JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

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

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Solution

Using $\Delta T_b = K_b \times \frac{w}{M} \times \frac{1000}{W}$

For PQ: $1.176 = 5 \times \frac{1}{M_{PQ}} \times \frac{1000}{50} = \frac{100}{M_{PQ}}$

$M_{PQ} = \frac{100}{1.176} = 85$ g/mol

For $PQ_2$ : $0.689 = \frac{100}{M_{PQ_2}}$

$M_{PQ_2} = \frac{100}{0.689} = 145$ g/mol

$P + Q = 85$ ... (i)

$P + 2Q = 145$ ... (ii)

Subtracting: $Q = 60$ g/mol

From (i): $P = 85 - 60 = 25$ g/mol

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Solution

Using $\Delta T_b = K_b \times \frac{w}{M} \times \frac{1000}{W}$

For PQ: $1.176 = 5 \times \frac{1}{M_{PQ}} \times \frac{1000}{50} = \frac{100}{M_{PQ}}$

$M_{PQ} = \frac{100}{1.176} = 85$ g/mol

For $PQ_2$ : $0.689 = \frac{100}{M_{PQ_2}}$

$M_{PQ_2} = \frac{100}{0.689} = 145$ g/mol

$P + Q = 85$ ... (i)

$P + 2Q = 145$ ... (ii)

Subtracting: $Q = 60$ g/mol

From (i): $P = 85 - 60 = 25$ g/mol

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry