Question

If the sum of the mole fraction of $NaOH$ in its aqueous solution and the mole fraction of $H_{2}O$ in an another aqueous solution of $KOH$ is equal to one, find the molality of $KOH$ solution. The molality of $NaOH$ solution is 'm'.
Report your answer as 'Y' where Y = $\frac{\left(molalityofKOHsolution\right)}{\left(molalityofNaOHsolution\right)}$

Answer 1

Let mole fraction of $NaOH$ in its aqueous solution be X
Therefore, mole fraction of $H_{2}O$ in an aqueous solution of $KOH$ = 1 - X
So,
mole fraction of $KOH$ in its aqueous solution be = 1 - (1 - X) = X
Thus, both aqueous $NaOH$ solution and aqueous $KOH$ solution have the same mole fraction of solute.
Now,
$\text{molality}=\frac{\text{moles of solute}\times 1000}{\text{mass}{}_{\text{solvent}}}$

$\text{molality}=\frac{X{}_{\text{solute}}\times 1000}{(1-X_{}\text{solute})\times \text{Molar mass}{}_{\text{solvent}}}$

Since both solutions have the same solvent ($H_{2}O$) and same mole fraction of solute, so they will have same molality.
Therefore,
Molality of aqueous $KOH$ solution = m
$Y=\frac{m}{m}=1$