Question

Equal weight of $NaCl$ and $KCl$ are dissolved separately in equal volumes of solutions, the molarity of two solutions will be:
A. Equal
B. That of $NaCl$ will be less than that of $KCl$
C. That of $NaCl$ will be more than that of $KCl$ solution
D. That of $NaCl$ will be half of that of $KCl$ solution

Answer 1

The amount of $NaCl$ and $KCl$ dissolved in a solution is equal. But they both have different molar mass. We will find the number of moles of each solution separately. Then we will find the molarity of each solution by assuming a fixed volume of solution. Thus we get the molarity of each solution respectively.
Formula Used:
${\text{Molarity = }}\dfrac{{{\text{moles of solute}}}}{{{\text{volume of solution}}}}$

Complete answer:
Since the weight of $NaCl$ and $KCl$ dissolved in solution are equal, therefore we cannot find the molarity of these solutions separately. We will assume this equal weight is equal to the molecular mass of the solute $NaCl$. The molecular mass of $NaCl$ can be found as,
Molecular mass of $NaCl{\text{ = 23 + 35}}{\text{.5 g}}$
Molecular mass of $NaCl{\text{ = 58}}{\text{.5 g}}$ which is now the equal weight of solute. Therefore we can say that for each solute the given mass is ${\text{ 58}}{\text{.5 g}}$. Now we will calculate molarity for each solute for one litre of solution.
Molarity of $NaCl$
Given mass ${\text{ = 58}}{\text{.5 g}}$
Molar mass ${\text{ = 58}}{\text{.5 g}}$
Number of moles ${\text{ = }}\dfrac{{{\text{58}}{\text{.5 g}}}}{{58.5{\text{ g}}}}{\text{ = 1}}$
${\text{Molarity = }}\dfrac{{{\text{moles of solute}}}}{{{\text{volume of solution}}}}$
${\text{Molarity = }}\dfrac{1}{1}{\text{ = 1 M}}$
Molarity of $KCl$
Given mass ${\text{ = 58}}{\text{.5 g}}$
Molar mass ${\text{ = 39 + 35}}{\text{.5 = 74}}{\text{.5 g}}$
Number of moles ${\text{ = }}\dfrac{{{\text{58}}{\text{.5 g}}}}{{74.5{\text{ g}}}}{\text{ = 0}}{\text{.78}}$
${\text{Molarity = }}\dfrac{{{\text{moles of solute}}}}{{{\text{volume of solution}}}}$
${\text{Molarity = }}\dfrac{{0.78}}{1}{\text{ = 0}}{\text{.78 M}}$
Thus we can observe that for equal weight of mass of both solute the molarity of $KCl$ is less than that of molarity of $NaCl$ , this is because for equal weight of solute and fixed volume of solution the molarity of solution is inversely proportional to molecular mass of solute.
Thus the correct option is C. That of $NaCl$ will be more than that of $KCl$ solution.

Note:
We can also find the relation between the molarity and molecular weight of a solution when weight of solute is equal and volume of solution is constant by using formulae of molarity. The relation will be inverse to each other. The molecular mass of compounds can vary as the atomic mass of all these elements vary. The mass of chlorine also depends on which isobars are taken. The volume of solution is taken in litres.