# How to calculate molarity of a solution by solving the percentage?

Molarity is number of moles per litre of solution and we calculate it by simply dividing moles by volume. But, if moles are not given and the w/w or w/v percentage is given then, how can we solve for the molarity?

For example, if we have 2 % w/v solution of $\ce{NaOH}$, how would we find its molarity? Do we solve by dividing 2 by 40 (molar mass of $\ce{NaOH}$)?

Both of them are pretty straightforward. Don't go by the formulae, but instead, internalize the steps required here.

$2\%$ w/v solution has $\pu{2g}$ of solute in $\pu{100 ml}$ solution. That implies $\pu{20g}$ of solute in $\pu{1L}$ solution. $\pu{20g}$ solute is 0.5 moles of $\ce{NaOH}$ (why?). So, molarity is simply moles divided by volume of solution.

The $2\%$ w/w solution case is trickier, and you'll also be given the density of solution in this case. $2\%$ w/w solution has $\pu{2g}$ solute in $\pu{100g}$ solution, or $\pu{20g}$ solute in $\pu{1000g}$ solution. If the density is $\pu{2gm L^-1}$ (assumed), then the volume of solution would be $\ce{500 mL}$. Now, you can proceed by the same logic as above.

In fact, the conversion of $\pu{100ml}$ volume to $\pu{1L}$ in case 1 wasn't even necessary, but I only did that to make the step-by-step calculations more approachable for you.

• Do you mean I have to divide 0.5 by 100ml?If so ,then answer comes 0.005M, and my book says answer is 0.5M?Can you please explain it in a bit more simple words? – Rabik John Apr 28 '18 at 12:16
• @RabikJohn As I said earlier, you either take the combination of (20g and 1 litre) or (2gram and 100ml). I took the former, and computed moles = 20g/40g=0.5 That gave me molarity = 0.5moles/1litre=0.5M. You can take the latter combination and will get the same answer. – Gaurang Tandon Apr 28 '18 at 12:20

For example, if the solution is said to be $10\%$ v/v, then it has $\pu{10 mL}$ of solute in $\pu{100 mL}$ of solution. That implies $\pu{100 mL}$ of solute in $\pu{1000 mL}$ or $\pu{1 L}$ of solution. If the density of solute is $x~ \mathrm{g\cdot mL^{-1}}$, then the mass of $\pu{100 mL}$ of solute is $(100~\mathrm{mL})\cdot (x~ \mathrm{g\cdot mL^{-1}})=100x~\mathrm{g}$. If molar mass of the solute is given (it should be), now you can find (how?) the number of moles of solute in $\pu{1 L}$ of solution, which is the molarity of your solution (moles per $\pu{1 L}$ of solution).