Ratio of C_P and C_V of a gas 'X' is 1.4. The number of atoms of the gas 'X' present in 11.2 litres of it at NTP is

The ratio of ${\mathrm{C}}_{\mathrm{P}}$ and ${\mathrm{C}}_{\mathrm{V}}$ of gas gives the atomicity of gas if 

$\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}=1.66--\mathrm{monoatomic}$

$\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}=1.40--\mathrm{diatomic}$

Since, here $\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}=1.4$, so the gas "X" is diatomic like ${\mathrm{H}}_2, {\mathrm{N}}_2$ etc.

Now, we know that, at NTP 1 mole of gas =22.4L 

So, 11.2 L of gas X = 0.5 mole of X gas

Now, 1 mole will have = $6.02\times {10}^{23}$molecules of gas = $6.02\times {10}^{23}\times 2$atoms of gas

So, 0.5 mole of gas X will have = $6.02\times {10}^{23}\times 2\times 0.5$ of atoms gas = $6.02\times {10}^{23}$ atoms.