# Question #516aa

Aug 11, 2016

${C}_{M} = 0.301 \frac{m o l}{L}$

#### Explanation:

The molarity (${C}_{M}$) is calculated by:

${C}_{M} = \frac{n}{V}$

where, $n$ is the number of mole of the solute ($N a O H$) and
$V$ is the volume of the solution.

To find the number of mole of $N a O H$ we can use the following relationship:

$n = \frac{m}{M M}$

where, $m$ is the mass of $N a O H$ and $M M$ is the molar mass of $N a O H$: $M M = 40.0 \frac{g}{m o l}$

$\implies n = \frac{m}{M M} = \frac{29.5 \cancel{g}}{40.0 \frac{\cancel{g}}{m o l}} = 0.738 m o l$

The molarity is then can be found as:

${C}_{M} = \frac{n}{V} = \frac{0.738 m o l}{2.45 L} = 0.301 \frac{m o l}{L} \left(\text{corrected to 3 significant figures}\right)$