# Question #0a1c0

Aug 27, 2016

1.71M

#### Explanation:

First you must find the amount of NaOH in the first solution, using
$\text{amount" = "concentration" xx "Volume}$
$n = c V = 6.00 m o l {L}^{-} 1 \times 0.100 L = 0.600 m o l$

Since the only thing added is water, the amount of NaOH does not change, so we can calculate the new concentration using the amount we just calculated and the new volume of the solution ($0.100 L + 0.250 L = 0.350 L$)

$\text{concentration" = "amount"/"Volume}$

$c = \frac{n}{V} = \frac{0.600 m o l}{0.350 L} = 1.33 m o l {L}^{-} 1 = 1.71 M \text{(3s.f.)}$