A solution of urea (mol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf, and Kb, for water are 1.86 and 0.512 K kg kg mol-1 respectively, the above solution will freeze at

A
0.654°C
B
—0.654°C
C
654°C
D
—654°C

Solution:

${ΔT}_{f} = K_{f} . m {ΔT}_{b} = K_{b} . m$

if the value of m is same, then
Hence, $\frac{{ΔT}_{f}}{K_{f}} = \frac{{ΔT}_{b}}{K_{b}}$
or ${ΔT}_{f} = {ΔT}_{b} \frac{K_{f}}{K_{b}}$
$[{ΔT}_{b} = 100 .18 - 100 = 0 {.18}^{\circ} C]$
$∆ T_{f} = 0 .18 \times \frac{1 .86}{0 .512} = 0 {.65}^{\circ} C$
As the Freezing Point of pure water is 0°C,
${ΔT}_{f} = T_{1} (temperature of pure solvent) - T_{2} (when solute is added) 0 .654 ° C = 0 - T_{2} T_{2} = - 0 .654 ° C$

A solution of urea (mol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If K_f, and K_b, for water are 1.86 and 0.512 K kg kg mol-¹respectively, the above solution will freeze at

Solution:

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