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The density of a 3 M sodium thiosulphate solution (Na 2 S 2 O 3 ) is 1.25 g per ml. Calculated (i) the percentage by weight of sodium thiosulphate, (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of Na + and S 2 O 3 2- ions.

The density of a 3 M sodium thiosulphate solution (Na2S2O3) is 1.25 g per ml. Calculated (i) the percentage by weight of sodium thiosulphate, (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of Na+ and S2O32- ions.

Grade:10

4 Answers

Jitender Pal
askIITians Faculty 365 Points
9 years ago
(i) Mole fraction = Moles of substance/Total moles
(ii) 1 mole of Na2S2O3 gives 2 moles of Na+ and 1 mole S2O32-
Molecular wt. of sodium thiosulphate solution (Na2S2O3) = 23 * 2 + 32 * 2 + 16 * 3 = 158
(i) The percentage by weight of Na2S2O3
= wt of Na2S2O3/wt of solution * 100 = 3 * 158 * 100/100 * 1.25 = 37.92
[wt. of Na2S2O3 = Molarity * Molwt]
(ii) Mass of 1 litre solution = 1.25 * 1000 g = 1250 g
[∵ density = 1.25g/l]
Mole fraction of Na2S2O3
= Number of moles of Na2S2O3/Total number of moles
Moles of water = 1250 – 158 *3/18 = 43.1
Mole fraction of Na2S2O3 = 3/3 + 43.1 = 0.065
(iii) 1 mole of sodium thiosulphate (Na2S2O3) yields 2 moles
of Na+ and 1 mole of S2O2-3
Molality of Na2S2O3 = 3 * 1000/776 = 3.87
Molality of Na+ = 3.87 * 2 = 7.74m
Molality of S2O2-3 = 3.87m
R Parker
40 Points
6 years ago
Calculate the mass of given solution using Density = mass /volumeMass of 1000 ml of 3M solution will be given by density x volume = 1.25 g ml-1 x 1000 ml = 1250 g​Mass of Na2S2O3 in 1000 ml solution = molar mass of Na2S2O3 x number of moles of solute = 158 g/mol x 3 mol​ = 474 g​Mass of the solvent = mass of solution - mass of Na2S2O3 = 1250 g - 474 g =776 g​Moles of Na2S2O3 = 3 mol ( 3 M) givenMoles of solvent water = mass of water/molar mass of water = 776 g/18 gmol-1​ = 43.1 molTotal number of moles in solution = 3 + 43.1 = 46.1 mol​Mole fraction of Na2S2O3 = moles of Na2S2O3 /total moles = 3/46.1 = 0.065 Molality of solution = moles of solute/solvent in kg = 3 mol / 0.776 kg = 3.865 mol·kg-11 mole of Na2S2O3 gives 2 moles of Na+ So, molality of Na+ is 2 x 3.865 = 7.73 mol.kg-1​​1mole of Na2S2O3 gives 1 moles of S2O32- So, molality of S2O32- is 3.865mol.kg-1
A Sinthamani
15 Points
4 years ago
3 molal solution of sodium thiosulphate means that there are 3 moles of it per 1000g of solvent (H₂O).
Hence, the number of moles of water would be 1000/18 = 55.55.
Mole fraction of sodium thiosulphate is the number of moles of it over total number of moles in the solution, which would be 3/(3+55.55) = 3/58.5 = 0.0513.
Ajeet Tiwari
askIITians Faculty 86 Points
3 years ago
hello students


Molecular wt. of sodium thiosulphate solution (Na2S2O3) = 23 * 2 + 32 * 2 + 16 * 3 = 158
(i) The percentage by weight of Na2S2O3
= wt of Na2S2O3/wt of solution * 100 = 3 * 158 * 100/100 * 1.25 =37.92
[wt. of Na2S2O3= Molarity * Molwt]
(ii) Mass of 1 litre solution = 1.25 * 1000 g = 1250 g
[∵ density = 1.25g/l]
Mole fraction of Na2S2O3
= Number of moles of Na2S2O3/Total number of moles
Moles of water = 1250 – 158 *3/18 = 43.1
Mole fraction of Na2S2O3= 3/3 + 43.1 = 0.065
(iii) 1 mole of sodium thiosulphate (Na2S2O3) yields 2 moles
of Na+and 1 mole of S2O2-3
Molality of Na2S2O3= 3 * 1000/776 = 3.87
Molality of Na+= 3.87 * 2 =7.74m
Molality of S2O2-3=3.87m


hope it helps
thankyou and regards

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