Answer : The vapor pressure (in atm) of a solution is, 0.679 atm
Explanation : Given,
Mass of
= 1.00 kg = 1000 g
Moles of
= 3.68 mole
Molar mass of
= 18 g/mole
Vapor pressure of water = 0.692 atm
First we have to calculate the moles of
.

Now we have to calculate the mole fraction of 

Now we have to partial pressure of solution.
According to the Raoult's law,

where,
= vapor pressure of solution
= vapor pressure of water = 0.692 atm
= mole fraction of water = 0.938



Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm