View Single Post
  #2 (permalink)  
Old 04-20-2008, 06:54 AM
rubidium.chloride rubidium.chloride is offline
Member
 
Join Date: Mar 2008
Posts: 5
Rep Power: 0
rubidium.chloride has a spectacular aura about
Well, the first thing you ought to do is to multiply 0.1 L by 0.10 M. That way, you'll find out how many moles CuSO4*5H2O were there.

0.1 L * 0.10 mol CuSO4*5H2O / 1 L

Then, multiply the thing by its molar mass.

0.1 L * 0.10 mol CuSO4*5H2O / 1 L * 249.68 g CuSO4*5H2O / 1 mol CuSO4*5H2O = 2.5 g

Reply With Quote