A volume of 100 ml of hydrogen were confined in a diffusion tube and exposed to air and at equilibrium a volume of 26 ml of air was measured in the tube. Again, when 100 ml of an unknown gas were plac

Question

A volume of 100 ml of hydrogen were confined in a diffusion tube and exposed to air and at equilibrium a volume of 26 ml of air was measured in the tube. Again, when 100 ml of an unknown gas were placed in the same tube and exposed to air, 130 ml of air were measured in the tube at the equilibrium. The molecular mass of the unknown gas is

TestHub · Accepted Answer

By Graham's law, the volumes exchanged in the same time are in the ratio of diffusion rates: V(air)/V(gas) = r(air)/r(gas) = √(M g as/M a ir). For hydrogen: 26/100 = √(2/M a ir), giving M a ir = 2 × (100/26) 2 = 29.59 g/mol. For the unknown gas: 130/100 = √(M x /29.59), so M x = 29.59 × (1.30) 2 = 50.0 g/mol. (Equivalently, directly: M x = 2 × (130/26) 2 = 50.) Hence the molecular mass is 0050.

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