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:

- Depression of freezing point for PQ₂: 0.8°C

- Depression of freezing point for PQ₃: 0.625°C

- Mass of benzene = 51 g

- K_f of benzene = 5.1 K·kg/mol

We can use the formula for depression in freezing point:

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

Where:

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

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

\(m\) is the mass of the solute (in g),

\(M\) is the molar mass of the solute (in g/mol).

First, for PQ₂:

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

Solving for \(M_{PQ_2}\):

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

Next, for PQ₃:

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

Solving for \(M_{PQ_3}\):

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

Now, for the molar masses of the individual atoms P and Q, assume:

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

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

Solving these two equations:

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

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

Subtract the first from the second:

\(M_Q = 35\)

Substitute \(M_Q = 35\) into the first equation:

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

\(M_P = 125 - 70 = 55\)

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