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QJEEAdv22Chccc8: JEE Advanced 2022 Chemistry - Class 12 JEEAdv22Chccc8 Solution: Dissolving of white phosphorous in boiling solution in an inert …

Dissolving 1.24 g1.24 \mathrm{~g} of white phosphorous in boiling NaOH\mathrm{NaOH} solution in an inert atmosphere gives a gas Q\mathbf{Q} . The amount of CuSO4\mathrm{CuSO}_{4} (in g) required to completely consume the gas Q\mathbf{Q} is _________.

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

Step-by-Step Verified Solution

P41.24gor0.01mole+3NaOH+3H2O→PH3+3NaH2PO2

As NaOH is present in excess. So, amount of phosphine formed is 0.01 mole (as P4 is limiting)

2PH30.01 mole+3CuSO4→Cu3P2+3H2SO4\mathop {2P{H_3}}\limits_{0.01\,mole} +3CuSO_4\to Cu_3P_2+3H_2SO_4

Amount of CuSO4 required = 3×0.012\frac{3\times0.01}{2} mole

Mass of CuSO4 (in g) required = 0.032×(63+32+16×4)\frac{0.03}{2}\times(63+32+16\times4)

=0.032×159=2.38=\frac{0.03}{2}\times159=2.38 g

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