Answer
The molarity of that $NaOH$ solution is equal to $2.14$ $M$ $NaOH$.
Work Step by Step
1. Determine the molar mass of this compound ($NaOH$), and setup the conversion factors:
Molar mass :
$Na: 22.99g $
$O: 16.00g $
$H: 1.008g$
22.99g + 16.00g + 1.008g = 40.00g
$ \frac{1 \space mole \space (NaOH)}{ 40.00 \space g \space (NaOH)}$ and $ \frac{ 40.00 \space g \space (NaOH)}{1 \space mole \space (NaOH)}$
2. Calculate the number of moles $(NaOH)$
$ 30.0 \space g \times \frac{1 \space mole}{ 40.00 \space g} = 0.750 \space moles$
3. Convert the volume to liters:
$350 \space mL \times \frac{1 \space L}{1000 \space mL} = 0.350 \space L$
3. Find the concentration in mol/L (Molarity) $(NaOH)$:
$C(M) = \frac{n(moles)}{volume(L)}$
$ C(M) = \frac{ 0.750 \space moles}{ 0.350 \space L} $
$C(M) = 2.14 \space M$
