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 :

(Delta T_ b)_ PQ= K_ b m 1.176=5 x frac1 M_1 x (1000 / 50) M_1=85.03 (Delta T_ b)_ PQ_2=5 x frac1 M_2 x (1000 / 50)=0.68 M_2=145.13 Let molar mass of P \ Q are M_ P and M_ Q respectively \ M_ P+ M_ Q=85.03 M_ P+2 M_ Q=145.13 M_ p=24.93 = 25 M_ Q=60.1 = 60

Elements P and Q form two types of non volatile non ion... | null (Chemistry)

JEE Mains 2026 21st January Shift 1

Chemistry

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 :

70,110

65, 145

60, 25

25, 60

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Explanation

\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{PQ}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}\\

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

\mathrm{M}_{1}=85.03\\

\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{PQ}_{2}}=5 \times \frac{1}{\mathrm{M}_{2}} \times \frac{1000}{50}=0.68\\

\mathrm{M}_{2}=145.13\\

Let molar mass of

\mathrm{P} \& \mathrm{Q}

are

\mathrm{M}_{\mathrm{P}}

and

\mathrm{M}_{\mathrm{Q}}

respectively

\\ \mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}=85.03\\

\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{Q}}=145.13\\

\mathrm{M}_{\mathrm{p}}=24.93 \approx 25\\

\mathrm{M}_{\mathrm{Q}}=60.1 \approx 60

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