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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2026 · 21 January · Morning Shift · Chemistry · Solutions

Elements P and Q form two types of non-volatile, non-ionizable...

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ2\mathrm{PQ}_{2}. When 1 g of PQ is dissolved in 50 g of solvent ′A′,ΔTb{ }^{\prime} \mathrm{A}^{\prime}, \Delta \mathrm{T}_{\mathrm{b}} was 1.176 K while when 1 g of PQ2\mathrm{PQ}_{2} is dissolved in 50 g of solvent ′A′,ΔTb{ }^{\prime} \mathrm{A}^{\prime}, \Delta \mathrm{T}_{\mathrm{b}} was 0.689 K. (Kb\mathrm{K}_{\mathrm{b}} of ' A′=5 K kg mol−1\mathrm{A}^{\prime}=5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}). The molar masses of elements P and Q (in gmol−1\mathrm{g} \mathrm{mol}^{-1}) respectively, are :

Options

  1. A70,11070,110
  2. B65,14565,145
  3. C25,6025,60
  4. D60,2560,25

Correct answer

C. 25,6025,60

Step-by-step solution

Using ΔTb=Kb×wM×1000W\Delta T_b = K_b \times \frac{w}{M} \times \frac{1000}{W}
For PQ: 1.176=5×1MPQ×100050=100MPQ1.176 = 5 \times \frac{1}{M_{PQ}} \times \frac{1000}{50} = \frac{100}{M_{PQ}}
MPQ=1001.176=85M_{PQ} = \frac{100}{1.176} = 85 g/mol
For PQ2PQ_2: 0.689=100MPQ20.689 = \frac{100}{M_{PQ_2}}
MPQ2=1000.689=145M_{PQ_2} = \frac{100}{0.689} = 145 g/mol
P+Q=85P + Q = 85 ... (i)
P+2Q=145P + 2Q = 145 ... (ii)
Subtracting: Q=60Q = 60 g/mol
From (i): P=85−60=25P = 85 - 60 = 25 g/mol
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