a solution containing $0.456\, g$ of camphor (molar mass $=152$ ) dissolved in $31.4\, g$ of acetone $\left(\right.$ b. $\left.p =56.30^{\circ} C \right)$, if the molecular elevation constant per $100\, g$ of acetone is $17.2^{\circ} C ?$

• A. $56.46^{\circ} C$
• B. $36.56^{\circ} C$
• C. $56.14^{\circ} C$
• D. $72.52^{\circ} C$

Answer 1

Answer: (A)

We know that
$\Delta T_{b}=\frac{100 k_{b} \times w}{W \times m}$
$=\frac{100 \times 17.2 \times 0.456}{31.4 \times 152} $
$=0.16^{\circ} C$
$\therefore $ Boiling point of solution
$\left(T_{s}\right)=T^{\circ}+\Delta T_{b} $
$=56.30+0.16$
$=56.46^{\circ} C$