Dissolving 1.24 g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q . The amount of CuSO 4 (in g ) required to completely consume the gas Q is . [Given: Atomic mass of H =1, O =16, Na =23, P =31, S =32, Cu =63 ]

Chemistry · p-Block Elements (Group 15-18)

JEE Advanced 2022 — Paper 1 — Question 39

Dissolving 1.24 g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas $\mathbf{Q}$ . The amount of $\mathrm{CuSO}_{4}$ (in g ) required to completely consume the gas $\mathbf{Q}$ is $\qquad$ .

[Given: Atomic mass of $\mathrm{H}=1, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{P}=31, \mathrm{~S}=32, \mathrm{Cu}=63$ ]

Answer: 2.38

Numerical answer — enter this value.

Step-by-step solution

Mole of $\mathrm{P}_{4}=\frac{1.24}{124}=0.01$

$\mathrm{P}_{4}+3 \mathrm{NaOH}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow 3 \mathrm{NaH}_{2} \mathrm{PO}_{2}+\underset{(\mathrm{Q})}{\mathrm{PH}_{3}}$

$2 \mathrm{PH}_{3}+3 \mathrm{CuSO}_{4} \longrightarrow \mathrm{Cu}_{3} \mathrm{P}_{2}+3 \mathrm{H}_{2} \mathrm{SO}_{4}$

(Q) mole $0.01$

$\therefore$ Mole of $\mathrm{CuSO}_{4}$ required $=0.01 \times \frac{3}{2}$

Weight of $\mathrm{CuSO}_{4}=0.01 \times \frac{3}{2} \times 159=2.385 \mathrm{~g}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2022
Paper: Paper 1
Subject: Chemistry
Chapter: p-Block Elements (Group 15-18)
Topic: Compounds of other Group 15 elements - preparation and properties