An unknown composition of a mixture of carbon disulphide and hydrogen sulphide was burnt in sufficient amount of oxygen. The resulting gases found to exert a pressure of 1.97 atm in a 20 l vessel at 4

Question

An unknown composition of a mixture of carbon disulphide and hydrogen sulphide was burnt in sufficient amount of oxygen. The resulting gases found to exert a pressure of 1.97 atm in a 20 l vessel at 400 K. The gaseous mixture required 2.8 M iodine solution and 250 ml of it was required to reach the end point, forming I⁻. Calculate the mole fraction of CS₂ in the original mixture.

TestHub · Accepted Answer

Total product gas = PV/RT = 1.97 x 20/(0.0821 x 400) = 1.2 mol. I₂ = 2.8 x 0.25 = 0.7 mol; only SO₂ reduces I₂ (SO₂ + I₂ + 2H₂O → H₂SO₄ + 2HI), so SO₂ = 0.7 mol. Let a = CS₂, b = H₂S: CS₂ → CO₂ + 2SO₂; H₂S → H₂O(g) + SO₂. Then 2a + b = 0.7 and CO₂ + SO₂ + H₂O = 3a + 2b = 1.2, giving a = 0.2, b = 0.3. Mole fraction of CS₂ = 0.2/0.5 = 0.4, option (b).

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