Question

`CH_(4)` diffuse two times faster than a gas X. The number of molecules present in 32 g of gas X is: (N is Avogardro number)
A. `N`
B. `(N)/(2)`
C. `(N)/(4)`
D. `(N)/(16)`

Answer 1

Correct Answer - B
`(r_(CH_(4))/(r_("gas"))) = sqrt((m_("gas"))/(m_(CH_(4))))`
`2 = sqrt((m_("gas"))/(16))`
`m_("gas") = 64`
Number of molecules `= (w)/("molar mass") xx N`
`= (32)/(64) xx N = (N)/(2)`