Question

Calculate the vapor pressure of a solution made by dissolving 109 grams of glucose (molar mass = 180.2 g/mol in 920.0 ml of water at 25 °c. the vapor pressure of pure water at 25 °c is 23.76 mm hg. assume the density of the solution is 1.00 g/ml.

Answer 1

Given data: molar mass = 180.2 g/mol in 920.0 ml of water at 25 °c.
                   the vapor pressure of pure water at 25 °c is 23.76 mm hg. 
Asked: the vapor pressure of a solution made by dissolving 109 grams of glucose

Solution:
moles glucose = 109 g/ 180.2 g/mol=0.605 
mass water = 920 mL x 1 g/mL = 920 g 
moles water = 920 g/ 18.02 g/mol=51.1 
mole fraction water = 51.1 / 51.1 + 0.605 =0.988 
vapor pressure solution = 0.988 x 23.76 = 23.47 mm Hg

Answer 2

We have that the vapor pressure of the solution is mathematically given as

vapor pressure solution = 23.47 mm Hg

Vapor pressure of solution

Question Parameters:

Dissolving 109 grams of glucose (molar mass = 180.2 g/mol

in 920.0 ml of water at 25 °c.

vapor pressure of pure water at 25 °c is 23.76 mm hg

density of the solution is 1.00 g/ml.

Generally the equation for the moles glucose  is mathematically given as

moles glucose = 109 g/ 180.2 g/mol

moles glucose=0.605

Where

mass water = 920 mL x 1 g/mL

mass water= 920 g

Hence  

moles water = 920 g/ 18.02 g/mol

moles water=51.1

Finally

mole fraction of water = 51.1 / 51.1 + 0.605

mole fraction of water = =0.988

vapor pressure solution = 0.988 x 23.76

vapor pressure solution = 23.47 mm Hg

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