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← JEE Advanced 2022 · Paper 1
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JEE Advanced 2022 · Paper 1, Chemistry question 5

Dissolving 1.24 g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q. The amount of CuSOX4\ce{CuSO4} (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]

Answer and worked solution

Official answer: 2.37

The gas. PX4\ce{P4} (M=124M=124): 1.24 g is 0.01 mol.
PX4+3 NaOH+3 HX2O→PHX3↑+3 NaHX2POX2\ce{P4 + 3NaOH + 3H2O -> PH3 ^ + 3NaH2PO2}
So Q is phosphine, 0.01 mol.

With copper sulphate.
3 CuSOX4+2 PHX3→CuX3PX2+3 HX2SOX4\ce{3CuSO4 + 2PH3 -> Cu3P2 + 3H2SO4}
This needs 0.015 mol CuSOX4\ce{CuSO4} (M=159M=159): 0.015×159=2.390.015\times159=2.39 g.