Question

Question #006a2

Answer 1

`0.18`

Explanation:

You know that molality is defined as moles of solute, in your case sucrose, divided by the mass of the solvent - expressed in kilograms.

If you have a `"1 molal"` solution of sucrose and water, you essentially have 1 mole of sucrose dissolved in 1 kilogram of water.

To get the mole fraction of sucrose in the solution, you need to first find the number of moles of water you get in one kilogram of water.

Use water's molar mass, which tells you what the mass of one mole of water is, to get

`1color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole water"/(18.02color(red)(cancel(color(black)("g")))) = "55.5 moles water"`

The mole fraction of sucrose will be the ratio between the number of moles of sucrose divided by the total number of moles in the solution.

The total number of moles will be

`n_"total" = "1 mole sucrose" + "55.5 moles water" = "56.5 moles"`

The mole farction will thus be

`chi_"sucrose" = n_"sucrose"/n_"total"`

`chi_"sucrose" = (1color(red)(cancel(color(black)("mole"))))/(56.5color(red)(cancel(color(black)("moles")))) = "0.0177"`

I'll round this off to two sig figs, although based on your values the result should only have one sig figs

`chi_"sucrose" = color(green)("0.18")`