## anonymous one year ago **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.

1. anonymous

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

2. anonymous

@sweetburger

3. anonymous

@beckie29

4. anonymous

@geny55 @sammixboo @omgitsjc @Cuanchi @jadaaaa

5. anonymous

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

6. aaronq

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

7. anonymous

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

8. aaronq

yup

9. aaronq

then use the formula

10. anonymous

Can you setup how to find the moles for me

11. anonymous

@aaronq

12. aaronq

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

13. anonymous

Ok I have 0.15996

14. anonymous

-0.15996* @aaronq

15. aaronq

what is that?

16. anonymous

Delta T @aaronq

17. aaronq

alright

18. anonymous

So that is the freezing point

19. anonymous

@aaronq

20. aaronq

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