JEE Advanced 2022 — Paper 1 — CHEM-I solution

JEE Advanced2022Paper 1CHEM-I

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]

Your answer

Solution

Molar mass of P $_{4}$ = 124 g/mol. Moles = $\frac{1.24}{124} = 0.01$ .

Step 1 — formation of phosphine:

$P_{4} + 3 NaOH + 3 H_{2} O \to PH_{3} + 3 NaH_{2} PO_{2}$

1 mol P $_{4}$ yields 1 mol PH $_{3}$ . So PH $_{3}$ formed = 0.01 mol.

Step 2 — reaction with CuSO $_{4}$ :

$2 PH_{3} + 3 CuSO_{4} \to Cu_{3} P_{2} + 3 H_{2} SO_{4}$

CuSO $_{4}$ needed = $\frac{3}{2} \times 0.01 = 0.015$ mol.

Molar mass of CuSO $_{4}$ = 63 + 32 + 64 = 159 g/mol.

Mass = $0.015 \times 159 = 2.385 \approx 2.38$ g.

← Back to full paper

Practice more CHEM-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →