Q47 / 180 · NEET 2019 (Odisha) · Chemistry
The volume occupied by $1.8 \mathrm{~g}$ of water vapour at $374^{\circ} \mathrm{C}$ and 1 bar pressure will be [Use $\mathrm{R}=0.083$ bar $\mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ]
- A$96.66 \mathrm{~L}$
- B$55.87 \mathrm{~L}$
- C$3.10 \mathrm{~L}$
- Correct answerD$5.37 \mathrm{~L}$
Solution
According to ideal gas equation,
$$
\begin{aligned}
& \mathrm{pV}=\mathrm{nRT} \\
& \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{p}}=\frac{\mathrm{w}}{\text { M.wt }} \frac{\mathrm{RT}}{\mathrm{p}} \ldots(\mathrm{i})\left[\because \mathrm{n}=\frac{\mathrm{w}}{\mathrm{M} \cdot \mathrm{wt}}\right]
\end{aligned}
$$
Given, $\mathrm{w}=1.8 \mathrm{~g}, \mathrm{~T}=374^{\circ} \mathrm{C}$
$$
=(374+273) \mathrm{K}=647 \mathrm{~K}
$$
$$
\mathrm{p}=1 \text { bar, } \mathrm{R}=0.083 \mathrm{bar} \mathrm{LK}^{-1} \mathrm{~mol}^{-1}
$$
On substituting the given values in Eq. (i), we get
$$
\begin{aligned}
\mathrm{V} & =\frac{1.8 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}} \times \frac{0.083 \mathrm{bar} \mathrm{LK}^{-1} \mathrm{~mol}^{-1} \times 647 \mathrm{~K}}{1 \mathrm{bar}} \\
& =5.37 \mathrm{~L}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.