Q63 / 180 · NEET 2019 (Odisha) · Chemistry
In water saturated air the mole fraction of water vapour is 0.02 . If the total pressure of the saturated air is $1.2 \mathrm{~atm}$, the partial pressure of dry air is
- A$1.18 \mathrm{~atm}$
- B$1.76 \mathrm{~atm}$
- Correct answerC$1.176 \mathrm{~atm}$
- D$0.98 \mathrm{~atm}$
Solution
Partial pressure of dry air $=$ total pressure $\times$ mole fraction of dry air
$$
\begin{array}{ll}
\Rightarrow \quad \mathrm{p}_{\text {dry air }} & =\mathrm{p}_{\text {total }} \times \lambda_{\text {dry air }} \\
\text { Given, } \lambda_{\text {saturated air }} & =0.02 \\
\lambda_{\text {dry air }} & =1-0.02=0.98 \\
\mathrm{p}_{\text {total }} & =1.2 \mathrm{~atm} \\
\therefore \quad \mathrm{p}_{\text {dry air }} & =1.2 \mathrm{~atm} \times 0.98=1.176 \mathrm{~atm}
\end{array}
$$
Hence, the correct answer is option: 'C'.