Chemistry q. from Kaplan book!!! HELP!

[aquafreshlovin]

So I am working on Chapter 9 of Kaplan blue book (gen chem)...and I just CANNOT figure these out! Sure, the answers are provided but I didn't understand them. I would like to see if others can work these problems in diff. ways.

1) Given that the molecular weight of ethyl alcohol, CH3CH2OH, is 46, and that of water is 18, how many grams of ethyl alcohol must be mixed with 100mL of water for the mole fraction (X) of ethyl alcohol to be 0.2?

answer: 64.4 grams

2) What are the concentrations of each of the ions in a saturated solution of PbBr2, given that the Ksp of PbBr2 is 2.1x10^-6?

answer: [Pb^2+] = 8.07e-3M and [Br-]=1.61e-2M

For this problem, I can do everything up 2.1e-6 = 4x^3. I am having hard time figuring out for X ... can somebody show the math in solving for X by hand? Of course it's easy to figure it out with a calculator but as ya'll know, it's not allowed...

THANKS!!!

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[osimsDDS]

I dunno the first one but i can help you out with the second...

PBr2 ----> P2- + 2Br -
Now in terms of ksp basically put an x unknown for both Phosphorous and Bromine.... therefore it will be (x) times (2x)^2 which gives you 4(x)^3.

So now just solve for x which will give you the concentrations but be careful because Br is 2 moles so multiply x by 2.

Here: 4(x)^3 = 2.1x10^-6
x = 8.06x10^-3

P2- = x
Br- = 2x so just multiple x by 2

 

[osimsDDS]

opps sorry its supposed to be Pb not P my bad just change it to Pb you knkow wut i mean

 

[smile101]

So I am working on Chapter 9 of Kaplan blue book (gen chem)...and I just CANNOT figure these out! Sure, the answers are provided but I didn't understand them. I would like to see if others can work these problems in diff. ways.

1) Given that the molecular weight of ethyl alcohol, CH3CH2OH, is 46, and that of water is 18, how many grams of ethyl alcohol must be mixed with 100mL of water for the mole fraction (X) of ethyl alcohol to be 0.2?


2) What are the concentrations of each of the ions in a saturated solution of PbBr2, given that the Ksp of PbBr2 is 2.1x10^-6?

I dunno is I am doing this right or not, but here it is:
Since 100 ml is ~ same as 100g
mole of H2O: 100g/18 = 5.56mole
Now, mole fraction = moles of ethyl alcohol/ (moles of ethyl alcohol + H2O)
Lets say that moles of ethyl alcohol is x, so 0.2=x/(5.6+x)
x= 1.37 mole
that means that the moles of ethyl alochol needed for 0.2mole fraction is 1.37moles, therefore the weight in grams would be 1.37*46= ~63g

Let me know if this right!

 

[daviddav]

Density of water = 1 g/ml

g of water = 100 ml * 1 g/ml = 100 g of water

mol of water = 100 g of water * 1 mol of H2O / 18 g H2O = 5.55 mol water

5.55 / X = 0.8 (mol fraction of water) X (total number of moles) = 5.55/.8 = 6.94

6.94 - 5.55 = 1.39 mol of ethyl alcohol

1.39 mol ETOH * 46 g/mol ETOH = 63.94 g ETOH that must be added to 100 ml of water

I hope this is right. I tried at least!!

 

[daviddav]

Haha...I think I posted mine right after you did..oh well..at least I know that I was right.

 

[Zerconia2921]

1) Given that the molecular weight of ethyl alcohol, CH3CH2OH, is 46, and that of water is 18, how many grams of ethyl alcohol must be mixed with 100mL of water for the mole fraction (X) of ethyl alcohol to be 0.2?


2) What are the concentrations of each of the ions in a saturated solution of PbBr2, given that the Ksp of PbBr2 is 2.1x10^-6?


Problem 1)

Mole fraction formula is Moles of element/total number of moles

Starting the problem we werent told that density of water is 1g/ml. This is needed for us to find the number of moles in the mixture.
So
1g/ml (100ml) = 100grams of water
100grams H20 x 1mole/18g H20 = 5.6 moles of H20

So 5.6moles of H20 are in the mixture

They gave us the mole fraction for ethyl alco to be 0.2, but we dont know the number of moles of ethyl alco. So let x = number of moles of ethyl alco

Since mole fraction is part over whole 5.6 moles of H20 belong to the whole and the other amount x belongs to ethyl alco.

0.2 = x/(x + 5.6)
Solve for x you get 1.4 moles of ethyl alco

Now you have the moles of ethyl alco then just use stoci to find grams
1.4 moles x 46g/mole = 64.6 g ethyl alco



First step is write out the equation for PbBr2
PbBr2 ---> Pb^2 + Br-

Second step is balance

PbBr2 ----> Pb^2 + 2Br-

Let x = unknown concentrations
PbBr2 ----> x (2x)^2 => 4x^2 x => 4x^3
Ksp = 4x^3
2.10x10^-6 = 4x^3
Solve for x. Once you solved for x plug back in into original concentration.