The density of a sodium sulfate solution is 1.07 g/cm3 The solution is 8.00% sodium sulfate by mass. How many cm3 of the solution are needed to supply 4.28 g of sodium sulfate?
2006-08-27 12:18:14 · 4 answers · asked by pilotmanitalia 5 in Science & Mathematics ➔ Chemistry
What you need to do is figure out the weight of the sodium sulfate in the solution. (assuming what you're asking is to obtain pure sodium sulfate out of a solution, this problem is confusingly worded)
so, at 8%, multiply the 1.07g by .08 and you get .0856 g.
take that number and divide 4.28 by it. You get 50 g/cm3.
I'm not sure if that was the question is asking, but I'm reasonably sure it is. I hope this helps you figure it out!
2006-08-27 12:29:51 · answer #1 · answered by solitusfactum 3 · 0⤊ 0⤋
Okay, first you have to know how much mass of solution you will need. To do this you divide 4.28 by 0.08 (8.00%). From this you will get total grams of solution. You also know the density of the solution so you then divide by density of the sodium sulfate solution. This gives you your cm3 because the grams cancel out.
4.28 g/ 0.08 = 53.5 g solution
53.5 g/ 1.07 g/cm^3 = 50 cm^3 of solution
2006-08-27 12:30:52 · answer #2 · answered by Magnetochemist 4 · 0⤊ 0⤋
4 x 1.07 g/cm3 = 4.28 g. But only 8.00% of that is sodium sulfate by mass. 8/100 = 12.5 . So... you'll need 12.5 * 4.28 cm3 of the solution, or 53.5 cm3.
2006-08-27 12:29:33 · answer #3 · answered by Spot! 3 · 0⤊ 0⤋
you know that 4.28g represent 8% of the total mass of the solution.
that you can use to compute the total mass of the solution.
Afterwards, having the mass and density of the solution, you can compute the volume of it.
density = mass/volume -> volume = mass/density
2006-08-27 12:30:28 · answer #4 · answered by Anonymous · 0⤊ 0⤋