JEE Main 2006 — Some Basic Concepts Of Chemistry Question with Solution

2.05M solution of acetic acid in water means in 1 litre solution 2.05 moles of CH₃COOH present.

Density of solution = 1.02 g/ml (Given)

Assume the volume of solution = 1 litre = 1000 ml

$$\therefore$$ Mass of solution = 1000 $$\times$$ 1.02 = 1020 gm

Molar mass of CH₃COOH = 60

So Mass of CH₃COOH (m_solute) = 2.05 $$\times$$ 60 = 123

$$\therefore$$ Mass of solvent (m_solvent) = 1020 - 123 = 897 gm = 0.897 kg

Formula of molality (m) = $$\frac{noofmolesofsolute}{weightofsolventinkg}$$

$$\therefore$$ m = $$\frac{2.05}{0.897}$$ = 2.28

Using Formula :

Molality (m) = $$\frac{1000\times M}{1000\times d-M\times M_{solute}}$$

Here M = molarity, M_solute = molecular mass of solute, d = density of solution

$$\therefore$$ m = $$\frac{1000\times 2.05}{1000\times 1.02-2.05\times 60}$$ = 2.28