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. (a)`6.02xx10^(23) `
B. (b)`1.2xx10^(23) `
C. (c )`3.01xx10^(23) `
D. (d)`2.01xx10^(23) `

Answer

A

Explanation

`C_(P)/C_(V)=1.4` so, given gas is diatomic
`11.2 L=3.10xx10^(23)` molecules
`:.` No. of atoms`=3.01xx10^(23)xx2=6.323xx10^(23)`