Answer
The molarity of this $KOH$ solution is equal to 0.0356 M
Work Step by Step
1. Determine the molar mass of this compound (KOH), and setup the conversion factors:
Molar mass :
$K: 39.10g $
$O: 16.00g $
$H: 1.008g$
39.10g + 16.00g + 1.008g = 56.11g
$ \frac{1 \space mole \space (KOH)}{ 56.11 \space g \space (KOH)}$ and $ \frac{ 56.11 \space g \space (KOH)}{1 \space mole \space (KOH)}$
2. Calculate the number of moles $(KOH)$
$ 4.00 g \times \frac{1 mole}{ 56.11g} = 0.0713 \space moles$
3. Find the concentration in mol/L $(KOH)$:
$C(M) = \frac{n(moles)}{volume(L)}$
$ C(M) = \frac{ 0.0713 \space moles}{ 2.00 \space L} $
$C(M) = 0.0356 \space M$
