Question

Ethylene glycol (C2H6O2) is used as an antifreeze in cars. If 400 g of ethylene glycol is added to 4.00 kg of water, what is the molality? Calculate how much the freezing point of water will be lowered. The freezing-point depression constant for water is Kf = –1.86°C/m. Show your work.

Answer 1

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Answer 2

ok thanks ...
how do i even do this?
im trying to help a friend with her homework and i dont understand it at all

Answer 3

There are two steps here. First, we want to find the molality. That is one of the ways to express concentration of a solute in a solvent (like molarity), by taking a ratio between moles of solute divided by kilograms of solvent.
\( \dfrac{\text{moles solute}}{\text{kilograms solvent}} \)
So if we find (1) moles of solute, and (2) kilograms of solvent, we can find molality by division.

We have 400 g of Ethylene Glycol. We have to convert this to moles because this is the solute (dissolved in). The water (solvent, dissolves) is already in kilograms. :)

Answer 4

ok ... i follow

Answer 5

To convert between grams and moles, we need to find the molar mass of Ethylene Glycol.( \( \rm C_2 H_6 O_2 \)). This amounts to adding up the molar mass of each element, multiplying by the number of atoms of each. Are you able to do that?

Answer 6

i think i can do that ... i just need to find a periodic table

Answer 7

An online periodic table can be found here: http://www.ptable.com/
But I don't know if the class provides one, so you might have the hard copy at hand there. Carbon, Hydrogen, and Oxygen are also some of the better known since they come up so often, they're good to memorize.

Answer 8

is it going to be 62.07?
or do i leave it un rounded? (62.06784)

Answer 9

You can keep it as 62.07. This is usually sufficient for mole conversions.

Starting from grams, we would divide by molar mass (which is grams/mole) to obtain moles:
\( \text{grams} \times \underbrace{\dfrac{\text{moles}}{\text{gram}}}_{1/molar\_mass} = \text{moles} \)

Answer 10

i do not know how to do that

Answer 11

400 g, we divide by 62.07 g/mole
400 g divided by (62.07 g/mole) = ? moles (calculate 400/62.07)

Answer 12

6.44 ?

Answer 13

Yup.
So now we have moles solute (6.44 moles ethylene glycol) and kilograms solvent (4 kg water) We just divide the two for moles/kilogram, which is molality. Often the unit is written as a lowercase m.

Answer 14

1.61 m

Answer 15

Looks good to me. That completes the first part of the problem.

The second requires a formula.
\( \Delta T = i \times K_f \times m \)
Where i is the number of solute particles (I'll explain this better in a moment), K_f is the molal freezing point depression constant (of water; given), and m is molality (which we found).

i is the most easily missed part. You have to consider how many particles that your solute can dissolve into. This is most often the case for soluble salts like NaCl, that dissolve into the ions Na+ and Cl-. Because two ions form for every one NaCl, the number of solute particles is 2.
For Ethylene Glycol, we have no ionic bonds. These compounds are much more stable, so they stay as one unit. That means i = 1 solute particle.

Answer 16

ok

Answer 17

So with i =1, Kf = 1.86 C/m, and m = 1.61
\( \Delta T = 1 \times 1.86 \times 1.61 \ \ ^o C\)
You can do this from here?

This will give the change in temperature, which the question asked for. It did not ask us to find the actual temperature, which requires us to subtract the result of this from \( 0^o C \) (freezing pt of water).

Answer 18

ok thank you , i do think that i can get it from here

Answer 19

thanks a lot

Answer 20

Glad to help! :)