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

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Solutions

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',

Δ T_{b}

was 1.176 K while when 1 g of

PQ_{2}

is dissolved in 50 g of solvent 'A',

Δ 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:

Answer: (D)

Boiling-point elevation gives

Δ T_{b} = K_{b} m = K_{b} \frac{w}{M} \frac{1000}{W}

. For PQ,

1.176 = 5 \times \frac{1}{M _{1}} \times \frac{1000}{50} = \frac{100}{M _{1}}

, so

M_{1} = 85.0 g mo l^{- 1}

; for PQ

_{2}

0.689 = \frac{100}{M _{2}}

, so

M_{2} = 145.1 g mo l^{- 1}

. Writing

M_{P} + M_{Q} = 85.0

and

M_{P} + 2 M_{Q} = 145.1

and subtracting gives

M_{Q} = 60.1

and

M_{P} = 24.9

. The molar masses of P and Q are 25 and 60 g mol

^{- 1}

respectively.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option