Question

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ\(_2\). When 1 g of PQ is dissolved in 50 g of solvent 'A', \(\Delta T_b\) was 1.176 K while when 1 g of PQ\(_2\) is dissolved in 50 g of solvent 'A', \(\Delta T_b\) was 0.689 K. (\(K_b\) of 'A' = 5 K kg mol\(^{-1}\)). The molar masses of elements P and Q (in g mol\(^{-1}\)) respectively, are:

• A. 70, 110
• B. 60, 25
• C. 25, 60
• D. 65, 145

Hint:

In problems involving colligative properties, always ensure your units are consistent. The mass of the solvent must be in kilograms when using the standard K_b or K_f values. Setting up and solving systems of linear equations is a common follow-up step in such questions.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:

This problem uses the colligative property of elevation in boiling point (ΔT_b) to determine the molar masses of two compounds, PQ and PQ₂. From these molar masses, the atomic masses of elements P and Q are to be calculated.

Step 2: Formula Used:

Elevation in boiling point is given by:
ΔT_b = K_b × m

where molality,
m = (mass of solute / molar mass of solute) ÷ mass of solvent (in kg)

Rearranging for molar mass (M):
M = (K_b × mass of solute) / (ΔT_b × mass of solvent in kg)

Step 3: Given Data:

Mass of solute (PQ or PQ₂) = 1 g
Mass of solvent = 50 g = 0.050 kg
Ebullioscopic constant, K_b = 5 K kg mol⁻¹
ΔT_b for PQ = 1.176 K
ΔT_b for PQ₂ = 0.689 K

Calculation of Molar Mass of PQ:

M_PQ = (5 × 1) / (1.176 × 0.050)
M_PQ = 5 / 0.0588 ≈ 85.03 g mol⁻¹

Calculation of Molar Mass of PQ₂:

M_PQ₂ = (5 × 1) / (0.689 × 0.050)
M_PQ₂ = 5 / 0.03445 ≈ 145.14 g mol⁻¹

Step 4: Determination of Atomic Masses:

Let atomic masses of P and Q be M_P and M_Q respectively.

From molar masses:
M_P + M_Q = 85.03
M_P + 2M_Q = 145.14

Subtracting the first equation from the second:
M_Q = 145.14 − 85.03 = 60.11 g mol⁻¹ ≈ 60 g mol⁻¹

Substituting back:
M_P = 85.03 − 60.11 = 24.92 g mol⁻¹ ≈ 25 g mol⁻¹

Final Answer:

Atomic mass of P ≈ 25 g mol⁻¹
Atomic mass of Q ≈ 60 g mol⁻¹

Correct option: (C)