The time taken for a certain volume of gas to diffuse through a small hole was $2$ min. Under similar conditions an equal volume of oxygen took $5.65$ minutes tp pass. The molecular mass of the gas is

32.0

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11.33

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4.0

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8.0

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Solution

The correct option is B $4.0$
According to the problem:
$\frac{r_{A}}{r_{O_{2}}} = \sqrt{\frac{M O_{2}}{M A}}$
Implies that
$= \frac{t_{O_{2}}}{t_{A}}$
Hence,
$\sqrt{\frac{32}{M_{A}}} = \frac{5.65}{2}$
Since, $M_{A} = 4$

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50 mL of $H_{2}$ gas diffuses through a small hole from a vessel in 20 min. Time taken by 40 mL of $O_{2}$ gas to diffuse under similar conditions will be :

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The time taken for a certain volume of gas to diffuse through a small hole was 2 min. Under similar conditions an equal volume of oxygen took 5.65 minutes tp pass. The molecular mass of the gas is

The correct option is B 4.0According to the problem:rArO2=√MO2MAImplies that=tO2tAHence,√32MA=5.652Since,MA=4

The time taken for a certain volume of gas to diffuse through a small hole was $2$ min. Under similar conditions an equal volume of oxygen took $5.65$ minutes tp pass. The molecular mass of the gas is

The correct option is B $4.0$
According to the problem:
$\frac{r_{A}}{r_{O_{2}}} = \sqrt{\frac{M O_{2}}{M A}}$
Implies that
$= \frac{t_{O_{2}}}{t_{A}}$
Hence,
$\sqrt{\frac{32}{M_{A}}} = \frac{5.65}{2}$
Since, $M_{A} = 4$