KVPY SA Exam 2013 Question Paper

Question : 38 of 80

Marks: +1, -0

A specific volume of

H_2

requires 24 s to diffuse out of a container. The time required by an equal volume of

O_2

to diffuse out under identical conditions, is

Solution:

\r_{\O_2}/\r_{\H_2}=√{\M_{\H_2}/\M_{\O_2}}

=√{2/32}=1/4

\r_{\O_2}:\r_{\H_2}=1:4

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