# A 295*mL volume of solution that is 1.00*mol*L^-1 with respect to "ammonium chloride" is diluted with a 2.25*L volume of fresh solvent. What is the resultant [NH_4Cl]?

Aug 14, 2016

Approx. $0.116 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

As with all these sorts of problems,

$\text{concentration"="moles of solute"/"volume of solution}$

So we just work out moles and volume appropriately:

$=$ $\frac{295 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 1.00 \cdot m o l \cdot {L}^{-} 1}{0.295 \cdot L + 2.25 \cdot L}$

$=$ ??mol*L^-1 with respect to $N {H}_{4} C l$.

The concentration makes sense, as you dilute the initial concentration almost 10fold.