JEE Advance - Chemistry (1983 - No. 4)
(i) the percentage by weight of sodium thiosulphate,
(ii) the mole fraction of sodium thiosulphate and
(iii) the molalities of Na+ and $$S_2O_3^{-2}$$ ions
Paaiškinimas
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 $${S_2}O_3^{ - 2} = 3.865$$