
Molal depression constant of water is \[1.86Kgmo{l^{ - 1}}\].0.02 mole of urea dissolved in 100g of water will freeze at:
A. 0.186$^\circ C$
B. -0.372$^\circ C$
C. 1.86$^\circ C$
D. 3.72$^\circ C$
Answer
562.8k+ views
Hint: This question is of depression in freezing point where we have molal depression constant and molality given so we can calculate freezing point with the help of formula that depression of freezing point is proportional to molality of the solution.
Complete step by step solution:
First let us discuss depression in freezing point before solving the question.
The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent .We know that at the freezing point of a substance, the solid phase is in dynamic equilibrium with the liquid phase. Thus, freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
According to Raoult’s law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.
Let ${T_f}^\circ $ be the freezing point of pure solvent and ${T_f}$ be its freezing point when non-volatile solute is dissolved in it. The decrease in freezing point
$\Delta {T_f} = {T_f}^\circ - {T_f}$ is known as depression in freezing point.
$\Delta {T_f} \propto m$,
Now we apply given data to our formula.
$
\Delta {T_f} = {K_f}m \\
\Rightarrow {T_f}^\circ - {T_f} = {K_f}\dfrac{n}{V} \\
$
Where n-number of moles of solute, V-volume of solvent
$\Rightarrow \Delta {T_f} = 1.86 \times \dfrac{{0.02}}{{100 \times {{10}^{ - 3}}}}$
$\Rightarrow \Delta {T_f} = - 0.372^\circ C$
Hence our correct option is B.
Note: Remember the formula carefully sometime students learn molarity instead of molality and solve incorrectly. Molal depression constant is also known as cryoscopic constant or freezing point depression constant.
Complete step by step solution:
First let us discuss depression in freezing point before solving the question.
The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent .We know that at the freezing point of a substance, the solid phase is in dynamic equilibrium with the liquid phase. Thus, freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
According to Raoult’s law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.
Let ${T_f}^\circ $ be the freezing point of pure solvent and ${T_f}$ be its freezing point when non-volatile solute is dissolved in it. The decrease in freezing point
$\Delta {T_f} = {T_f}^\circ - {T_f}$ is known as depression in freezing point.
$\Delta {T_f} \propto m$,
Now we apply given data to our formula.
$
\Delta {T_f} = {K_f}m \\
\Rightarrow {T_f}^\circ - {T_f} = {K_f}\dfrac{n}{V} \\
$
Where n-number of moles of solute, V-volume of solvent
$\Rightarrow \Delta {T_f} = 1.86 \times \dfrac{{0.02}}{{100 \times {{10}^{ - 3}}}}$
$\Rightarrow \Delta {T_f} = - 0.372^\circ C$
Hence our correct option is B.
Note: Remember the formula carefully sometime students learn molarity instead of molality and solve incorrectly. Molal depression constant is also known as cryoscopic constant or freezing point depression constant.
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