Question

P and Q combine to form two compounds PQ₂ and PQ₃. If 1 g of PQ₂ is dissolved in 51 g of benzene, the depression of freezing point becomes 0.8°C. On the other hand, if 1 g of PQ₃ is dissolved in 51 g of benzene, the depression of freezing point becomes 0.625°C. The atomic mass of P and Q are (K_f of benzene = 5.1 K·kg/mol):

• A. 35, 55
• B. 45, 45
• C. 55, 35
• D. 55, 45

Hint:

Tip for solving molar mass problems In problems related to freezing point depression, use the formula and solve for the molar mass of the solute. By setting up a system of equations, you can solve for the individual atomic masses of the elements involved.

Answer 1

The Correct Option is C

Given:

- Freezing point depression for PQ₂: 0.8°C

- Freezing point depression for PQ₃: 0.625°C

- Mass of benzene: 51 g

- Benzene's K_f: 5.1 K·kg/mol

We use the freezing point depression formula:

\(\Delta T_f = \frac{K_f \times m}{M}\)

Where:

\(\Delta T_f\) is the freezing point depression,

\(K_f\) is the cryoscopic constant (5.1 K·kg/mol),

\(m\) is the solute's mass (g),

\(M\) is the solute's molar mass (g/mol).

For PQ₂:

\(0.8 = \frac{5.1 \times 1}{51 \times M_{PQ_2}}\)

Solving for \(M_{PQ_2}\):

\(M_{PQ_2} = \frac{5.1 \times 1}{51 \times 0.8} = 0.125 \, \text{kg/mol} = 125 \, \text{g/mol}\)

For PQ₃:

\(0.625 = \frac{5.1 \times 1}{51 \times M_{PQ_3}}\)

Solving for \(M_{PQ_3}\):

\(M_{PQ_3} = \frac{5.1 \times 1}{51 \times 0.625} = 0.16 \, \text{kg/mol} = 160 \, \text{g/mol}\)

Assuming atomic masses:

\(M_{PQ_2} = M_P + 2M_Q = 125\)

\(M_{PQ_3} = M_P + 3M_Q = 160\)

Solving the equations:

1. \(M_P + 2M_Q = 125\)

2. \(M_P + 3M_Q = 160\)

Subtracting equation 1 from 2:

\(M_Q = 35\)

Substituting \(M_Q = 35\) into equation 1:

\(M_P + 2(35) = 125\)

\(M_P = 125 - 70 = 55\)

Therefore, the atomic masses of P and Q are 55 and 35, respectively.