Answer
0.30 kg.
Work Step by Step
Table 13–3 gives the saturated vapor pressure at different temperatures. The saturated vapor pressure at $85^{\circ}C$ is the average of the values for $80^{\circ}C$ and $90^{\circ}C$, or $5.87\times10^4 Pa$. For a relative humidity of 10 precent, the vapor pressure will be $5.87\times10^3 Pa$.
Use the ideal gas law to calculate the amount of water. Assume that the water vapor is an ideal gas and that the temperature remains constant.
$$PV=nRT$$
$$n=\frac{PV}{RT}=\frac{ (5.87\times10^3Pa) (8.5m^3)}{(8.314 J/(mol \cdot K))(273+85K)}=16.76mol$$
$$16.76mol\frac{0.018kg}{1mol}\approx 0.30\;kg$$
