What is the vapor pressure when 56.7 g of CuSO4 are added to 700 mL of H2O at 330K.
The vapor pressure of pure H2O at 330K is 14kPa.
The density of H2O at 330K is 1.00 g/mL.
In order to be able to find the vapour pressure of the solution the mole fraction of both the water and the CuSO4.
1 mole water = 18g and 1 mole CuSO4 = 160g
Therefore the moles of each component are:
\begin {aligned}
moles\;water = \frac{700}{18}=38.89\;moles
\end {aligned}\begin {aligned}
moles\;CuSO_4 = \frac{56.7}{160}=0.354\;moles
\end {aligned}
This means that the mole fraction of water can be found:
\begin {aligned}
mf_{(water)}=\frac{moles_{(water)}}{total\;moles}=\frac{38.89}{38.89+0.354}=0.9910
\end {aligned}This would make the vapour pressure of the solution as:
\begin {aligned}
VP_{(solution)}=mf_{(water)}*VP{(pure\;water)}=0.9910*14=13.87kPa
\end {aligned}