Question

An aqueous solution freezes at $-0.186^{\circ} C \left(k_{f}=1.86^{\circ} ; k_{b}=0.512^{\circ}\right)$ What is the elevation in boiling point?

• A. $ 0.186 $
• B. $ 0.512 $
• C. $ \frac{0.512}{1.86} $
• D. $ 0.0512 $

Answer 1

Answer: (D)

$\Delta T_{f}=k_{f}'\times m$
$0.186=1.86 \times m$
$m=\frac{0.186}{1.86}=0.1 \,mol$
$\therefore \Delta T_{b}=k_{b} \times$ molality
$\Delta T_{b}=0.512 \times 0.1$
$\Delta T_{b}=0.0512$