Question

The element X and Y form compounds having molecular formula $X{Y}_2$ and $X{Y}_4$. When dissolved in 20 gm of benzene, 1 gm $X{Y}_2$ lowers the freezing point by ${2.3}^{\circ },$ whereas 1gm of $X{Y}_4$ lowers the freezing point by ${1.3}^{\circ }C.$ The molal depression constant for benzene is 5.1. Calculate the atomic masses of X and Y.

• A. $x=25.6,y=42.6$
• B. $x=42.6,y=25.6$
• C. $x=35.89,y=54.9$
• D. $x=54.9,y=35.89$

Answer 1

The correct option is A $x=25.6,y=42.6$

Freezing point depression is a colligative property observed in solutions that result from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.

$\Delta T_f={T_f}^0-T_f=K_b.m$.

$\Delta T_f$ is freezing point depression.

$T_f$ is freezing point of solution.

$K_b$ is depression constant.

Molality is $m=\frac{n}{M_1}=\frac{\frac{m_1}{M}}{M_1}$

M is molecular weight of solute.

Depression constant $K=5.1$

Let molecular weight of $X$ and $Y$ be $x$ and $y$.

For $X{Y}_2$

$\Delta T_f={T_f}^0-T_f=K_b.m$.

$2.3=\frac{\frac{1}{x+2y}}{\frac{20}{1000}}\times K$

$⟹x+2y=110.8$

For $X{Y}_4$,

$\Delta T_f={,T_f}^0-T_f=K_b.m$.

$1.3=\frac{\frac{1}{x+4y}}{\frac{20}{1000}}\times K$

$⟹x+4y=196.14$

Solving both equations, we get,

$x=25.6$ and $y=42.6$