The density of a 3 M Na S203 (sodium thiosulphate) solution is 1.25 g cm . Calculate (1) the percentage by
weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate (iii) the molality of Na and
,0z- ions.
Answers
Answer:
(i) Mass of 1000 mL of Na
2
S
2
O
3
solution = 1.25×1000=1250g
Mass of Na
2
S
2
O
3
in 1000 mL of 3 M solution
=3×MolmassofNa
2
S
2
O
3
=3×158=474g
Mass percentage of Na
2
S
2
O
3
in solution
=
1250
474
×100=37.92
Alternatively, M=
m
A
x×d×10
3=
158
x×1.25×10
x=37.92
(ii) No. of moles of Na
2
S
2
O
3
=
158
474
=3
Mass of water =(1250−474)=776g
No. of moles of water =
18
776
=43.1
Mole fraction of Na
2
S
2
O
3
=
43.1+3
3
=
46.1
3
=0.065
(iii) No. of moles of Na
+
ions
=2×No.ofmolesofNa
2
S
2
O
3
=2×3=6
Molality of Na
+
ions =
Massofwaterinkg
No.ofmolesofNa
+
ions
=
776
6
×1000
=7.73m
No. of moles of S
2
O
3
2−
ions = No. of mole of Na
2
S
2
O
3
=3
Molality of S
2
O
3
2−
ions =
776
3
×1000=3.86mNa2S2O3
Explanation: