A solution of urea in water has a boiling point of 100.128 *C.

A solution of urea in water has a boiling point of 100.128 *C. Calculate the freezing point of the same solution: Given Kb and K: for water are 0.512 ‘C/m and 1.86 “C/m respectively: Answer: T; =-0.465 %

The Correct Answer and Explanation is:1

To determine the freezing point of the urea solution, begin by calculating the molality using the boiling point elevation, then apply that to determine the freezing point depression.

Step 1: Use boiling point elevation to find molality

Boiling point elevation formula:ΔTb=i⋅Kb⋅m\Delta T_b = i \cdot K_b \cdot mΔTb​=i⋅Kb​⋅m

Where:

  • ΔTb=Tboiling,solution−Tboiling,pure water=100.128∘C−100.000∘C=0.128∘C\Delta T_b = T_{boiling, solution} – T_{boiling, pure\ water} = 100.128^\circ C – 100.000^\circ C = 0.128^\circ CΔTb​=Tboiling,solution​−Tboiling,pure water​=100.128∘C−100.000∘C=0.128∘C
  • iii is the van’t Hoff factor for urea, which is 1 (non-electrolyte)
  • Kb=0.512∘C/mK_b = 0.512^\circ C/mKb​=0.512∘C/m
  • mmm is molality

Solving for molality mmm:m=ΔTbi⋅Kb=0.1281⋅0.512=0.25 mol/kgm = \frac{\Delta T_b}{i \cdot K_b} = \frac{0.128}{1 \cdot 0.512} = 0.25\ \text{mol/kg}m=i⋅Kb​ΔTb​​=1⋅0.5120.128​=0.25 mol/kg

Step 2: Use freezing point depression formula

Freezing point depression formula:ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot mΔTf​=i⋅Kf​⋅m

Where:

  • Kf=1.86∘C/mK_f = 1.86^\circ C/mKf​=1.86∘C/m
  • i=1i = 1i=1 (urea does not dissociate)
  • m=0.25 mol/kgm = 0.25\ mol/kgm=0.25 mol/kg

ΔTf=1⋅1.86⋅0.25=0.465∘C\Delta T_f = 1 \cdot 1.86 \cdot 0.25 = 0.465^\circ CΔTf​=1⋅1.86⋅0.25=0.465∘C

Step 3: Calculate the freezing point

The freezing point of pure water is 0.000∘C0.000^\circ C0.000∘C. Since freezing point is depressed:Tf=0.000∘C−0.465∘C=−0.465∘CT_f = 0.000^\circ C – 0.465^\circ C = -0.465^\circ CTf​=0.000∘C−0.465∘C=−0.465∘C

Conclusion:

The freezing point of the urea solution is −0.465∘C-0.465^\circ C−0.465∘C.

This outcome demonstrates how a solute such as urea affects the colligative properties of a solvent like water. Despite being a non-electrolyte, urea still lowers the freezing point proportionally to its molality. The calculation involves the use of fundamental relationships between temperature changes and solution concentration, using the known constants KbK_bKb​ and KfK_fKf​ for water.To determine the freezing point of the urea solution, begin by calculating the molality using the boiling point elevation, then apply that to determine the freezing point depression.

Step 1: Use boiling point elevation to find molality

Boiling point elevation formula:ΔTb=i⋅Kb⋅m\Delta T_b = i \cdot K_b \cdot mΔTb​=i⋅Kb​⋅m

Where:

  • ΔTb=Tboiling,solution−Tboiling,pure water=100.128∘C−100.000∘C=0.128∘C\Delta T_b = T_{boiling, solution} – T_{boiling, pure\ water} = 100.128^\circ C – 100.000^\circ C = 0.128^\circ CΔTb​=Tboiling,solution​−Tboiling,pure water​=100.128∘C−100.000∘C=0.128∘C
  • iii is the van’t Hoff factor for urea, which is 1 (non-electrolyte)
  • Kb=0.512∘C/mK_b = 0.512^\circ C/mKb​=0.512∘C/m
  • mmm is molality

Solving for molality mmm:m=ΔTbi⋅Kb=0.1281⋅0.512=0.25 mol/kgm = \frac{\Delta T_b}{i \cdot K_b} = \frac{0.128}{1 \cdot 0.512} = 0.25\ \text{mol/kg}m=i⋅Kb​ΔTb​​=1⋅0.5120.128​=0.25 mol/kg

Step 2: Use freezing point depression formula

Freezing point depression formula:ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot mΔTf​=i⋅Kf​⋅m

Where:

  • Kf=1.86∘C/mK_f = 1.86^\circ C/mKf​=1.86∘C/m
  • i=1i = 1i=1 (urea does not dissociate)
  • m=0.25 mol/kgm = 0.25\ mol/kgm=0.25 mol/kg

ΔTf=1⋅1.86⋅0.25=0.465∘C\Delta T_f = 1 \cdot 1.86 \cdot 0.25 = 0.465^\circ CΔTf​=1⋅1.86⋅0.25=0.465∘C

Step 3: Calculate the freezing point

The freezing point of pure water is 0.000∘C0.000^\circ C0.000∘C. Since freezing point is depressed:Tf=0.000∘C−0.465∘C=−0.465∘CT_f = 0.000^\circ C – 0.465^\circ C = -0.465^\circ CTf​=0.000∘C−0.465∘C=−0.465∘C

Conclusion:

The freezing point of the urea solution is −0.465∘C-0.465^\circ C−0.465∘C.

This outcome demonstrates how a solute such as urea affects the colligative properties of a solvent like water. Despite being a non-electrolyte, urea still lowers the freezing point proportionally to its molality. The calculation involves the use of fundamental relationships between temperature changes and solution concentration, using the known constants KbK_bKb​ and KfK_fKf​ for water.

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