Calculate the molality of a 1.5 M aqueous salt solution in a solvent X, if the mole fraction of the salt in the solution is 0.25.
(Molar mass of solvent = 40 g/mol)

The correct option is C $8.33m$
1.5 molar solution means 1.5 mol of soluble salt is present in 1 L of the solution.
$Molefractionofsolute=x_{solute}=\frac{molesofsolute}{molesofsolute+molesofsolvent}$
⇒ $0.25=\frac{1.5}{1.5+n_X}$
On solving,
Moles of solvent X $=n_X=4.5mol$
Weight of solvent $=4.5\times 40=180g$
$Molality\left(m\right)=\frac{molesofsolute}{massofsolventing}\times 1000$
$Molality=\frac{1.5}{180}\times 1000=8.33m$