734 mL of a 1.13 M calcium sulfate (#CaSO_4#) solution is diluted with 2.56 L of water. What is the new concentration in molarity?

1 Answer
Apr 20, 2016

Answer:

#C_2=(0.734*Lxx1.13*mol*L^-1)/(0.734*L+2.56*L)# #=# #??*mol*L^-1#

Explanation:

#C_2=(0.734*Lxx1.13*mol*L^-1)/(0.734*L+2.56*L)# #=# #??*mol*L^-1#

How was the calculation performed?

You have a known volume of calcium sulfate with known concentration. We work out the initial molar quantity:

#"Moles"# #=# #"Concentration"xx"Volume"#.

Given the molar quantity, we divide this by the new volume:

#"Concentration"# #=# #"Moles"/"Volume"#

Note that the question said it is diluted with (and not to) #2.56*L#; this is why the volumes are additive. I don't think this question is entirely realistic in that calcium sulfate is not particularly soluble in water. They should have made it calcium nitrate or calcium acetate.