Can ideal gas law be applied to water vapour?
Yes, for many practical purposes, you may use the ideal gas law also as an approximation for water vapour.

However, you need two independent quantities to describe the state of the gas, for example temperature and pressure. The given temperature of $T=298\ \mathrm K$ alone is not enough. The ideal gas law does not give you a value for the second quantity.

For example, assuming equilibrium conditions for saturated steam at a temperature of $T=298.00\ \mathrm K$, the corresponding pressure is $p=3141.7\ \mathrm{Pa}$.

Using the ideal gas law yields a concentration of

$$\frac nV=\frac p{RT}=\frac{3141.7\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298.00\ \mathrm K}=1.2680\ \mathrm{mol\ m^{-3}}$$

The actual value for water vapour at this temperature and pressure is

$$\frac nV=1.2701\ \mathrm{mol\ m^{-3}}$$

Thus, using the ideal gas law introduces an error of about $0.2\ \%$ in this case, which could be acceptable for many practical purposes.