Question

**Please Help Me**A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Answer 1

@dan815 @Luigi0210 @zepdrix @welshfella @BTaylor @Cuanchi @Mehek14 @eyesac @aaronq

Answer 2

@sweetburger

Answer 3

@beckie29

Answer 4

@geny55 @sammixboo @omgitsjc @Cuanchi @jadaaaa

Answer 5

@geny55 @pamela.g @radar @sammixboo @foreverthebeast @eyesac @Lokid @Luigi0210 @Mertsj @MrNood @Crazypizzalover @Xaze @Megaforte8600 @razor99 @sleepyjess

Answer 6

use the formula \(\Delta T=i*m*K_f\)

where \(\Delta T\) is the change in temp from the pure solvent's f.p.

i is the van't hoff constant, 1 in the case for sugar

m is the molality of the solution

\\(K_f\) is the f.p. depression constant

Answer 7

wait don't we got to find the morality first @aaronq

Answer 8

yup

Answer 9

then use the formula

Answer 10

Can you setup how to find the moles for me

Answer 11

@aaronq

Answer 12

divide the mass by the molar mass of glucose (which is 180 g/mol)

Answer 13

Ok I have 0.15996

Answer 14

-0.15996* @aaronq

Answer 15

what is that?

Answer 16

Delta T @aaronq

Answer 17

alright

Answer 18

So that is the freezing point

Answer 19

@aaronq

Answer 20

idk, it mean it sounds right. you found moles, then molality, then multiplied by the constant?

if so, then it should be right.