The density of NaOH solution is 1.2 gm cm⁻³. The molality of this solution is _______ m. (Round off to the Nearest integer) [Use: Atomic masses: Na 23.0u O: 16.0 u H:1.0 u Density of H₂O: 1.0g cm⁻³]

? 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

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)