What is the vapor pressure (in mm Hg) of a solution of 17.5 g of glucose (C6H12O6) in 82.0 g of methanol (CH3OH) at 27∘C? The vapor pressure of pure methanol at 27∘C is 1.40*102 mm H

134.8 mmHg is the vapor pressure for solution

Explanation:

We must apply the colligative property of lowering vapor pressure, which formula is: P° - P' = P°. Xm

P° → Vapor pressure of pure solvent

P' → Vapor pressure of solution

Xm → Mole fraction for solute

Let's determine the moles of solute and solvent

17.5 g. 1 mol/180 g = 0.0972 moles

82 g. 1mol / 32 g = 2.56 moles

Total moles → moles of solute + moles of solvent → 2.56 + 0.0972 = 2.6572 moles

Xm → moles of solute / total moles = 0.0972 / 2.6572 = 0.0365

We replace the data in the formula

140 mmHg - P' = 140 mmHg. 0.0365

P' = - (140 mmHg. 0.0365 - 140mmHg)

P' = 134.8 mmHg