`CH_(4)` diffuses two times faster than a gas X The number of molecules present in 32 g of gas X is (N is Avogadro number)

Options

A. N
B. `N/2`
C. `N/4`
D. `N/16`

Answer

(b)

Explanation

According to Graham's law of diffusion, `r alpha sqrt(1/m)`
`:. r_(CH4)/r_(x) = sqrt(M_(x)/M_(CH4))`
`rArr 2 = sqrt(M_(x)/16)`
`rArr sqrt(M_(x)) = 8`
`rArr M_(x) = 64`
`:.`Molecular mass of gas X `(M_(x))= (8)^(2) = 64`
.`:.`Number of molecules of gas X in 32 g gas
`= 32//64 xx N`
`= N/2`