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Solved Examples on Solutions

Question: 1.The density of a 3M sodium thiosulphate solution is 1.25 gm cm–3. Calculate

i)   the molalities of Na+ and S2O32–  ions

ii)   percentage of weight of solution

iii)  mol-fraction of sodium thiosulphate.                                 

Solution:       

3 M Na2S2O3 (Sodium thiosulphate) solution means

3 moles Na2S2O3 is present in 1 L or, 1000 ml solution

Wt. of solute Na2S2O3 = 3×158

wt. of solution = v × d

= 1000 ml × 1.25 gm/ml

= 1000 × 1.25 gm

Wt. of solvent = (1000 × 1.25 – 3 ×158) gm H2O

Molality = no. of moles of solute per 1000 gm solvent

= 3.865 mol kg–1 solvent

Now, Na2S2O3  2Na+ +S2O3-

a) Hence molality of Na+ = 2 ×3.865 mol kg-1

= 7.73 mol kg–1

Hence molality  = 1× 3.865 mol kg–1

= 3.865 mole/kg

b) % of wt. of solution

1 L i.e. 1000 ml solution containing 3 moles Na2S2O3

1000 × 1.25 gm solution containing 3 ×158 gm Na2S2O3

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Question:2. 8.0575 ×  10–2  kg of glauber’s salt (Na2SO4.10H2O) is dissolved in water to obtain 1 dm3 of a solution of density 1077.2 kgm–3. Calculate the molality, molarity and mole fraction of Na2SO4 in the solution.

Solution:       

Wt. of Glauber’s satt = w1 = 8.0575 × 10–2 kg

= 80.575 gm

density of solution = d = 1077.2 kg m–3


 

Question:3. A liquid mixture of benzene (mole fraction 0.33) and toluene (mole fraction 0.67) is taken. What will be the composition of vapours over the liquid mixture? This vapours is collected, condensed to a liquid and then allowed to evaporate so as to come into equilibrium with its vapour. What is composition of...

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