Ratio of `C_(p)` and `C_(v)` of a gas X is 1.4, the number of atom of the gas 'X' present in 11.2 litres of it at NTP will be

Options

A. `6.02xx10^(23)`
B. `1.2xx10^(24)`
C. `3.0xx10^(23)`
D. `2.01xx10^(23)`

Answer

A

Explanation

Since `(C_(P))/(C_(V))=1.4`, the gas should be diatomic.
If volume is `11.2L`, then no. of moles =`(1)/(2)`
`:.` No. of molecules `=(1)/(2)xx"Avogadro's No."`
No. of atoms`=2xx "no. of molecules"`
`2xx(1)/(2) xx "Avogadro's No." =6.0223xx10^(23)`