Answer
\begin{equation}
6165 \mathrm{g}
\end{equation}
Work Step by Step
Assuming ideal gas assumption to hold valid, then we know,
$$
p \cdot V=n \cdot R \cdot T \quad\quad\quad\quad\quad(1)
$$
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 $27^{\circ} \mathrm{C}$ the equilibrium water vapour pressure $=0.03513$ atm.
Hence
$\begin{aligned} \text { Relative Humidity } &=\frac{\text { Partial Pressure of water vapour }}{\text { Equilibrium Pressure }} \\ \text { Partial Pressure of water vapour } &=0.6 \times 0.03513=0.0211 \mathrm{atm} \end{aligned}$
we know,
Volume $=\mathrm{V}=400 \mathrm{m}^{3}=4 \times 10^{5}$ litre
Temperature $=\mathrm{T}=27^{\circ} \mathrm{C}=300.15 \mathrm{K}$
Pressure $=\mathrm{p}=0.0211 \mathrm{atm}$
Substituting the values in the ideal gas equation we get,
$$
\begin{aligned} p \cdot V &=n \cdot R \cdot T \\ n &=\frac{p \cdot V}{R \cdot T} \\ n &=\frac{0.0211 \times 4 \times 10^{5}}{0.0821 \times 300.15} \mathrm{moles} \\ &=342.5 \text { moles } \end{aligned}
$$
$\begin{aligned} \frac{\text { Mass of water }}{\text { Molar mass of water }} &=342.5 \text { moles } \\ \text { Molar mass of water } &=342.5 \times \text { Molar mass of water } \\ &=342.5 \times 18 \mathrm{g} \\ &=6165 \mathrm{g} \end{aligned}$
Hence the mass of water vapour present in the room is 6165 $\mathrm{g}$
