Question

Question #497d6

Answer 1

The mole fraction of sucrose in your solution is 0.018.

Explanation:

Molality is defined as moles of solute, in your case sucrose, per kilograms of solvent, in your case water.

To make the calculations easier, you can assume that your 1 molal solution contains 1 mole of sucrose in 1 kg of water.

To get the mole fraction of sucrose, you need to determine how many total moles are present in your solution. Since you know that you have 1 mole of sucrose, all you need to determine is how many moles of water you get in 1 kg, or 1000 g, of water.

To do that, use water's moalr mass

`1000cancel("g") * "1 mole water"/(18.02cancel("g")) = "55.49 moles"` `H_2O`

Thus, the mole fraction of sucrose in your solution will be

`chi_"sucrose" = (1cancel("mole"))/((1 + 55.49)cancel("moles")) = 1/56.49 = "0.0177"`

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the molality of the solution

`chi_"sucrose" = color(green)("0.018")`