Question

Question #92cc8

Answer 1

`0.13`

Explanation:

Your strategy here will be to pick a sample of this solution and use its percent concentration by mass, `"% m/m"`, to figure out the mass of sucrose and the mass of water present in the sample.

Once you know that, use the molar masses of the two compounds to find how many moles of each you have in the sample.

So, a `"20% m/m"` solution will contain `"20 g"` of solute, which in your case is sucrose, for every `"100 g"` of solution.

To make the calculations easier, pick a `"100 g"` sample of this solution. This sample will contain `"20 g"` of sucrose and

`m_"water" = "100 g" - "20 g" = "80 g"`

Use the molar masses of sucrose and water to convert the masses to moles

`20 color(red)(cancel(color(black)("g"))) * "1 mole sucrose"/(342.3 color(red)(cancel(color(black)("g")))) = "0.05843 moles sucrose"`

`80 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.441 moles water"`

The total number of moles present in solution will be

`n_"total" = 0.05843 + 4.441 = "4.499 moles"`

The mole fraction of sucrose, `chI_"sucrose"`, can be calculated by taking the number of moles of sucrose and the total number of moles present in solution

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

In your case, you will have

`chi_"sucrose" = (0.05843 color(red)(cancel(color(black)("moles"))))/(4.499 color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.013)color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs.