A solution of urea in water has boiling point of 100.15°C. Calculate the freezing point of the same solution if Kf and Kb for water are 1.87 K kg mol-1 and 0.52 K kg mol-1 respectively

Question

A solution of urea in water has boiling point of 100.15°C. Calculate the freezing point of the same solution if Kf and Kb for water are 1.87 K kg mol-1 and 0.52 K kg mol-1 respectively (A) - 0.54°C (B) - 0.44°C (C) - 0.64°C (D) - 0.34°C

Tardigrade Admin · Accepted Answer

Δ Tb=(100.15-100)=0.15°C We know that , Δ Tb = molality × Kb Molality =(Δ Tb/Kb)=(0.15/0.52) =0.2884 Δ Tf = molality × Kf =0.2884 × 1.87=0.54°C Thus, the freezing point of the solution =-0.54°C

Q. A solution of urea in water has boiling point of $100.1 5^{\circ} C$ . Calculate the freezing point of the same solution if $K_{f}$ and $K_{b}$ for water are $1.87 K k g m o l^{- 1}$ and $0.52 K k g m o l^{- 1}$ respectively

$- 0.5 4^{\circ} C$

$- 0.4 4^{\circ} C$

$- 0.6 4^{\circ} C$

$- 0.3 4^{\circ} C$

Q. A solution of urea in water has boiling point of 100.15∘C. Calculate the freezing point of the same solution if Kf​ and Kb​ for water are 1.87Kkgmol−1 and 0.52Kkgmol−1 respectively

−0.54∘C

−0.44∘C

−0.64∘C

−0.34∘C

ΔTb​=(100.15−100)=0.15∘C We know that , ΔTb​ = molality ×Kb​ Molality =Kb​ΔTb​​=0.520.15​ =0.2884 ΔTf​ = molality ×Kf​ =0.2884×1.87=0.54∘C Thus, the freezing point of the solution =−0.54∘C

Q. A solution of urea in water has boiling point of $100.1 5^{\circ} C$ . Calculate the freezing point of the same solution if $K_{f}$ and $K_{b}$ for water are $1.87 K k g m o l^{- 1}$ and $0.52 K k g m o l^{- 1}$ respectively

$- 0.5 4^{\circ} C$

$- 0.4 4^{\circ} C$

$- 0.6 4^{\circ} C$

$- 0.3 4^{\circ} C$