Answer
$0.93\space mol/m^{3}$
Work Step by Step
Let's use equation 12.6 & graph in problem 75 in chapter 12 to find the partial pressure of water vapor.
Partial pressure = $\frac{(55)(4250\space Pa)}{100}=2337.5\space Pa$
of water vapor
Let's apply the ideal gas law $PV=nRT$ to find the number of moles of water vapor per cubic meter of air.
$PV=nRT=>\frac{n}{V}=\frac{P}{RT}$
Let's plug known values into this equation.
$\frac{n}{V}=\frac{2337.5\space Pa}{(8.31\space J/mol\space K)(303\space K)}=0.93\space mol/m^{3}$
Number of moles of water vapor = $0.93\space mol/m^{3}$