At identical temperature and pressure the rate of diffusion of hydrogen gas is `3sqrt3` times that of a hydrocarbon having molecular formula `C_(n)H_(2n-n)` What is the value of n ? .

Options

A. 1
B. 4
C. 3
D. 8

Answer

B

Explanation

`(r_(H_(2)))/(r_(C_(n) H_(2n - 2))) = sqrt((M_(C_(n) H_(2n - 2)) = 3sqrt3 = sqrt(27)`
`implies M_(C_(n) H_(2n - 2)) = 27 xx 2 = 54`
Hence , `12 n + (2n - 2) xx 1 = 54`
`implies 14 n = 56 implies n = 4`
Thus , hydrocarbon is `C_(4) H_(6)` .