Answer
\begin{equation}
2717\ \mathrm{g}
\end{equation}
Work Step by Step
Assuming ideal gas assumption to hold valid, then we know,
$$
p \cdot V=n \cdot R \cdot T \quad\quad\quad\quad(1)
$$
Relative humidity of a gas is defined as the ratio of the partial pressure of water vapour to that of the equilibrium vapour pressure of water at a given
temperature.
At $23^{\circ} \mathrm{C}$ the equilibrium water vapour pressure $=0.0277$ atm.
Hence
$$
\text { Relative Humidity }=\frac{\text { Partial Pressure of water vapour }}{\text { Equilibrium Pressure }}
$$
$$
\text { Partial Pressure of water vapour }=0.53 \times 0.0277=0.01468 \text { atm }
$$
we know,
Volume $=\mathrm{V}=250 \mathrm{m}^{3}=2.5 \times 10^{5}$ litre
Temperature $=\mathrm{T}=23^{\circ} \mathrm{C}=296.15 \mathrm{K}$
Pressure $=\mathrm{p}=0.01468 \mathrm{atm}$
Substituting the values in the ideal gas equation we get,
$\begin{aligned} p \cdot V &=n \cdot R \cdot T \\ n &=\frac{p \cdot V}{R \cdot T} \\ n &=\frac{0.01468 \times 2.5 \times 10^{5}}{0.0821 \times 296.15} \mathrm{moles} \\ &=150.94 \text { moles } \\ \frac{\text { Mass of water }}{\text { Mass of water }} &=150.94 \text { moles } \\ &=150.94 \text { molar mass of water } \\ &=150.94 \times 18 \mathrm{g} \\ &=2717 \mathrm{g} \end{aligned}$
Hence the mass of water vapour present in the room is 2717 $\mathrm{g}$
