Question

Calculate the vapour pressure lowering of a 1M aqueous sucrose (sugar) solution at 25 oC where po(H2O) = 23 mm Hg. The density of 1M sucrose at 25 oC is 1.129 g/mL.

The answer to this question is 0.51 mm Hg

Can you please explain how ti find this?

We know that vapour pressure lowering is DeltaP = X_b * P_A

but how do i find the mole fraction?

Can some one please explain this in detail

Thank you

Answer 1

We have, Mole fraction of sucrose = [ moles of sucrose / ( Moles of sucrose + moles of water ) ]

Let's calculate moles sucrose & water in 1 L of 1 M sucrose solution

We know that, Molarity = No. of moles of solute / volume of solution in L

No. of moles of solute = Molarity volume of solution in L

No. of moles of sucrose = 1 mol / L 1 l = 1 mol

We have, no. of moles = Mass / Molar mass

Mass of sucrose = no. of moles Molar mass

Molar mass of sucrose = 342.30 g /mol

Mass of sucrose = 1 mol 342.30 g / mol = 342.30 g

We have relation, density = Mass / volume

Mass of solution = density volume of solution

Mass of solution = 1.129 g /ml 1000 ml = 1129 g

Mass of solution = Mass of sucrose + Mass of water

Mass of water = Mass of solution - Mass of sucrose

Mass of water = 1129 g - 342.30 g = 786.7 g

No. of moles of water = Mass of water / Molar mass of water

No. of moles of water =  786.7 g / 18.02 g /mol = 43.66 mol

Mole fraction of sucrose = [ moles of sucrose / ( Moles of sucrose + moles of water ) ]

Mole fraction of sucrose = [ 1 / ( 1+ 43.66 ) ] = 0.02239

We have relation, P = X _B ( P ⁰_A )  

Where P ⁰_A is vapor pressure of solvent &   X _B is mole fraction of solute B.

P = 0.02239 ( 23 mm Hg )

P = 0.51