Question

If 1.00 mol of CHO is added to 0.150 kg of water and the water is heated to 100°C. What is the vapor pressure of the solution at this temperature? a. 81.4 mm Hg b. 679 mm Hg c. 712 mm Hg d. 760 mm Hg e. 814 mm Hg

Answer 1

According to Raoult's law,

for a solution containing non-volatile solute

(as C₁₂H₂₂O₁₁(sucrose) is a non-volatile solute)

P_solution= X_solvent * P⁰_solvent

P_solution = vapor pressure of the solution

X_solvent = mole fraction of the solvent(i.e water)

P⁰_solvent = vapor pressure of pure solvent(i.e water)

Lets calculate

mole fraction of the solvent(X_solvent) = moles of the solvent/ moles of the solution

given

1 mole C₁₂H₂₂O₁₁,(solute)

0.15 kg of water (solvent)

Moles of water =Mass of water/Molar Mass of water

= 0.15 *1000 g / 18.01528 g/mol

= 8.33 moles

moles of the solvent = 8.33 moles

moles of the solute = 1 mole

moles of the solution = 8.33 moles+1 mole = 9.33 moles

mole fraction of the solvent = 8.33 moles/9.33 moles

X_solvent = 0.893

vapor pressure of pure solvent(P⁰_solvent) water at 100⁰C = 1 atm = 760 mm Hg

so

P_solution= X_solvent * P⁰_solvent

= 0.893 *760 mm Hg

= 678.7

Approximately 679 mm Hg