hard gen chem problem

[polarmolar]

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How many grams of Al2(SO4)3 are needed to make 87.5 g of 0.3 m Al2(SO4)3 solution?

 

[joonkimdds]

I got 8.14 but that's the accurate answer.
If it's on DAT and I want to round up everything to 1~2 decimal places, then I would end up getting 8.5~9.

could u double check the answer for me?

 

[Isradoc1983]

can you please explain how you got this answer?
cause i am getting weired results
thanks

 

[ZenoVT]

I got about 8.977 so 9 grams.

You have 2 equations to solve for grams of Al2(SO4)3
1) grams Al2(SO4)3 + grams liquid = 87.5 g
2) .3 molality = mol of Al2(SO4)3/ (kg of Al2(SO4)3 + kg of liquid)

The rest is just conversion factors and rearrangement of the terms to isolate grams of Al2(SO4)3 on one side of the equation. Then use substituting to solve the problem.

 

[joonkimdds]

I got about 8.977 so 9 grams.

You have 2 equations to solve for grams of Al2(SO4)3
1) grams Al2(SO4)3 + grams liquid = 87.5 g
2) .3 molality = mol of Al2(SO4)3/ (kg of Al2(SO4)3 + kg of liquid)

The rest is just conversion factors and rearrangement of the terms to isolate grams of Al2(SO4)3 on one side of the equation. Then use substituting to solve the problem.

I believe u missed one step.
After u get 8.9775, u should divide it by 1.1026.

 

[Doa110]

I believe u missed one step.
After u get 8.9775, u should divide it by 1.1026.

Can you please tell how you solved the problem because I don't know how to got to 1.1026? It doesn't seem that I am on the right track! Thanks.

 

[ZenoVT]

I believe u missed one step.
After u get 8.9775, u should divide it by 1.1026.

Yeah i did a lot of rounding, so my answer isn't too precise. But it's good enough to pick an answer =)
X = g of Al2(SO4)3
Y= g of liquid
G = molar mass of Al2(SO4)3 = 342
.3 m = .0003 mol/g
.0003 mol/g = X / [ G* (X + Y)]
.0003 * [ G* (X + Y)] = X
.0003*G*X + .0003 G* Y = X
.0003*G*X - X = -.0003 G* Y
X ( .0003 * G - 1) = -.0003 G* Y
X = -.0003 G* Y / ( .0003 * G - 1) = -.0003 *342* Y / (.0003 * 342 -1) = .1026Y / .8974
X= .1143 Y
X + Y = 87.5
X + X/.1143 = 87.5
X ( 1 + 1/.1143) = 87.5
X ( 9.74) = 87.5
X = 87.5 / 9.74 = 9

 

[IWantSmile]

How many grams of Al2(SO4)3 are needed to make 87.5 g of 0.3 m Al2(SO4)3 solution?

Wao~. I have a no idea... .

 

[dcl222]

I got about 8.977 so 9 grams.

You have 2 equations to solve for grams of Al2(SO4)3
1) grams Al2(SO4)3 + grams liquid = 87.5 g
2) .3 molality = mol of Al2(SO4)3/ (kg of Al2(SO4)3 + kg of liquid)

The rest is just conversion factors and rearrangement of the terms to isolate grams of Al2(SO4)3 on one side of the equation. Then use substituting to solve the problem.

wouldn't molality be mol solute/ kg solvent (not kg solution as you have)?

.3molal x 1kg solvent = mol Al2(SO4)3
.3 mol = X g / MW g/mol ---> .3mol = Xg / 342g ---> X g Al2SO43 = .3 x 342 = 102.6 g Al2(SO4)3 (in a .3 molal solution)
So in .3molal solution there is a total of 1000g(solvent) + 102.6g (solute)= 1102.6 g
Given 87.5 g solution, how much solute (Al2SO43) are there?
87.5g solution/ 1102.6 g solution = Y g Al2SO43 / 102.6 g Al2SO43
Y g = 87.5 x 102.6 / 1102.6

 

[joonkimdds]

all u guys have to know to solve this problem is the
fact that molarity = mol of solute / Kg of solvent.
so your first goal is to set it up that way.

0.3m = mol of solute / Kg of solvent

ok let's start

0.3m = [?g*(1mol/342g)] / [(87.5g solution - ?g solute) / 1000]

let me pause and explain
[?g* (1mol/342g)] will give us mol of solute.
[(87.5g solution - ?g solute) / 1000] will give us Kg of solvent. Solution is solute + solvent is what you need to know, and you also need to know that we should convert it into Kg and that's why I divided it by 1000.

ok? so far so good?

next, just solve for ? g.

0.3m = [?g* (1mol/342g)] / [0.0875g – (?g/1000)]
[0.02625 – (3x10^-4)X] = [X / 342]
x = 8.9775 – 0.1026X
1.10256 X = 8.9775
X = 8.14

 

[ZenoVT]

wouldn't molality be mol solute/ kg solvent (not kg solution as you have)?

.3molal x 1kg solvent = mol Al2(SO4)3
.3 mol = X g / MW g/mol ---> .3mol = Xg / 342g ---> X g Al2SO43 = .3 x 342 = 102.6 g Al2(SO4)3 (in a .3 molal solution)
So in .3molal solution there is a total of 1000g(solvent) + 102.6g (solute)= 1102.6 g
Given 87.5 g solution, how much solute (Al2SO43) are there?
87.5g solution/ 1102.6 g solution = Y g Al2SO43 / 102.6 g Al2SO43
Y g = 87.5 x 102.6 / 1102.6

you are right. sorry bout that

 

[pbure9]

all u guys have to know to solve this problem is the
fact that molarity = mol of solute / Kg of solvent.
so your first goal is to set it up that way.

0.3m = mol of solute / Kg of solvent

ok let’s start

0.3m = [?g*(1mol/342g)] / [(87.5g solution - ?g solute) / 1000]

let me pause and explain
[?g* (1mol/342g)] will give us mol of solute.
[(87.5g solution - ?g solute) / 1000] will give us Kg of solvent. Solution is solute + solvent is what you need to know, and you also need to know that we should convert it into Kg and that’s why I divided it by 1000.

ok? so far so good?

next, just solve for ? g.

0.3m = [?g* (1mol/342g)] / [0.0875g – (?g/1000)]
[0.02625 – (3x10^-4)X] = [X / 342]
x = 8.9775 – 0.1026X
1.10256 X – 8.9775
X = 8.14

I'm a little unsure about how the problem is phrased, but if 87.5 grams is the weight of the entire solution, then thats about what i got too!

 

[IWantSmile]

all u guys have to know to solve this problem is the
fact that molarity = mol of solute / Kg of solvent.
so your first goal is to set it up that way.

0.3m = mol of solute / Kg of solvent

ok let’s start

0.3m = [?g*(1mol/342g)] / [(87.5g solution - ?g solute) / 1000]

let me pause and explain
[?g* (1mol/342g)] will give us mol of solute.
[(87.5g solution - ?g solute) / 1000] will give us Kg of solvent. Solution is solute + solvent is what you need to know, and you also need to know that we should convert it into Kg and that’s why I divided it by 1000.

ok? so far so good?

next, just solve for ? g.

0.3m = [?g* (1mol/342g)] / [0.0875g – (?g/1000)]
[0.02625 – (3x10^-4)X] = [X / 342]
x = 8.9775 – 0.1026X
1.10256 X = 8.9775
X = 8.14

Thanks! you make it so easy.

 

[joonkimdds]

Thanks! you make it so easy.

Haha, thanks 🙂