Question

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?

Answer 1

`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.