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 :

12 min

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64 min

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8 min

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32 min

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Solution

The correct option is B
64 min

$r_{H_{2}} = \frac{50}{20} = 2.5$

$r_{O_{2}} = \frac{40}{t}$

$\frac{r_{H_{2}}}{r_{O_{2}}} = \frac{2.5}{\frac{40}{t}} = \sqrt{\frac{M_{O_{2}}}{M_{H_{2}}}} = \sqrt{\frac{32}{2}} = 4$

2.5 t = 160

t = 64 min.

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Similar questions

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 :

Q. 50 ml of hydrogen diffuse through a small hole from a vessel in 20 minutes. Time taken for 40 ml of oxygen to diffuse under similar conditions through a same hole will be:

Q. 50

m l

of hydrogen diffuses through a small hole in 20 minutes. The time taken for 40

m l

of oxygen to diffuse under the same conditions in minutes is:

5 mL

H_{2}

gas diffuses out in

1

second from a hole. Find the volume of

O_{2}

that will diffuse out from the same hole under identical conditions in

2

seconds.

50 ml of hydrogen diffuses out through a small hole from a vessel in 20 minutes. The time needed for 40 ml of oxygen to diffuse out is........

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50 mL of H2 gas diffuses through a small hole from a vessel in 20 min. Time taken by 40 mL of O2 gas to diffuse under similar conditions will be :

The correct option is B 64 min rH2=5020=2.5 rO2=40t rH2rO2=2.540t=√MO2MH2=√322=4 2.5 t = 160 t = 64 min.

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 :

The correct option is B
64 min

$r_{H_{2}} = \frac{50}{20} = 2.5$

$r_{O_{2}} = \frac{40}{t}$

$\frac{r_{H_{2}}}{r_{O_{2}}} = \frac{2.5}{\frac{40}{t}} = \sqrt{\frac{M_{O_{2}}}{M_{H_{2}}}} = \sqrt{\frac{32}{2}} = 4$

2.5 t = 160

t = 64 min.