Reviewing for final (this should be basic but...I am stuck):

Question

Describe how you would prepare 500.0 grams of a solution of sucrose, C12H22O11, in water which the mole fraction of sucrose is 0.125.

I feel like somehow this is a molality question too :(

Thank you for any input.

Answer 1

Xsucrose = 0.125
XH2O = 0.875
0.125 x molar mass sucrose = about 43g (but you can do it better than that, and should).
0.875 x molar mass H2O = about 15.75g
Just to check, 43/342 = 0.125
15.75/18 = 0.875 and
0.125/(0.125+0.875) = 0.125 etc.
Now how much will all of that weigh?
43 g + 15.75g = about 58.5 g which is far short of the 500 g you want. So scale it up. The easy way to do that is to see what percent 43 is of 58.5 and what percent 15.75 is of 58.5. Find grams then of each at that percentage of 500 g and check to see that you still have the mole fractions right. I checked it out and it works fine.

Answer 2

Oh thank you!

43 is 73.11% of 58.52 and
15.76 is 26.93% of 58.52

so when we scale it up to 500g we got 365g of sucrose and 135g of water!

Answer 3

When I worked it I came out with 368.5 g sucrose and 134.61 g H2O and I expect that is close enough since we probably didn't round the same and we may have used slightly different values for molar mass. I used 342 for sucrose and 18 for water. I used 58.5 for the initial mass.