The mole fraction of a solute in a 100 molal aqueous solution is _______ ×10^-2. (Round off to the nearest integer).
[Given : Atomic masses : H : 1.0u, O : 16.0u]

? 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 = moles of solute / mass of solvent (kgs) ⇒ 1 = 0.5 / W (kgs)
W_solvent (kgs) = 0.5 = 500 g

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