An aqueous $KCl$ solution of density $1.20$ ${gmL}^{-1}$ has a molality of $3.30 mol{ kg}^{-1}$ the molarity of the solution in $mol{L}^{-1}$ is _____.

(Nearest integer) [Molar mass of $KCl=74.5$ ]

$1000 kg$ solvent has $3.3$ moles of $KCl$

$1000 kg$ solvent $\to 3.3\times 74.5gmKCl$

$\to 245.85$

Weight of solution $=1245.85gm$

Volume of solution $=\frac{1245.85}{1.2}=1.0382 L$

Molarity=moles of $KCl/$ Vol. of Sol. in it $=3.3/1.0382=3.17 mol/L$

Hence the correct answer is 3.17.