Question

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)

• A. $18.52m$
• B. $12.85m$
• C. $8.33m$
• D. $15.24m$

Answer 1

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 $=4.5\times 40=180g$
$Molality\left(m\right)=\frac{molesofsolute}{massofsolventing}\times 1000$
$Molality=\frac{1.5}{180}\times 1000=8.33m$