Posted by STACY on Monday, December 13, 2010 at 7:56pm.
Consider an aqueous solution prepared from 250.0 mL of water and 1.27 g of potassium bromide. Express the amount of potassium bromide in this solution in terms of
(a) mass percent
(b) mole fraction
(c) molarity
assume the density of the solution is equal to that of water, 1.00 g mL-1.........help!
CHEMISTRY - DrBob222, Monday, December 13, 2010 at 8:54pm
Surely you don't want me to work all three. Surely you know SOMETHING about SOME of them. Just remember the definitions:
1. mass percent. grams solute/100 g solution.
2. mole fraction. moles solute/total mules.
3. molarity = moles/L of solution.
Apply those three definitions and you will have your answer(s).
Post your work if you get stuck.
CHEMISTRY - STACY, Tuesday, December 14, 2010 at 4:40pm
I surely do, and don't call me Shirley! LOL yessss i'm sorry I don't know much about chemistry that's why i'm here!!
1. Mass percent= grams solute/100solution 1.27g/250.0mL = 5.08x10^-03 is my answer but the correct one is 0.505%
2. Mole fraction= moles solute/total moles ?/119.002
3. Molarity 1.27g/119.002= 0.0107moles
than 0.0107/0.25L = 0.048 that's my answer but the correct one is 4.25x10^-2
If your name is really Stacy, that's a pretty name, and I wouldn't think of calling you Shirley.
We welcome you back any time but we want to know how you would approach the problems and/or what you know and what you don't understand.
1. mass percent = mass solute/100 g solution.
We have 1.27 g KBr and we have that in 250 mL H2O(250 g).
So the total mass of solution is 250g + 1.27 g = 251.27 g.
mass percent = (1.27/251.27)*100 = 0.505% w/w. (You get 0.508 if you forget to add 250 gH2O + 1.27 g KBr = total mass 251.27. That's easy to forget (I know because I sometimes forget) so be careful with these.
2. moles KBr = 1.27g/119 = 0.01067
moles H2O in 250 g = 250/18.015 = 13.877.
mole fraction KBr = moles KBr/total moles. You can finish.
3. M = moles/L. I don't get your answer. You just read your calculator wrong. If I do the math with your numbers I obtained 0.0428; I think you just skipped the 2 when you read it.
If the density of the solution is 1.00 g/mL, then the volume of the solution is volume = mass/density = 251.27/1.00 = 251.27 (go back and re-read how we arrived at that), so M = moles/L and M = 0.01067/0.25127 = 0.04246 M which rounds to 0.0425 M.
Thanks! and No Stacy is not my real name, although Shirley is my mothers name LOL!
To find the amount of potassium bromide in the solution in terms of mass percent, you need to divide the mass of the solute (1.27 g) by the mass of the solution (250.0 mL of water), and multiply by 100%.
So, the calculation would be:
Mass percent = (1.27 g / 250.0 mL) x 100% = 0.508% (rounded to 3 decimal places)
However, the correct answer is 0.505% according to the provided information. It's possible that there might be a rounding error or small discrepancy in the data.
For mole fraction, you need to know the number of moles of the solute and the total number of moles in the solution. The number of moles of the solute can be calculated by dividing the mass of the solute (1.27 g) by the molar mass of potassium bromide (119.002 g/mol).
The calculation would be:
Mole fraction = (1.27 g / 119.002 g/mol) / total moles in solution
However, in this case, the total moles in the solution are not provided, so it's not possible to calculate the mole fraction accurately.
For molarity, you need to know the number of moles of the solute and the volume of the solution. The number of moles of the solute can be calculated in the same way as for the mole fraction.
The calculation would be:
Molarity = moles of solute / volume of solution
Given that the volume of the solution is 250.0 mL (or 0.25 L), and the number of moles of the solute is calculated as 0.0107 moles, the molarity works out to be:
Molarity = 0.0107 moles / 0.25 L = 0.0428 M (rounded to 4 decimal places)
However, the correct answer is 4.25x10^-2 M, which is the same value but expressed in scientific notation.