If equal weights of NaCl and KCl are dissolve | vedclass

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(C) Molarity $(M)$ is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V} = \frac{w}{MW 	imes V}$.Since the weight $(w)$

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Molarity of $NaCl$ is greater than $KCl$

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(C) Molarity $(M)$ is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V} = \frac{w}{MW 	imes V}$.Since the weight $(w)$ and volume $(V)$ are the same for both,$M \propto \frac{1}{MW}$.The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \ g/mol$.The molar mass of $KCl$ is $39 + 35.5 = 74.5 \ g/mol$.Since the molar mass of $NaCl$ $(58.5 \ g/mol)$ is less than the molar mass of $KCl$ $(74.5 \ g/mol)$,the molarity of the $NaCl$ solution will be greater than that of the $KCl$ solution.

If equal weights of $NaCl$ and $KCl$ are dissolved in the same volume of solution,how will the molarity of the two solutions compare?

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