Question

An aqueous $KCl$ solution of density $1.20 \,g \,mL ^{-1}$ has a molality of $3.30\, mol\, kg ^{-1}$. The molarity of the solution in $mol\, L ^{-1}$ is _____ (Nearest integer)
$[$ Molar mass of $KCl =74.5]$

Answer 1

$1000\, kg$ solvent has $3.3$ moles of $KCl$
$1000 \,kg $ solvent $ \longrightarrow 3.3 \times 74.5 \, gm\, KCl$
$\longrightarrow 245.85$
Weight of solution $=1245.85\, gm$
Volume of solution $=\frac{1245.85}{1.2} ml$
So molarity $=\frac{3.3 \times 1.2}{1245.85} \times 1000=3.17$