Question

Question #46792

Answer 1

You can go from `"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

`"%m/v" = "grams of solute"/"100 mL solution" * 100`

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

`5 = m_(NaOH)/"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 = n/V = "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.