Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m?

Answer 1

You ready?

Answer 2

I had to write this one out....., which I rarely do

Answer 3

You gonna learn today

Answer 4

xD

Answer 5

Okay

Answer 6

Molality (m) = grams of solutes in/ 1000g of water

Answer 7

25.5 grams of glucose (C6H12O6) in 398 grams of water

25.5*1000/398 = m

Answer 8

Tf = -(tc)* m

tc = -1.86 *m

tc = freezing-point depression

Answer 9

m = 64.0703
tc = -119.045

Answer 10

uhhhh let me calculate it

Answer 11

oops i skipped a step!

Answer 12

25.5 grams of glucose (C6H12O6) in 398 grams of water

25.5*1000/398 = x

x/1000 = m

Answer 13

m = 0.0640
tc = -0.119
tf = 0.00761

Answer 14

right?

Answer 15

@oOKawaiiOo

Answer 16

Winner winner chicken dinner

Answer 17

So is that the answer then?

Answer 18

Your answer is tf

Answer 19

okay, thanks!