Question

Can I please get some Chemistry help?

Answer 1

Last two questions, I promise haha.

Answer 2

Urea (CON2H4), 60.056 g/mol) is a nondissociating solute. How many grams of urea would have to be added to 50.0 grams of water to give it a freezing point of -1oC? The freezing point depression constant of water is 1.86 oC/m.
Group of answer choices

Answer 3

Cyclohexane normally freezes at 6.6oC and has freezing point depression constant of 20.4 oC/m. Suppose that 0.157 grams of an unknown solute are added to 10.0 grams of cyclohexane. If the resulting solution freezes at 3.9oC, calculate the molar mass of the unknown solute.

Answer 4

@AZ

Answer 5

wsp

Answer 6

whatchu need help wit

Answer 7

So for your first question

\(\Delta T_f = \Delta K_f ~m\)

\(\Delta T_f \) is our freezing point depression
\(\Delta K_f\) is our freezing point depression constant for the solvent
and m is molality which is moles of solute/ kg of solvent

Answer 8

You can also have your van't Hoff factor in the equation but since urea is a non-electrolyte, i = 1

Answer 9

What is the normal freezing temperature of water?

We're then trying to decreasing it to -1

so what would that change in temperature be?

Answer 10

32 degrees F

Answer 11

What would that be in Celsius?
We always want to keep it in SI units

Answer 12

0 degrees

Answer 13

Good and so what's the difference between 0 and -1

Answer 14

1

Answer 15

I also noticed I made a typo when writing the formula haha

It's just \( K_f\) which is not reflected below:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ
So for your first question

\(\Delta T_f =K_f ~m\)

\(\Delta T_f \) is our freezing point depression
\(K_f\) is our freezing point depression constant for the solvent
and m is molality which is moles of solute/ kg of solvent
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 16

So \(\Delta T_f\) is going to be 1

They already told us the freezing point depression constant is 1.86 oC/min

now

molality = moles of solute/ kg of solvent

The question is asking us for the grams of urea (which is our solute)
So first we have to calculate the moles of solutes and then use the molar mass of urea, which they graciously provided us with, to calculate the grams of urea needed

Our solvent or solution is water. Remember molality is moles/kg
So we need to convert 50.0 grams to kg first

Answer 17

0.05 kg

Answer 18

Let's plug things in now!


\(\Delta T_f =K_f ~m\)

\( 1 = 1.86 \times \dfrac{\text{moles urea}}{0.05}\)

can you solve for moles urea?

Answer 19

0.02688174

Answer 20

I also want to say that all the units work out too if you check

Answer 21

It is looking for the units in grams

Answer 22

\(\color{#0cbb34}{\text{Originally Posted by}}\) @kamachavis
0.02688174
\(\color{#0cbb34}{\text{End of Quote}}\)

So this is moles of urea

multiply it by the molar mass 60.056 g/mol

and we'll get grams of urea

Answer 23

1.61

Answer 24

Which the closest answer choice is 1.64g

Answer 25

We've been using the numbers with a lot of digits so I'm guessing they probably rounded off in an earlier answer which is why we're slightly off

Answer 26

Yes, that helped a lot!

Answer 27

You have taught me a lot that my professor failed to cover! I am taking notes of everything.

Answer 28

\(\color{#0cbb34}{\text{Originally Posted by}}\) @kamachavis
Cyclohexane normally freezes at 6.6oC and has freezing point depression constant of 20.4 oC/m. Suppose that 0.157 grams of an unknown solute are added to 10.0 grams of cyclohexane. If the resulting solution freezes at 3.9oC, calculate the molar mass of the unknown solute.
\(\color{#0cbb34}{\text{End of Quote}}\)

Now for this question, it's a similar concept

\(\Delta T_f =K_f ~m\)

They've given you Kf
You have the temperature it normally freezes at and then the temperature the new solution freezes at. So find the difference between that and you'll have you \(\Delta T_f\)

what they want you to solve for is molar mass of this unknown substance

Answer 29

so first calculate \(\Delta T_f\) and then plug that value in along with the Kf value.

Solve for m

Answer 30

\(\color{#0cbb34}{\text{Originally Posted by}}\) @kamachavis
You have taught me a lot that my professor failed to cover! I am taking notes of everything.
\(\color{#0cbb34}{\text{End of Quote}}\)

I'm glad to hear that I've been able to help :D
Not all professors are the best at teaching. Based on these questions, I'm assuming this is the second semester of an introductory chemistry course.

This guy had some great lectures if your professor is inadequate (or possibly incompetent LOL)-
https://www.youtube.com/playlist?list=PLMG7wphV7HYW-6SHqV4c0NsaV08RfO5B3

Answer 31

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ
so first calculate \(\Delta T_f\) and then plug that value in along with the Kf value.

Solve for m
\(\color{#0cbb34}{\text{End of Quote}}\)

So remember m is molality

and molality = moles of solute / kg of solvent

The question says `0.157 grams of an unknown solute are added to 10.0 grams of cyclohexane`

What you have to remember is that moles = grams / molar mass

So first solve for moles and then we can use the above to calculate the molar mass
And don't forget to convert 10.0 grams to kg before you try solving for moles

Answer 32

I am a bit confused on this one, this was the hardest one to try and understand for me.

Answer 33

Okay, let's go step by step

`Cyclohexane normally freezes at 6.6oC and has freezing point depression constant of 20.4 oC/m. Suppose that 0.157 grams of an unknown solute are added to 10.0 grams of cyclohexane. If the resulting solution freezes at 3.9oC, calculate the molar mass of the unknown solute`

What is the change in temperature from 6.6 to 3.9
i.e. 6.6 - 3.9 = ?

Answer 34

2.7

Answer 35

that's our \(\Delta T_f\)

now we know our Kf is 20.4 oC/m

so let's solve for molality

2.7 = 20.4 * m

what is m = ?

Answer 36

0.13

Answer 37

good so that's molality and remember

molality = moles solute / kg solvent

and now we have to calculate moles solute
because we know grams, we can then use that to calculate the molar mass

so
molality = 0.1323 mol/kg
moles solute is what we're solving for now
kg solvent is our cyclohexane which is 10.0 grams

what is 10.0 grams in kg?

Answer 38

0.01

Answer 39

good so now let's calculate moles solute and just call it `x`

molality = moles solute / kg solvent

0.1323 mol/kg = x / 0.01 kg

x = ?

Answer 40

0.001323

Answer 41

Good so we have 0.001323 moles of this unknown solute

and this 0.001323 moles is equivalent to `0.157 grams` (from the question)

molar mass of a substance is how many grams are in 1 mole

so basically if you divide 0.157 by 0.001323, you would get the number of grams in 1 mole

and that would be your molar mass

Answer 42

0.00842675?

Answer 43

And the answer choices are in g/mol

Answer 44

no

molar mass = grams/ mole

so 0.157 / 0.001323

Answer 45

119!

Answer 46

and yes the answer choices should be in g/mol

remember when we converted the g to kg earlier? If you put all the units back in, it got cancelled out because molality has a kg in there as well

Answer 47

There you go!

Answer 48

Yes, Thank you!

Answer 49

A student prepares a solution by dissolving 30.0 grams of urea (CON2H4, 60.056 g/mol) in 36.0 grams of water. What is the mole fraction of water in this solution?

Group of answer choices

0.80

0.014

0.55


0.20

Answer 50

We did not finish this problem earlier.

Answer 51

I got 0.014 by dividing 30/60.056/36

Answer 52

mhmm that doesn't look properly set up

we first want to find the number of moles of water

water's molar mass is 18.015 g/mol

so how many moles would that be?

Answer 53

3?

Answer 54

Or 2

Answer 55

we have 36 grams

divide it by the molar mass

Answer 56

2

Answer 57

you can use the units as a way to help you

36 grams / 18.015 grams/mol

grams would cancel out and moles would be on top

Answer 58

but yes

Answer 59

now similarly, 30 grams of urea
molar mass is 60.056 grams / moles

How many moles is 30 grams of urea?

Answer 60

I'm not sure.

Answer 61

and then finally

\( \text{mole fraction of water} = \dfrac{\text{moles of water}}{\text{moles of water + moles of urea}}\)

Answer 62

60

Answer 63

To go from grams to moles

You have to take your grams and divide it by the molar mass
So just like we did for water

30 / 60.056 =

Answer 64

So that would be 3/13 or .230769

Answer 65

where did you get 3 and 13 from

Answer 66

I'm not sure I messed up

Answer 67

so 2/2+0.499

Answer 68

There you go!

Answer 69

0.80 would be the answer

Answer 70

That's right

Answer 71

Thank you!

Answer 72

You're welcome!