# Question 46792

May 17, 2015

You can go from $\text{m/v%}$ to molarity by using the number of moles of solute, in your case sodium hydroxide.

A mass by volume percent concentration solution is defined as grams of solute per 100 mL of solution, and multiplied by 100

$\text{%m/v" = "grams of solute"/"100 mL solution} \cdot 100$

If you assume a volume of 100 mL of solution, which would make the calculations easier, you'd get

$5 = {m}_{N a O H} / \text{100 mL" * 100 => m_(NaOH) = (5 * 100)/100 = "5 g NaOH}$

This means that each 100 mL of solution contains 5 grams of sodium hydroxide. Use sodium hydroxide's molar mass to see how many moles you'd get

5cancel("g") * "1 mole NaOH"/(40.0cancel("g")) = "0.125 moles NaOH"#

This would make the solution's molarity be

$C = \frac{n}{V} = \text{0.125 moles"/(100 * 10^(-3)"L") = "1.25 M}$

SIDE NOTE Don't forget that molarity is defined as moles of solute per liters of solution, that's why I converted 100 mL to L in the above equation.