3 M Na₂S₂O₃ solution means 3 moles of Na₂S₂O₃ present in 1 litre of solution.

$$\therefore$$ Volume of solution = 1 litre = 1000 ml

Density of solution = 1.25 g/ml

$$\therefore$$ Mass of solution = 1000 $$\times$$ 1.25 = 1250 gm

Molar mass of Na₂S₂O₃ = 158 gm

$$\therefore$$ Weight of solute Na₂S₂O₃ = 3 $$\times$$ 158 = 474 gm

$$\therefore$$ Weight of water = 1250 $$-$$ 474 = 776 gm

(i) % weight of $$N{a_2}{S_2}{O_3} = {{Weight\,of\,N{a_2}{S_2}{O_3}} \over {Weight\,of\,solution}} \times 100$$

$$ = {{474} \over {1250}} \times 100$$

$$ = 37.92$$

(ii) Mole fraction of $$N{a_2}{S_2}{O_3} = {{Moles\,of\,N{a_2}{S_2}{O_3}} \over {Moles\,of\,N{a_2}{S_2}{O_3} + Moles\,of\,{H_2}O}}$$

$$ = {3 \over {3 + {{776} \over {18}}}}$$

$$ = 0.065$$

(iii) Molality of solution (m) $$ = {{Moles\,of\,solute} \over {Weight\,of\,solvent\,in\,kg}}$$

$$\therefore$$ $$m = {3 \over {{{776} \over {1000}}}}$$

$$ = 3.865$$

$$N{a_2}{S_2}{O_3}\buildrel {} \over \longrightarrow 2N{a^ + } + {S_2}O_3^{ - 2}$$

$$\therefore$$ Molality of $$N{a^ + } = 2 \times 3.865 = 7.732$$

and Molality of $${S_2}O_3^{ - 2} = 3.865$$