# A 75*mL volume of 3.80*mol*L^-1 "iron(II) sulfate" is diluted to 4.0*L. What is the new concentration?

Feb 1, 2017

$\text{Molarity}$ $=$ $\text{Moles of solute"/"Volume of solution}$

#### Explanation:

And thus we use the above equation to give us (i), the initial moles of solute:

$\text{Moles of}$ $F e S {O}_{4}$ $=$ $\text{Volume}$ $\times$ $\text{Molarity}$

$75 \times {10}^{-} 3 L \times 3.8 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.285 \cdot m o l$

But this is diluted to a volume of $4.0 \cdot L$, and so to (ii), the new concentration:

$\text{Concentration}$ $=$ $\frac{0.285 \cdot m o l}{4.0 \cdot L} = 0.071 \cdot m o l \cdot {L}^{-} 1$