The density of a 3 M sodium thiosulphate solution (Na2S2O3) is 1.25 g per mL. Calculate
(i) The percentage by weight of sodium thiosulphate
(ii) The mole fraction of sodium thiosulphate and
(iii) The molalities of Na+ and S2O32– ions.
Text Solution
Verified by ExpertsCHECK THE SOLUTION
Given, molarity of Na 2 S 2 O 3 = 3 mol/L
∴ Moles of Na 2 S 2 O 3 = 3
Thus, weight of Na 2 S 2 O 3 = 3 × 158 = 474 g and volume (V) of solution = 1L = 1000 mL
∴ wt. of solvent = 1250 – 474 = 776 g
(i) % by weight of Na 2 S 2 O 3
= 
= 37.92
(ii) Mole fraction of Na 2 S 2 O 3
Moles of Na 2 S 2 O 3 = 3 and moles of H 2 O 
∴ Mole fraction of Na 2 S 2 O 3 
(iii) Molality of Na + ion
= 7.732
and molality of 
ions
= 3.865
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