On heating, 8.2 g of H3PO3 disproportionates completely according to 4H3PO3 -> 3H3PO4 + PH3. Volume of PH3 at STP is (H = 1, P = 31, O = 16; molar volume 22.4 L/mol)

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JEE Advanced levelSingle correct optionDifficulty 3 of 3pn144

On heating, $8.2 g$ of $H X_{3} PO X_{3}$ disproportionates completely according to $4 H X_{3} PO X_{3} 3 H X_{3} PO X_{4} + PH X_{3}$ . Volume of $PH X_{3}$ at STP is (H = 1, P = 31, O = 16; molar volume 22.4 L/mol)

A $0.28 L$
B $2.24 L$
C $1.68 L$
D $0.56 L$

Answer: D

Molar mass of H3PO3 = 3 + 31 + 48 = 82, so n = 8.2/82 = 0.10 mol. Four moles of H3PO3 give one mole of PH3, so n(PH3) = 0.025 mol. Volume = 0.025 x 22.4 = 0.56 L.

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On heating, 8.2 g of HX3​POX3​ disproportionates completely according to 4HX3​POX3​​3HX3​POX4​+PHX3​. Volume of PHX3​ at STP is (H = 1, P = 31, O = 16; molar volume 22.4 L/mol)

Answer: D

More questions on Phosphorus allotropes, phosphine, halides and oxyacids of phosphorus

On heating, $8.2 g$ of $H X_{3} PO X_{3}$ disproportionates completely according to $4 H X_{3} PO X_{3} 3 H X_{3} PO X_{4} + PH X_{3}$ . Volume of $PH X_{3}$ at STP is (H = 1, P = 31, O = 16; molar volume 22.4 L/mol)