Ans
Q. Assuming that dry air contain 79 % $N_2$ and 21 % $O_2$ by volume, what is the density of moist air at 25$°$C and 1.0 atm when the relative humidity is 60 %. The vapor pressure of water at 25$°$C is 23.76 mm Hg.
 

Dear Student,

As given the relative humidity is 60 % i.e the partial pressure of water at which it is equal to the vapour pressure of water.
Hence the partial pressure of water in vapour at this condition will be p_v = 
                                                $$\begin{gathered} = 0.6 \times 23.76 mm of Hg \\ = 14.256 mm of Hg = 0.01875 atm \end{gathered}$$
As the atmospheric pressure = 760 mm of Hg 
The rest of the pressure is the pressure of the dry air p_d = 
                                               = 760 -  14.256 mm of Hg 
                                               = 745.744 mm of Hg  = 0.981 atm 

Molar mass of dry air = 28.94 g/ mol
Molar mass of water vapour = 18 g / mol 

Now, putting these values in formula : 
​                          $$\begin{gathered} {\rho }_{humid air }= \frac{{\rho }_d{M}_d+ {\rho }_v{M}_v}{RT} \\ = \frac{0.981\times 28.94 +\hspace{0.17em}0.01875\times 18 }{0.0821\times 298 } \\ = \frac{28.71}{24.77} \\ = 1.16 g / L \end{gathered}$$

So, the density of moist air is 1.16 g/L 

Regards.