# 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?

Apr 20, 2016

${C}_{2} = \frac{0.734 \cdot L \times 1.13 \cdot m o l \cdot {L}^{-} 1}{0.734 \cdot L + 2.56 \cdot L}$ $=$ ??*mol*L^-1

#### Explanation:

${C}_{2} = \frac{0.734 \cdot L \times 1.13 \cdot m o l \cdot {L}^{-} 1}{0.734 \cdot L + 2.56 \cdot L}$ $=$ ??*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:

$\text{Moles}$ $=$ $\text{Concentration"xx"Volume}$.

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

$\text{Concentration}$ $=$ $\text{Moles"/"Volume}$

Note that the question said it is diluted with (and not to) $2.56 \cdot L$; 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.