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The vapour of a substance effuses throug...

The vapour of a substance effuses through a small hole at the rate 1.3 times faster than `SO_(2)` gas at 1 atm pressure and 500 K. The molecular weight of the gas is

A

49.2

B

37.9

C

41.6

D

83.2

Text Solution

Verified by Experts

The correct Answer is:
B
`(r_(SO_(2)))/(r_("vap")) = sqrt(("Mol mass of vapour")/("Molar mass of "SO_(2)))`
`:.` Mol mass of vapour
`= ((r_(SO_(2))))/((r_("vap"))) xx` Mol mass of `SO_(2)`
`= ((1)^(2))/((1.3)^(2)) xx 64 = 37.9`
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