An aqueous KCl solution of density $1.2 \mathrm{~g} \mathrm{~mL}^{-1}$ 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 KCl = 74.5]

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Some Basic Concepts of Chemistry

An aqueous KCl solution of density

1.2 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: 3

Take the portion of solution that contains

1 kg

of water, which by the given molality holds

3.30 mol

of KCl. The mass of solute is

3.30 \times 74.5 = 245.85 g

, so the solution weighs

1000 + 245.85 = 1245.85 g

and occupies

V = \frac{1245.85}{1.2} = 1038.2 mL = 1.0382 L

Molarity = \frac{3.30}{1.0382} = 3.18 mol L^{- 1}

To the nearest integer the molarity of the solution is

3 mol L^{- 1}

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Integer answer