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Section: Chemistry

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Question : 6 of 29

Marks: +1, -0

A solution of urea in water has a boiling point of

100.18^{\circ}\mathrm{C}

. What is the freezing point of the same solution, if

K_{f}

K_{b}

of water are 1.86 and

0.52\,\mathrm{K}\,\mathrm{kg}\,\mathrm{mol}^{-1}

, respectively? (Boiling point of water

=100^{\circ}\mathrm{C}

[AP EAPCET 23 May 2025 S1]

$\;-0.34^{\circ}\mathrm{C}$
$\;-0.22^{\circ}\mathrm{C}$
$\;-0.64^{\circ}\mathrm{C}$
$\;-0.32^{\circ}\mathrm{C}$

Solution:

Elevation in boiling point,

\Delta T_b=100.18^{\circ}\mathrm{C}-100^{\circ}\mathrm{C}=0.18^{\circ}\mathrm{C}

Formula for elevation in boiling point is

\Delta T_b\;=K_b \cdot m

m\;=\;\frac{0.18}{0.52}=0.346\,\mathrm{mol/kg}

Depression in freezing point is,

\Delta T_f=K_f \cdot m=1.86 \times 0.346=0.643\,\mathrm{K}

Freezing point of sol

0^{\circ}\mathrm{C}-0.643^{\circ}\mathrm{C}=-0.64^{\circ}\mathrm{C}

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