# What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M KOH to 250 mL?

Apr 8, 2017

As always, $\text{concentration"="Moles of solute"/"Volume of solution} \ldots \ldots \ldots .$

#### Explanation:

And here (i) we calculate the $\text{moles of solute}$ in the initial volume:

$= 45 \times {10}^{-} 3 L \times 4.2 \cdot m o l \cdot {L}^{-} 1 = 0.189 \cdot m o l$ with respect to $K O H \left(a q\right)$.

And then (ii) we divide this value by the NEW volume of solution:

$\text{Concentration} = \frac{0.189 \cdot m o l}{0.250 \cdot L} \cong 0.8 \cdot m o l \cdot {L}^{-} 1$ with respect to $K O H$.

The volume has increased approx. fivefold, and hence the concentration decreases approx. fivefold........