Question

3. The density of CH₃CN 0.786 g/ml and the density of CH₃OH is 0.791g/ml. A solution is made by dissolving 22.5ml of CH₃OH in 98.7ml. Of CH₃CN.

a. What is the mole fraction of methanol in the solution?

b. What is the molality of the solution?

c. Assuming that the volumes are additive, what is the molarity of CH3OH in the solution?

4. Consider two solutions, one formed by adding 10 g of glucose to 1L of water and the other formed by adding 10g of sucrose to 1L of water. Calculate the vapor pressure for each solution at 20oZC; the vapor pressure of pure water at this temperature is 17.5 torr.

Answer 1

Moles of Methanol:
Mass of methanol = D x V = 0.791 g/ml x 22.5 = 17.7975 g
Moles of methanol = mass/ molar mass
= 17.7975 g / 32 g methanol/mole
= 0.55617 mole

Moles of CH3CN :
Mass of CH3CN = D x V = 0. 786 g/ ml x 98.7 ml = 77.57 g
Moles of CH3CN = 77.57 g / 41 g CH3CN / mole
= 1.89 mol

Mole fraction of methanol = moles of methanol / total moles of component
= 0.55617 / (0.55617 + 1.8896) = 0.227

molality CH3OH = moles methanol / mass of CH3CN (in kilograms)
   = 0.55617 / 77.57 x 10^-3 = 7.16991 molal

Molarity is moles solute per L of solution. Divide moles methanol by total volume of solution

molarity of CH3OH =  0.55617 / (22.5+98.7)x10^-3 = 4.588 molar

4. Vapor pressure drop is a colligative property, that is it depends on number of molecules, not their nature

10g of sucrose contains almost twice fewer molecules than 10 g of glucose, so pressure drop created by sucrose is smaller, and vapor pressure of its solution will be consequently higher