2 grams of Helium diffuses from a porous plate in 4min. How many grams of $C H_{4}$ would diffuse through the same plate in same time under similar conditions?

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16g

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Solution

The correct option is A
4g

$\frac{r_{1}}{r_{2}} = \frac{W_{1}}{W_{2}} \times \frac{M_{2}}{M_{1}} \times \frac{t_{2}}{t_{1}} = \sqrt{\frac{M_{2}}{M_{1}}}$

= $\frac{2}{W_{2}} \times \frac{16}{4} \times \frac{4}{4} = \sqrt{\frac{16}{4}}$

$W_{2} = 4 g m$

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2 grams of Helium diffuses from a porous plate in 4min. How many grams of CH4 would diffuse through the same plate in same time under similar conditions?

The correct option is A 4g r1r2=W1W2×M2M1×t2t1=√M2M1 = 2W2×164×44=√164 W2=4 gm

2 grams of Helium diffuses from a porous plate in 4min. How many grams of $C H_{4}$ would diffuse through the same plate in same time under similar conditions?

The correct option is A
4g

$\frac{r_{1}}{r_{2}} = \frac{W_{1}}{W_{2}} \times \frac{M_{2}}{M_{1}} \times \frac{t_{2}}{t_{1}} = \sqrt{\frac{M_{2}}{M_{1}}}$

= $\frac{2}{W_{2}} \times \frac{16}{4} \times \frac{4}{4} = \sqrt{\frac{16}{4}}$

$W_{2} = 4 g m$