Question

The density of NaOH solution is 1.2 g cm^–3 . The molality of this solution is ...... m. 

(Round off to the Nearest Integer) 

[Use : Atomic masses : Na : 23.0 u O : 16.0 u H : 1.0 u 

Density of H₂O : 1.0 g cm^–3]

Answer 1

Let volume of solution is x ml

So mass of solution = 1.2x

& mass of water (solvent) = x gram

So mass of solute = 0.2x gram

Molality = \(\frac{W_{solute} \times 1000}{GMM_{solute} \times W_{solvent}}\)

\(= \frac{0.2x \times 1000}{40\times x}\)

\(= \frac{200}{40}\)

\(= 5\, m\)

Answer 2

Answer is : 5

Consider 1l solution 

mass of solution = (1.2 × 1000)g = 1200 gm 

Neglecting volume of NaOH 

Mass of water = 1000 gm

Mass of NaOH = (1200 – 1000)gm 

= 200 gm

Moles of NaOH = \(\frac{200g}{40g/mol}\) = 5 mol

Molality = \(\frac{5\,mol}{1\,kg}\) = 5 m