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Standard XII

Chemistry

Real Gases

50 mL of a ga...

Question

50 mL of a gas A diffuse through a membrane in the same time as for the diffusion of 40 mL of a gas B under identical conditions. If the molecular weight of A is 64, that of B would be

100

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250

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200

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Solution

The correct option is A 100
$\frac{V_{1}}{V_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
$M_{2} = \frac{50 \times 50 \times 64}{40 \times 40} = 100$

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50 mL of a gas A diffuse through a membrane in the same time as for the diffusion of 40 mL of a gas B under identical conditions. If the molecular weight of A is 64, that of B would be

The correct option is A 100V1V2=√M2M1 M2=50×50×6440×40=100

The correct option is A 100
$\frac{V_{1}}{V_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
$M_{2} = \frac{50 \times 50 \times 64}{40 \times 40} = 100$