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? Show all of the work needed to solve this problem.

Answer 1

you would use \(\Delta T=i*m*K_f\) where m is molality.

If you want i can expand this formula \(\Delta T=\dfrac{m*K_f}{M*kg_{solvent}}\)

m= mass, M=molar mass

Answer 2

Wait so m=25.5 how do I get molar mass?

Answer 3

You would add up the individual molar masses of C, O and H in the right proportions (6 times carbon, 6 times oxygen and 12 times hydrogen).

Answer 4

Okay so carbon would be 12*6=72, oxygen would be 16*6=96, and hydrogen would be 12*1=12 so it would total 180
2.24*K_f / 180*kg_solvent ?

Answer 5

you've got the mass wrong \(\Delta T=\dfrac{(25.5~g)*K_f}{ (180~g/mol)*kg_{solvent}} \)

Answer 6

Oh .-.i didn't mean for it to be a 4. Wait, what stands for K_f and kg?

Answer 7

Kf is the freezing point depression constant, in this case it's -1.86 °C/m
Kg is the kilograms of solvent (in this case water)

Answer 8

Okay so it's 25.5*-1.86 / 180*398
23.64/71640

Answer 9

you didn't convert the mass of water to kilograms

Answer 10

Oh okay so that would be 0.398*180=71.64
So 23.64/71.64

Answer 11

That's a really weird number.... So I guess it's 0.33 @aaronq

Answer 12

i got \(\Delta T\)=-0.6620603015075377 =-0.66

Answer 13

Because the freezing point of water is 0 Celsius, your answer is -0.7 Celsius

Answer 14

Wait did I calculate it wrong? Because 23.64/71.64=0.33

Answer 15

you got the numerator wrong, it's 25.5*-1.86=-47.43

Answer 16

you subtracted, you were supposed to multiply it

Answer 17

Thank you c:

Answer 18

no problem !