Question

Elements X and Y combine to form two non-electrolyte compounds having molecular formula XY₂, and XY₄. When dissolved in 20 g benzene, 1 g XY₂ lowers the freezing point by 2.3°C and 1g XY₄ lowers the freezing point by 1.3°C. If the cryoscopic constant for benzene is 5.1 K kg mol⁻¹.

• 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

Answer: (A)

Atomic mass of X = 25.6, Atomic mass of Y = 42.6

Answer 2

The problem involves the application of the colligative property of depression in freezing point. The formula relating freezing point depression ($\Delta T_f$), molality ($m$), and cryoscopic constant ($K_f$) is:

$\Delta T_f = K_f \cdot m$

Molality ($m$) is defined as:

$m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{\text{mass of solute / molar mass of solute}}{\text{mass of solvent (kg)}}$

Rearranging the formula to find the molar mass ($M$) of the solute:

$M = \frac{K_f \cdot \text{mass of solute}}{\Delta T_f \cdot \text{mass of solvent (kg)}}$

If the mass of the solvent is given in grams ($W_{solvent(g)}$), the formula becomes:

$M = \frac{K_f \cdot \text{mass of solute} \cdot 1000}{\Delta T_f \cdot W_{solvent(g)}}$

Given data:

$K_f = 5.1 \text{ K kg mol⁻¹}$ Mass of solute (both XY₂ and XY₄) = 1 g Mass of benzene (solvent) = 20 g

1. Calculate the molar mass of XY₂ ($M_{XY_2}$):

$\Delta T_f = 2.3 \text{ °C}$ $M_{XY_2} = \frac{5.1 \text{ K kg mol⁻¹} \times 1 \text{ g} \times 1000}{2.3 \text{ K} \times 20 \text{ g}}$ $M_{XY_2} = \frac{5100}{46} \approx 110.87 \text{ g/mol}$

2. Calculate the molar mass of XY₄ ($M_{XY_4}$):

$\Delta T_f = 1.3 \text{ °C}$ $M_{XY_4} = \frac{5.1 \text{ K kg mol⁻¹} \times 1 \text{ g} \times 1000}{1.3 \text{ K} \times 20 \text{ g}}$ $M_{XY_4} = \frac{5100}{26} \approx 196.15 \text{ g/mol}$

3. Set up equations for atomic masses:

Let the atomic mass of X be $x$ and the atomic mass of Y be $y$. For XY₂: $x + 2y = M_{XY_2} \approx 110.87 \quad \text{(Equation 1)}$ For XY₄: $x + 4y = M_{XY_4} \approx 196.15 \quad \text{(Equation 2)}$

4. Solve the system of equations:

Subtract Equation 1 from Equation 2: $(x + 4y) - (x + 2y) = 196.15 - 110.87$ $2y = 85.28$ $y = \frac{85.28}{2}$ $y = 42.64$

Substitute the value of $y$ into Equation 1: $x + 2(42.64) = 110.87$ $x + 85.28 = 110.87$ $x = 110.87 - 85.28$ $x = 25.59$

Rounding to one decimal place, the atomic mass of X is approximately 25.6 and the atomic mass of Y is approximately 42.6.