Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ₂. When 1 g of PQ is dissolved in 50 g of solvent 'A', Tb was 1.176 K while when 1 g of PQ₂ is dissolved in 50 g of…

Given: - Compound PQ gives Tb = 1.176 K - Compound PQ₂ gives Tb = 0.689 K - Mass of solute in each case = 1 g - Mass of solvent = 50 g = 0.050 kg - Kb = 5 K kg mol^-1 Find: Atomic masses of elements P and Q. For a non-volatile, non-ionizable solute, Tb = Kb m and m = mass of solute/M mass of solvent in kg So, M = Kb mass of solute Tb mass of solvent in kg For PQ, MPQ = 5 11.176 0.050 = 50.0588 85.03 g mol^-1 For PQ₂, MPQ2 = 5 10.689 0.050 = 50.03445 145.14 g mol^-1 Let atomic masses be MP and MQ. Then, MP + MQ = 85.03 MP + 2MQ = 145.14 Subtracting the first equation from the second, MQ = 145.14 - 85.03 = 60.11 g mol^-1 So, MQ 60 g mol^-1 Now substitute into the first equation: MP = 85.03 - 60.11 = 24.92 g mol^-1 Thus, MP 25 g mol^-1 Therefore, the atomic masses of P and Q are 25 and 60 respectively. The correct option is C.

Elements P and Q form two types of non-volatile, non-ionizable co… — Solution | JEE Chemistry

MCQMediumJEE 2026Colligative Properties

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ₂. When $1 \, \text{g}$ of PQ is dissolved in $50 \, \text{g}$ of solvent 'A', $\Delta T_b$ was $1.176 \, \text{K}$ while when $1 \, \text{g}$ of PQ₂ is dissolved in $50 \, \text{g}$ of solvent 'A', $\Delta T_b$ was $0.689 \, \text{K}$ . ( $K_b$ of 'A' = $5 \, \text{K kg mol}^{-1}$ ). The molar masses of elements P and Q (in $\text{g mol}^{-1}$ ) respectively, are:

A
$70$ , $110$
B
$60$ , $25$
C
$25$ , $60$
D
$65$ , $145$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

Compound PQ gives $\Delta T_b = 1.176 \, \text{K}$
Compound PQ₂ gives $\Delta T_b = 0.689 \, \text{K}$
Mass of solute in each case = $1 \, \text{g}$
Mass of solvent = $50 \, \text{g} = 0.050 \, \text{kg}$
$K_b = 5 \, \text{K kg mol}^{-1}$

Find: Atomic masses of elements P and Q.

For a non-volatile, non-ionizable solute,

\Delta T_b = K_b m

and

m = \frac{\text{mass of solute}/M}{\text{mass of solvent in kg}}

So,

M = \frac{K_b \times \text{mass of solute}}{\Delta T_b \times \text{mass of solvent in kg}}

For PQ,

M_{PQ} = \frac{5 \times 1}{1.176 \times 0.050} = \frac{5}{0.0588} \approx 85.03 \, \text{g mol}^{-1}

For PQ₂,

M_{PQ_2} = \frac{5 \times 1}{0.689 \times 0.050} = \frac{5}{0.03445} \approx 145.14 \, \text{g mol}^{-1}

Let atomic masses be $M_P$ and $M_Q$ . Then,

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 \, \text{g mol}^{-1}

So,

M_Q \approx 60 \, \text{g mol}^{-1}

Now substitute into the first equation:

M_P = 85.03 - 60.11 = 24.92 \, \text{g mol}^{-1}

Thus,

M_P \approx 25 \, \text{g mol}^{-1}

Therefore, the atomic masses of P and Q are $25$ and $60$ respectively. The correct option is C.

Common mistakes

Using the mass of solvent as $50 \, \text{kg}$ or $50 \, \text{g}$ directly in the molality formula is incorrect because $K_b$ is given per kilogram of solvent. Convert $50 \, \text{g}$ to $0.050 \, \text{kg}$ before substitution.
Treating PQ and PQ₂ as ionizable species is incorrect because the question explicitly states they are non-ionizable. Therefore, no van't Hoff factor is used; take $i = 1$ .
Interchanging the identities of P and Q after solving the linear equations leads to the reversed pair $60, 25$ . From $M_P + M_Q = 85.03$ and $M_P + 2M_Q = 145.14$ , the extra Q in PQ₂ means the difference gives the mass of Q first.

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