JEE MAIN - Chemistry (2021 - 26th August Morning Shift - No. 22)
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
3
Explanation
1000 kg solvent has 3.3 moles of KCl
1000 kg solvent $$\to$$ 3.3 $$\times$$ 74.5 gm KCl $$\to$$ 245.85
Weight of solution = 1245.85 gm
Volume of solution = $${{1245.85} \over {1.2}}$$ ml
So molarity = $${{3.3 \times 1.2} \over {1245.85}} \times 1000 = 3.17$$
1000 kg solvent $$\to$$ 3.3 $$\times$$ 74.5 gm KCl $$\to$$ 245.85
Weight of solution = 1245.85 gm
Volume of solution = $${{1245.85} \over {1.2}}$$ ml
So molarity = $${{3.3 \times 1.2} \over {1245.85}} \times 1000 = 3.17$$
Comments (0)
