Question

a solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. what is the freezing point depression of the solvent if the freezing point constant is -1.86 °C/m?

Answer 1

@Vocaloid pls help :>

Answer 2

freezing point depression: ∆T = i*Kf *m
where i is the # of ions the compound dissociates into (in this case, i = 1 since this is glucose and it doesn't dissociate into ions)
Kf is freezing point depression constant (given as -1.86 °C/m? )
and m is the molality (not molarity) of the solution

recall that molality = moles solute/kg of solvent

so conver 21.5g of glucose to moles and convert 255g water to kg, then calculate molality, then go back to the freezing point depression formula

Answer 3

i am very lost

Answer 4

start by calculating the molality of your solution, which is

moles glucose/kg of water

Answer 5

how do i convert grams to moles eek

Answer 6

the kg is 0.255

Answer 7

divide grams by molar mass

Answer 8

so 21.5/180.156

Answer 9

so molarity is 0.119/0.255

Answer 10

*molality

good, then plug this, and the freezing point depression constant, into the formula

∆T = i*Kf *m

(as I explained before, i = 1 so it doesn't change anything)

Answer 11

∆T = (1)(-1.86)(0.47)

Answer 12

right (in the future, you should try to do all the rounding at the end so you don't lose precision, though)

Answer 13

okay thank you!

Answer 14

so for my final answer i got -0.87 degrees Celsius

Answer 15

almost, -0.870 because we have 3 sig figs

Answer 16

yay thank you for the help