Calculate the molality of a $2M$ aqueous salt solution in a solvent X, if the mole fraction of the salt in the solution is $0.5$.
(Molar mass of solvent = 40 g/mol)

The correct option is D $25m$
$2$ molar solution means $2mol$ of soluble salt is present in $1L$ of the solution.
$\text{Mole fraction of solute}=x_{solute}=\frac{\text{moles of solute}}{\text{moles of solute + moles of solvent}}$
⇒ $0.5=\frac{2}{2+n_X}$
On solving,
Moles of solvent X $=n_X=2mol$
$\text{Weight of solvent}=2\times 40=80g$
$\text{Molality (m)}=\frac{\text{moles of solute}}{\text{mass of solvent in g}}\times 1000$
$Molality=\frac{2}{80}\times 1000=25m$