Question

The number of nitrogen molecules present in $1c.c$ of gas at NTP is:

• A. $6.02\times {10}^{23}$
• B. $2.67\times {10}^{22}$
• C. $3.01\times {10}^{21}$
• D. $2.67\times {10}^{19}$

Answer 1

The correct option is D $2.67\times {10}^{19}$

The volume of the gas is $1c.c={10}^{-3}L$

$\therefore$ Number of moles of the gas is$=\frac{{10}^{-3}}{22.4}$

Now,

Number of molecules$=n\times 6.022\times {10}^{23}$

$\therefore$ Number of Nitrogen Molecules = $\frac{{10}^{-3}}{22.4}\times 6.023\times {10}^{23}$

= $2.67\times {10}^{19}$

Hence, option $D$ is correct.