Ratio of `C_(p) and C_(v)` of a gas `'X' is 1.4`. The number of atoms of the gas `'X'` presents in `11.2 "litres"` of it a `NTP` is

Options

A. `6.02 xx 10^(23)`
B. `1.2 xx 10^(24)`
C. `3.01 xx 10^(23)`
D. `2.01 xx 10^(23)`

Answer

A

Explanation

Since `(C_(p))/(C_(v))=1.4` , the gas should be diatomic.
If volume is 11.2 It then , no. of moles `=(1)/(2)`
` :. ` no. of molecules `=(1)/(2) xx` Avogadro's No.
no. of atoms `= 2 xx `no. of molecules
`2 xx (1)/(2) xx `Avagadro's No.
`= 6.023 x 10^(23)`