Questions & Answers

Question

Answers

a.) A

b.) 3A

c.) 4A

d.) 2A

Answer

Verified

92.4k+ views

$\Delta {T_f} = {K_f} \times m$

Where $\Delta {T_f}$ is the change in freezing point.

${K_f}$ is the molal depression constant

‘m’ is the molality of the solution.

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

Students Also Read