Question:
An aqueous $\mathrm{KCl}$ solution of density $1.20 \mathrm{~g} \mathrm{~mL}^{-1}$ has a molality of $3.30 \mathrm{~mol} \mathrm{~kg}^{-1}$. The molarity of the solution in $\mathrm{mol} \mathrm{L}^{-1}$ is (Nearest integer)
$[$ Molar mass of $\mathrm{KCl}=74.5]$
Solution:
$1000 \mathrm{~kg}$ solvent has $3.3$ moles of $\mathrm{KCl}$
$1000 \mathrm{~kg}$ solvent $\longrightarrow 3.3 \times 74.5 \mathrm{gm} \mathrm{KCl}$
$\longrightarrow \quad 245.85$
Weight of solution $=1245.85 \mathrm{gm}$
Volume of solution $=\frac{1245.85}{1.2} \mathrm{ml}$
So molarity $=\frac{3.3 \times 1.2}{1245.85} \times 1000=3.17$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.