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Rook said:
Molality = mol/Kg
(1000g/1Kg)*(100g/15g)*(85g/100g)*(1mol/98g) = 58 mol/Kg
You dont actually need the density because it cancel out.
R
Rook
Gasedo said:
Thanks for the reply!! 👍 👍
I am still confused though... 😕
So first,
Based on the question we Assume there are 85 g of H3PO4 and 15g of H20, right?
Then I thought just convert the 15g of H20 to Kg = .015 Kg
Then Convert 85g of H3PO4 to Moles = 85g / 98 g and divide 0.015 kg
This gives you 58 mol/Kg!!
But?? How can you even use density in this question. The answer is given as:
1000 * (1/1.70) * (100.0 / 15.0) * (1.70/1) * (85.0 / 100.0) * (1/98) m
why is it 100/15? and 85/100?
Which makes NO sense to me!! Please help!
Thanks!!
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well, molality= mol/kg of solvent
you have 100mL of solution and 85mL(not 85g) is H3PO4 and 15ml of water. So given density you can calculate mass of H3PO4 (m=85L * 1.70g/ml ) and water (15mL * 1.70g/ml).
plug in this numbers into the formula now. molality = (85*1.70/98) / (15*1.70/1000), which basically density will cancel out.
I hope this helped. 🙂
you have 100mL of solution and 85mL(not 85g) is H3PO4 and 15ml of water. So given density you can calculate mass of H3PO4 (m=85L * 1.70g/ml ) and water (15mL * 1.70g/ml).
plug in this numbers into the formula now. molality = (85*1.70/98) / (15*1.70/1000), which basically density will cancel out.
I hope this helped. 🙂
Rook said:
R
Rook
mahGOL said:
Hey Mahgol,
I was going over the DAT achiever answers for a different question but quite similar and it is in fact 85g of solute out of 100g total (meaning 15g of H20)
For example:
What is the molarity of a 70.0% concentrated nitric acid, HNO3, if the density of solution is 1.42 g/ml?
Answer
[1000 mL / 1 L ] * [1.42g of Solution / 1 mL] * [70.0g HNO3 / 100.0g of Solution] * [ 1mol HNO3 / 63g] M = mol/L
So when the solution is in %. it is g of solute / g of total solution! Hope that clarifies things for others. 👍
Appreciate the help!! Gasedo was right all along 👍
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Rook, pesare khoob
If you look closer at the questions you see one is asking about molarity and the other one is molality. 🙂
However, it does not make a difference if you use %g or %ml. I just wanted to show you how the density was cancelled out in the molality question.
let me know if you have any other questions. 🙂
If you look closer at the questions you see one is asking about molarity and the other one is molality. 🙂
However, it does not make a difference if you use %g or %ml. I just wanted to show you how the density was cancelled out in the molality question.
let me know if you have any other questions. 🙂
Rook said:
R
Rook
mahGOL said:
Khanoomeh Gol,
When given percents I think you use grams and not mLs regardless if asking for molarity or molality. I am not doubting your chemistry abilities 😛 you have demonstrated that you are a chem wiz already! But just trying to be as crystal clear as possible. Ultimately, I believe you are right. Whether you use one or the other it won't make a difference in your solutions. Thanks for the input! 👍
R
Rook
mahGOL said:
😛 I was shocked 😱 that you spoke farsi to me.. (in a good way of course 😀 ) Khoshvachtam Khanoomeh Gol! Vali azizehman
Solutions = mL, you are right. EXCEPT when given %
i.e. you have a 56% solution of (Gol) This means...56g of Gol to 44g of H20 (note thought: 44g of H20 = 44 mL of H20 considering density of H20 is 1 g/mL). That's how i'm going to do it on my DAT 😉 👍
R
Rook
Another question:
What will be the concentration of OH - at equilibrium if a mixture consisting 0.20 mole of NH4Cl and 0.30 mole of NaOH is diluted to 1.0 liter? (Assume ionization constant for ammonia, Kb = 1.8 x 10-5)
NH3 + H2O <==> NH4+ + OH -
And what is the concentration of NH4 at equilibrium
Answer for OH = .1
Answer for NH4 = 3.6*10^-5
For NH4 solution please break down as much as possible. thanks!!!
mahGOL make me proud! 😉
What will be the concentration of OH - at equilibrium if a mixture consisting 0.20 mole of NH4Cl and 0.30 mole of NaOH is diluted to 1.0 liter? (Assume ionization constant for ammonia, Kb = 1.8 x 10-5)
NH3 + H2O <==> NH4+ + OH -
And what is the concentration of NH4 at equilibrium
Answer for OH = .1
Answer for NH4 = 3.6*10^-5
For NH4 solution please break down as much as possible. thanks!!!
mahGOL make me proud! 😉
Rook said:
Salam mahGol, rook and Ms Homeyra
rokh, here is the solution:
Kb is a very small number. So that means the following reaction is a lot more favorable:
NH4+ + OH- ------> NH3 + H2O
0.2 M + 0.3 M ---->
0.......+ 0.1 M ----> 0.2 M + (H2O is liq, don't worry about this one)
Now use the above numbers to solve the problem:
NH3 + H2O <==> NH4+ + OH -
0.2 M........<---> 0......+ 0.1 M
0.2 M - x...<---> x......+ 0.1 M + x
because x is a tiny number (remember Kb is a very small number)
0.2 M........<---->x......+0.1 M
Kb = 1.8 * 10^-5 = (x)(0.1) / (0.2)
x = (1.8 * 10^-5)(0.2)/(.1) = 3.6 * 10^-5
[OH-] = approx. 0.1 M
[NH4+] = approx. 3.6 * 10^-5 M
bachehaa movafagh bashid 🙂
