An aqueous solution of sodium chloride is prepared by dissolving 10kg of sodium chloride in 50 kg of water find (i) Weight % (ii) Mole % of solution. [Atomic weight of Na = 23, Cl = 35.5] 

1 Answer

Answer :

Weight of NaCl = 10 kg

Weight of H2O = 50 kg

Total weight = 60 kg

Weight % of NaCl = (10/ 60) * 100 =16.67%

Weight % of H2O = (50/ 60) * 100 = 83.33%

Molecular weight of NaCl = 58.5

k moles of NaCl = 10/58.5 = 0.171

Molecular weight of H2O = 18

k moles of H2O = 50/18 = 2.78

Total moles = 0.171+2.78 = 2.949

Mol % of NaCl = (moles of NaCl / Total moles)*100

 = (0.171/ 2.949)*100 = 5.79%

Mol % of H2O = (moles of H2O / Total moles)*100

 = (2.78/ 2.949)*100 = 94.26%

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