SOLUTION: If the density of a 500.0g H2SO4 solution is 1.84 g/cm^3 what are its molarity, mole fraction, and molality? Don't know if you can help me. This is the only site I could find.

Algebra ->  Test -> SOLUTION: If the density of a 500.0g H2SO4 solution is 1.84 g/cm^3 what are its molarity, mole fraction, and molality? Don't know if you can help me. This is the only site I could find.      Log On


   



Question 1001501: If the density of a 500.0g H2SO4 solution is 1.84 g/cm^3 what are its molarity, mole fraction, and molality?

Don't know if you can help me. This is the only site I could find. It's chemistry. Do you know of a chemistry site like this one?

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The website yeahchemistry.com will help you with chemistry questions.
If the solution was 100% sulfuric acid, you would have 500g sulfuric acid and 0g water.
When you buy a bottle of sulfuric acid it would have a label with analysis data,
and it could read "assay = 98.6%", or something like that.
However, you may assume, it is 100%, because 1.84g/mL is the densityof 100% sulfuric acid at 10%5Eo .
(Warmer acid, and/or less concentrated acid has a lower density).
Then, %28500g%29%2A%281mol%2F%2298+g%22%29=5.1mol would give you the number of moles of sulfuric acid.
You can use the density to find the volume of those 500g solution,
500g%2A%281mL%2F1.84g%29=272mL=0.272L%29 .
Then, 5.1mol%2F%220.272+L%22=18.8mol/L is the molarity, often written as 18.8M , and read as "18.8 molar".
Since 500g=0.5kg , 5.1mol%2F%220.5+kg%22=10.2mol/kg is the molality,
which your instructor may express as 10.2molal or 10.2m .
The mole fraction would be
moles acid/{moles acid + moles water) ,
and since we are assuming 0% water,
that molar fraction would be 1.0%22 .