In one molal solution that contains 0.5 mole of a solute, there is:

? Solvent is H? O, which is in excess

So using m ( molality ) = (x? ×1000)/ (x? × (M? )? )

? x? = 0.74 (Mol? = 18 g)

x? = 1 – 0.74 = 0.26 ∴ m = (0.26 × 1000)/ (0.74 × 18) = 19.5

ppm of O? = (wt. of O? ) / (wt. of H? O) × 10?
= (10.3 mg) / (1.03 × 10? mg) × 10?
= 10ppm

Molality = (mole of solute * 1000) / wt of solvent (gm)
100 = (n_solute * 1000) / [ (1 - n_solute) * 18]
(1 - n_solute) / n_solute = 1000 / (100 * 18) = 10/18
18 (1 - n_solute) = 10 n_solute
18 - 18 n_solute = 10 n_solute
18 = 28 n_solute
n_solute = 18 / 28? 0.6428 = 64.28 * 10? ²
Ans = 64 (Rounded off)

Let volume of solution = x ml
So mass of solution = 1.2x
And mass of water = x gm
Mass of solute = 0.2x
Molality = (W_solute * 1000) / (M_solute * W_solvent) = (0.2x * 1000) / (40 * x) = 5 m