Question

Freezing point of a 4 % aqueous solution of X is equal to freezing point of 12 % aqueous solution of Y. If molecular weight of X is A, then the molecular weight of Y is :
a.) A
b.) 3A
c.) 4A
d.) 2A

Answer 1

Hint : Freezing point of a substance is the temperature of liquid at which it changes its state from liquid to solid state at atmospheric pressure. The freezing point of a solution is given by -
$\Delta {T_f} = {K_f} \times m$
Where $\Delta {T_f}$ is the change in freezing point.
${K_f}$ is the molal depression constant
‘m’ is the molality of the solution.

Complete step by step answer :
Let us start by writing what is given to us and what we need to find.
Thus, Given :
Freezing point of X = Freezing point of Y
X = 4 % aqueous solution = 4 g of solute in 100 g of water
Y = 12 % aqueous solution = 12 g of solute in 100 g of water
Molecular weight of X is A
To find :
Molecular weight of X is B
We know, the freezing point of a solution is given by -
$\Delta {T_f} = {K_f} \times m$
Where $\Delta {T_f}$ is the change in freezing point.
${K_f}$is the molal depression constant
‘m’ is the molality of the solution.
We have, Freezing point of X = Freezing point of Y
So, $\Delta {T_f}(x)$=$\Delta {T_f}(y)$
${({K_f} \times {m_x})_x}$=${({K_f} \times {m_y})_y}$
Thus, ${m_x}$=${m_y}$
We know molality of the solution is given by -
‘m’ = $\dfrac{{weight{\text{ of solute}} \times 1000}}{{weight{\text{ of solvent}} \times {\text{Molecular weight of solvent}}}}$
So, for ${m_x}$=${m_y}$
${(\dfrac{{weight{\text{ of solute}} \times 1000}}{{weight{\text{ of solvent}} \times {\text{Molecular weight of solvent}}}})_x}$=${(\dfrac{{weight{\text{ of solute}} \times 1000}}{{weight{\text{ of solvent}} \times {\text{Molecular weight of solvent}}}})_y}$
$\dfrac{{4 \times 1000}}{{100 \times {M_1}}}$=$\dfrac{{12 \times 1000}}{{100 \times {M_2}}}$
$\dfrac{4}{{{M_1}}}$=$\dfrac{{12}}{{{M_2}}}$
Molecular weight of X is A
So, $\dfrac{4}{A}$=$\dfrac{{12}}{{{M_2}}}$
${M_2}$= 3A

Thus, the correct option is option b.).

Note: It must be noted that the depression in freezing point is a colligative property and it depends on the number of solute particles. Addition of non-volatile solute leads to decrease in freezing point of a solid.