Q.

A 10 g mixture of hydrogen and helium is contained in a vessel of capacity 0.0125 m3 at 6 bar and 27°C. The mass of helium in the mixture is _______ g. (nearest integer)

Given : R = 8.3 JK–1mol–1 (Atomic masses of H and He are 1u and 4u, respectively)

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answer is 8.

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Detailed Solution

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PV=nmix RTnmix =6×12.50.083×3003

Let mole of He = x

Mole of H2 = 3 – x

4x + 2(3 – x) = 10

x = 2 mol

Mass of He = 8g

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