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 mol-1 the above solution will freeze at

Question

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 mol-1 the above solution will freeze at (A) -6.54°C (B) 6.54°C (C) 0.654°C (D) -0.654°C

Tardigrade Admin · Accepted Answer

Δ Tb = 100.18 - 100 = 0.18 (Δ Tb/Δ Tf) = (Kbm/Kfm) = (Kb/Kf) (0.18/Δ Tf) = (0.512/1.86) Δ Tf = (0.18 × 1.86/0.512) = 0.654 Tf = (0 - 0.654) ° C = - 0.654° C

Q. A solution of urea (mol mass $56 g m o l^{- 1})$ boils at $100.1 8^{\circ} C$ at the atmospheric pressure. If $K_{f}$ and $K_{b}$ for water are $1.86$ and $0.512 K k g$ $m o l^{- 1}$ the above solution will freeze at

-6.54 $^{\circ}$ C

6.54 $^{\circ}$ C

0.654 $^{\circ}$ C

-0.654 $^{\circ}$ C

$Δ T_{b} = 100.18 - 100 = 0.18$
$\frac{Δ T _{b}}{Δ T _{f}} = \frac{K _{b} m}{K _{f} m} = \frac{K _{b}}{K _{f}}$
$\frac{0.18}{Δ T _{f}} = \frac{0.512}{1.86}$
$Δ T_{f} = \frac{0.18 \times 1.86}{0.512} = 0.654$
$T_{f} = (0 - 0.654)^{\circ} C = - 0.65 4^{\circ} C$

Q. A solution of urea (mol mass 56gmol−1) boils at 100.18∘C at the atmospheric pressure. If Kf​ and Kb​ for water are 1.86 and 0.512Kkg mol−1 the above solution will freeze at

-6.54∘C

6.54∘C

0.654∘C

-0.654∘C