The vapor pressure of pure water at 25 °c is 23.8 torr. what is the vapor pressure (torr) of water above a solution prepared by dissolving 18.0 g of glucose (a nonelectrolyte, mw = 180.0 g/mol) in 95.0 g of water?

The vapour pressure of the solution is 23.4 torr.

Use Raoult’s Law to calculate the vapour pressure:  

p₁ = χ₁p₁°  

where  

χ₁ = the mole fraction of the solvent  

p₁ and p₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δp is  

Δp = p₁° - p₁  

Δp = p₁° - χ₁p₁° = p₁°(1 – χ₁) = χ₂p₁°  

where χ₂ is the mole fraction of the solute.  

Step 1. Calculate the mole fraction of glucose

n₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

n₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = n₂/(n₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

Step 2. Calculate the vapour pressure lowering  

Δp = χ₂p₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

Step 3. Calculate the vapour pressure  

p₁ = p₁° - Δp = 23.8 torr – 0.4430 torr = 23.4 torr

"the periodic table is arranged in order of atomic number."

dynamic equilibrium is the answer.