# 295 mL of a 1.00 M ammonium chloride (NH_4Cl) solution is diluted with 2.25 L of water. What is the new concentration in molarity?

May 24, 2016

Approx. $0.13 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$\text{Concentration}$ $=$ $\text{Number of moles"/"Volume of solution}$.

Initially, we have $0.295 \cdot L \times 1.00 \cdot m o l \cdot {L}^{-} 1$ $N {H}_{4} C l$ $=$ ??*mol

Finally this molar quantity is dissolved in a $2.25 \cdot L$ volume.

$\text{Concentration}$ $=$ $\frac{0.295 \cdot L \times 1.00 \cdot m o l \cdot {L}^{-} 1}{2.25 \cdot L}$ $=$ ??mol*L^-1