Question

1 g of \( XY_2 \) is dissolved in 20 g of \( C_6H_6 \). The \( \Delta T_f \) of the resultant solution is 2.318 K. When 1 g of \( XY_4 \) is dissolved in 20 g of \( C_6H_6 \), its \( \Delta T_f \) is found to be 1.314 K. What are the atomic masses of X and Y respectively?
(\( k_f \) of \( C_6H_6 \) is 5.1 K kg \( mol^{-1} \))

• A. 42 u, 26 u
• B. 38 u, 30 u
• C. 30 u, 38 u
• D. 26 u, 42 u

Hint:

To solve problems involving freezing point depression, use the relationship between the change in freezing point and the molality of the solution, considering the number of solute particles.

Answer 1

The Correct Option is D

We are tasked with finding the atomic masses of \(X\) and \(Y\) based on the freezing point depression (\(\Delta T_f\)) of solutions of \(XY_2\) and \(XY_4\) in benzene (\(C_6H_6\)). The given data is:

Mass of \(XY_2 = 1 \, \text{g}\),
Mass of benzene = 20 g,
\(\Delta T_f\) for \(XY_2\) solution = 2.318 K,
Mass of \(XY_4 = 1 \, \text{g}\),
\(\Delta T_f\) for \(XY_4\) solution = 1.314 K,
\(k_f\) of benzene = 5.1 K kg mol\(^{-1}\).

Step 1: Freezing point depression formula

The freezing point depression is given by:

\[ \Delta T_f = k_f \cdot m \]

where:

\( k_f \) is the cryoscopic constant,

\( m \) is the molality of the solution.

The molality \(m\) is given by:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] Step 2: Calculate molality for \(XY_2\)

For \(XY_2\):

\[ \Delta T_f = 2.318 \, \text{K} \] \[ m = \frac{\Delta T_f}{k_f} = \frac{2.318}{5.1} = 0.4545 \, \text{mol/kg} \]

The mass of benzene is 20 g = 0.02 kg. Thus, the moles of \(XY_2\) are:

\[ \text{moles of } XY_2 = m \times \text{mass of solvent in kg} = 0.4545 \times 0.02 = 0.00909 \, \text{mol} \]

The molar mass of \(XY_2\) is:

\[ M_{XY_2} = \frac{\text{mass of } XY_2}{\text{moles of } XY_2} = \frac{1}{0.00909} = 110 \, \text{g/mol} \] Step 3: Calculate molality for \(XY_4\)

For \(XY_4\):

\[ \Delta T_f = 1.314 \, \text{K} \] \[ m = \frac{\Delta T_f}{k_f} = \frac{1.314}{5.1} = 0.2576 \, \text{mol/kg} \]

The mass of benzene is 20 g = 0.02 kg. Thus, the moles of \(XY_4\) are:

\[ \text{moles of } XY_4 = m \times \text{mass of solvent in kg} = 0.2576 \times 0.02 = 0.005152 \, \text{mol} \]

The molar mass of \(XY_4\) is:

\[ M_{XY_4} = \frac{\text{mass of } XY_4}{\text{moles of } XY_4} = \frac{1}{0.005152} = 194 \, \text{g/mol} \] Step 4: Set up equations for atomic masses

Let the atomic mass of \(X\) be \(M_X\) and the atomic mass of \(Y\) be \(M_Y\).

For \(XY_2\):

\[ M_X + 2M_Y = 110 \]

For \(XY_4\):

\[ M_X + 4M_Y = 194 \] Step 5: Solve the equations

Subtract the first equation from the second:

\[ (M_X + 4M_Y) - (M_X + 2M_Y) = 194 - 110 \] \[ 2M_Y = 84 \implies M_Y = 42 \, \text{u} \]

Substituting \(M_Y = 42\) into the first equation:

\[ M_X + 2(42) = 110 \implies M_X + 84 = 110 \implies M_X = 26 \, \text{u} \] Step 6: Match with the options

The atomic masses are \(M_X = 26 \, \text{u}\) and \(M_Y = 42 \, \text{u}\), which matches option (4).

Final Answer: \[ \boxed{4} \]