Asked by STACY
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!
(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!
Answers
Answered by
DrBob222
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.
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.
Answered by
STACY
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
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
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