Question

Freezing point of urea solution is $-0.6^{\circ} C$. How much urea $( m . wt .=60\, g / mol )$ will be required to dissolve in $3\, kg$ water ?
$\left(k_{f}=1.5^{\circ} C\, kg\, mol ^{-1}\right)$

• A. 24 g
• B. 36 g
• C. 60 g
• D. 72 g

Answer 1

Answer: (D)

$\Delta T_{f}=\frac{k_{f} \times 1000\times w}{M\times W}$
$w =\frac{\Delta T_{f} \times M \times W}{k_{f} \times 1000}$
$\Delta T_{f} =0-(-0.6)=0.6^{\circ} C$
$M =60\, g / mol$
$k_{f} =1.5^{\circ} C\, kg\, mol ^{-1}$
$W =3 \times 10^{3} g$
$w =\frac{0.6 \times 60 \times 3 \times 10^{3}}{1.5 \times 1000}$
$=72\, g$