K-'mystery'

Chemistry Level 2

Find the molality of 4 molar aqueous pure nitric acid solution. Consider that the density of the solution is 1 gm/ml 1\text{ gm/ml} for the sake of this question.


The answer is 5.35.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Apr 27, 2016

There are 4 4 moles of nitric acid in 1000 1000 ml of solution.
= 4 × 63 = 252 4 × 63 = 252 g of nitric acid in 1000 × 1 1000 × 1 g of solution. (because molecular weight of nitric acid is 63 and density of solution is 1 gm/ml)
= 252 252 g of nitric acid in 1000 1000 gm of solution.
So, weight of solvent = (1000 - 252 = 748)g


Molality \text{Molality} = Moles of solute Weight of solvent × 1000 = 4 748 × 1000 = 1000 187 = 5.35 \dfrac{\text{Moles of solute}}{\text{Weight of solvent}} × 1000\\ = \dfrac{4}{748} × 1000\\ = \dfrac{1000}{187}\\ = \boxed{5.35}

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