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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 $\ce{NaOH}$)?

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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.

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  • $\begingroup$ 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? $\endgroup$ – Rabik John Apr 28 '18 at 12:16
  • $\begingroup$ @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. $\endgroup$ – Gaurang Tandon Apr 28 '18 at 12:20
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I'd like to add one more situation to @Gaurang Tandon answer:

If you are dealing with liquid solute and a solvent (also a liquid), then there would be another way of giving percent concentrations: %v/v. This is rare because sometimes, volumes reduce when a solute and a solvent mixed, specifically when both are liquids. Still, it's worth considering:

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).

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