Question

What is the vapor pressure in atm of a solution at 25 ^oC produced by dissolving 120.1 g of dextrose (C₆H₁₂O₆) in 395.2 g of water? Vapor pressure of water at 25 ^oC = 0.0313 atm

Answer 1

Solution-

Given-

Mass of dextrose = 120.1 g

Molar mass of dextrose = 180.15 g/mol

Mass of water = 395.2 g

Molar mass of water = 18 g/mol

Vapor pressure of water at 25 ^oC = 0.0313 atm

Let’s calculate the moles of dextrose and water

# mole = Mass/molar mass
Moles dextrose = 120.1g/180.15 gmol^-1 = 0.66 moles of dextrose
Moles water = 395.2 g/18 gmol^-1 = 21.95 moles of water

Total moles = 0.66 mol + 21.95 mol = 22.61 moles

Now let’s calculate the mole fraction of water

Mole fraction = Moles/total mole

Mole fraction water = 21.95 moles water / 22.61 total moles = 0.97

Now we have to multiply this mole fraction of water times the vapor pressure of pure water at 25 degrees C

vapor pressure in atm of a solution at 25 ^oC = 0.97 X 0.0313 atm = 0.03 atm.

Answer- vapor pressure in atm of a solution at 25 ^oC = 0.03 atm