Question

If the molality of an aqueous solution of cane sugar is 0.4445, what is the mole fraction of cane sugar?

Answer 1

Given: Molality of cane sugar m = 0.445 mol/kg water

Assume 1 kg (1000 g) of water as a solvent for simplicity.

Molar mass of water = 18 g/mol

Moles of water = `1000/18`

= 55.56 mol

Molality means moles of solute per kg of solvent, so:

Moles of cane sugar = 0.4445mol

`chi_"sugar" = "Moles of sugar"/("Moles of sugar" + "Moles of water")` 

= `0.4445/(0.445 + 55.56)`

= `0.4445/56.0045`

= 0.00794

Mole fraction of cane sugar is 0.00794.