Question

Can I please get some Chemistry help.

Answer 1

What is the hypothetical van't Hoff factor of magnesium nitrate, Mg(NO3)2?

Answer 2

Pentane and heptane are two hydrocarbon liquids present in gasoline. At 20 degrees, the vapor pressure of pentane is 420 torr and the vapor pressure of heptane is 36 torr. Calculate the vapor pressure of a solution made by mixing 0.60 moles of pentane and 0.90 moles of hexane.

Answer 3

An aqueous solution of H2SO4 has a concentration of 0.750 M and a density if 1.046 g/mL. What is the mass percent of H2SO4 in this solution?

Answer 4

@aZ

Answer 5

One question at a time, shall we?

To find the van't Hoff factor, how many ions will magnesium nitrate dissociate into?

\(\text{Mg(NO}_3)_2 \to \text{__ Mg}^{2+} + \text{__ NO}_{3}^{-}\)

balance the reaction, what will the coefficients be?
If you add the two coefficients that go in the blanks ____
you'll have your hypothetical van't Hoff factor

Answer 6

0 and 2?

Answer 7

So it would be 2?

Answer 8

Are you sure? We have one magnesium one the left side, so how many would we need on the other side?

Answer 9

We have to keep the equation balanced

Answer 10

One on both sides

Answer 11

You are correct about the 2 for the second blank

Answer 12

Yes!

Answer 13

So what's our ideal van't Hoff factor?

Answer 14

So 3?

Answer 15

You got it!

Answer 16

Awesome thank you.

Answer 17

Of course!

Answer 18

Could you help with the other two problems as well?

Answer 19

For your second question:

You have to know Raoult's Law

\(\Large P_{\text{solution}}= X_{\text{solvent}} \times P_{\text{solvent}}\)

vapor pressure of a solution = mole fraction of the solvent * vapor pressure of the pure solvent

And when you have multiple solutes, then we get

\( P_{\text{total}}=P_1 + P_2 + P_3 + ... \)

so using that first formula I wrote above, we would get

\( P_{\text{total}}=(X_AP_A) + (X_BP_B) + (X_CP_C) + ... \)

But in our question we only have two solutes so we'll just use

\( P_{\text{total}}=(X_AP_A) + (X_BP_B) \)

so it's important for us to first calculate the mole fraction of heptane and pentane
and that's not very difficult to do!

\(\text{mole fraction} = \dfrac{\text{moles of solute}}{\text{total number of moles}} \)

Answer 20

does that make sense so far?

They gave you
the vapor pressure of pentane = 420 torr
vapor pressure of heptane = 36 torr

0.60 moles of pentane
0.90 moles of heptane (you wrote hexane but I'm assume it's a typo)

Answer 21

So first we have to calculate the mole fraction for each of them

What's the total number of moles?
0.60 + 0.90 = ?

Answer 22

0.4 for pentane
0.6 for heptane? for mole fractions

Answer 23

1.5 total

Answer 24

That's correct! You got the mole fractions and you have the vapor pressures so all you do is put it into the formula

\( P_{\text{total}}=(X_AP_A) + (X_BP_B)\)

\(P_{\text{total}}=(0.4\times420) + (0.6\times 36)\)

Answer 25

Just multiply inside the parenthesis first and then add them

Answer 26

189.6 torr?

Answer 27

And that's your answer!

Answer 28

Thank you, now the last question shouldn't be too hard to work out but I still need an explanation kind of.

Answer 29

Sure!

So we have H2SO4 in an aqueous solution. That means it's in water and they tell us that the concentration of the solution is 0.750 M

Percent mass would be mass of H2SO4 / mass of solution * 100%

Since we know the concentration is 0.750 M
(remember M is for molarity which is moles of solute/liters of solution)

Let's assume that the solution is 1 L
(since we're looking for percent mass, it doesn't matter because it's going to be a ratio of the mass but 1 L is easiest to work with)

So if we have 1 L, then we have 0.750 moles of H2SO4

Answer 30

So do you know how we go from moles to gram?
Because percent mass is about mass and mass is in grams not moles

Answer 31

(and are you following along still, haha)

Answer 32

Yes haha, would it be 13.5 grams?

Answer 33

How did you get 13.5 grams?

To go from moles to grams, we have to multiply the number of moles by the molar mass of H2SO4

Answer 34

You can calculate molar mass of H2SO4 by finding each element on the periodic table and adding it all up or if you're lazy (like me), you can google it too :b

Answer 35

73.55925?

Answer 36

Now that's more like it!

So we have 73.559 grams of H2SO4

So we used the molarity to calculate the mass of H2SO4 because like we said

molarity = moles of SOLUTE/ liters of solution

But now we need to find how much the solution weights and for that they gave us the density!

Remember that
density = mass/volume

The solutions density is 1.046 g/mL
And like we said earlier, let's assume we have 1 L of solution

And let's not get tripped up by units so I'll just convert 1 L to 1000 mL, can you solve for the mass of the solution?
1.046 g/mL = MASS / 1,000 mL

Answer 37

1046?

Answer 38

Great!

So earlier we found that we have 73.559 grams of H2SO4
and now we found that have 1046 grams of the entire solution

And percent mass of H2SO4 would just be that ratio as a percentage

percent mass of H2SO4 = 73.559 grams/ 1046 grams * 100%

Answer 39

7.04?

Answer 40

7.04% yes

(although I calculated 7.03% but there could have been a rounding error somewhere along the way so it's close enough haha)

Answer 41

Anyway you could hang around to help me with a few more questions?

Answer 42

I'll try! But I've got a lot of studying of my own to do with a big exam coming up that I haven't started studying yet for 🙃u

But you can tag me @AZ on your next post and I'll check it out when I'm online!

Answer 43

Alright thank you. Anyway you can do one more?