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If $1$mole of ethyl alcohol in $1000{\text{grams}}$of water depresses the freezing point by ${1.86^ \circ }C$, what will be the freezing point of the solution of $1$mole of ethyl alcohol in $500{\text{grams}}$of water?
A. $ - {0.93^ \circ }C$
B. $ - {1.86^ \circ }C$
C. $ - {2.79^ \circ }C$
D. $ - {3.72^ \circ }C$
E. $ - {5.58^ \circ }C$

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Last updated date: 13th Jun 2024
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Answer
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Hint: Colligative properties: It is defined as the ratio of number of solute particles to the number of solvent particles. It only depends on the number of ions and not on the nature of the solute.
Depression in freezing point: It is defined as the decrease in freezing point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.

Complete step by step answer:
First of all we will talk about the colligative properties.
Colligative properties: It is defined as the ratio of number of solute particles to the number of solvent particles. It only depends on the number of ions and not on the nature of the solute.
There are four colligative properties: Depression in freezing point, elevation in boiling point, osmotic pressure and lowering in vapour pressure.
Depression in freezing point: It is defined as the decrease in freezing point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.
Elevation in boiling point: It is defined as an increase in boiling point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.
Osmotic pressure: It is defined as the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.
Relative lowering in vapour pressure: When non- volatile solutes are added to a solution then there is a decrease in its vapour pressure, which is known as relative lowering in vapour pressure.
In this question we are given that if $1$ mole of ethyl alcohol in $1000{\text{grams}}$ of water depresses the freezing point by ${1.86^ \circ }C$. And we know that molality is defined as the moles of solute per unit kilograms of solvent. So in this case the molality of solution is equal to $\dfrac{{1000}}{{1000}} = 1m$. In the second case where $1$ mole of ethyl alcohol in $500{\text{grams}}$ of water then molality will be $\dfrac{{1000}}{{500}} = 2m$.
We know that depression in freezing point is directly proportional to the molality i.e. $\Delta {T_f}\alpha m$ . In this question as molality doubles so the value of depression in freezing point also doubles. In the first case depression in freezing point was $ - {1.86^ \circ }C$ so the depression in freezing point for the final case will be $ - 1.86 \times 2 = - {3.72^ \circ }C$.

So, the correct answer is Option D.

Note:
Van’t Hoff factor: It is defined as the ratio of actual number of particles in the solution after dissociation or association to the number of particles for which no ionization takes place. It is represented by $i$.