Question

I would appreciate very much If you guys could show me how to approach this problem.
A solution is prepared by dissolving 10.8g of ammonium sulfate in enough water to make 100.0mL of stock solution.
A 10.00mL sample of this stock solution is added to 50.00mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
the way I approached this problem was by getting the molarity of ammonium sulfate(of corse after changing the grams to moles and the mL to L.)then I multiply the concentration by .01L since that is the amount of the solution that will be used. Then I calculated the number of moles again with the new liter and molarity, to obtain .01 moles of ammonium sulfate, I did the mole to mole ratio and got .01 moles of ammonium. Then I took the numberof moles and divide bi the water being add which is .05L which gave me .2 of the concentration of ammonium, which is wrong since the answer is .272M for ammonium and .136M for sulfate. Please help me.

Answer 1

calculate the molarity of the first solution. Now, .01L*XXM= new moles of ammonium sulfate.

Concentrated is now the new moles ammoniumslufate/.06

Overall, you have

Concentration= 2*(10.8/molmass)/.1 *.01/.06
Of course, the 2 is because there are two ammonium molecules per molecule of ammonium sulfate.

Answer 2

I do not understand where you got the .06 from?

Answer 3

when one adds 10 ml to 50 ml, the new volume is .06L.