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Calculate (a) molality(b) molarity and (c ) mole fraction of KI if the density of 20 % (mass / mass) aqueous KI is 1.202 g mL^(-1). |
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Answer» Solution :(a)Molar mass of KI `= 39+127=16 g mol^(-1)` 20% (mass/mass) aqueous solution of KI means 20 kg of KI is present in 100 g of solution. That is, 20 g of KI is present in `(100 - 20)` g of water = 80 g of water THEREFORE, molality of the solution `= ("Moles of KI")/("Mass of water in kg")=((20)/(166))/(0.08)=1.506 m` = 1.51 m (approximately). (b)It is GIVEN that the density of hte solution `= 1.202 g mL^(-1)` `therefore` Volume of 100 g solution `= ("Mass")/("Density")=(100 g)/(1.202 g mL^(-1))` = 83.19 mL `= 83.19xx10^(-3)L` Therefore, molarity of the solution `= ((20)/(160))/(83.19xx10^(-3)L)=1.45M` (c )Moles of KI `= (20)/(166)=0.12 mol` Moles of water `= (80)/(18)=4.44 mol` Mole fraction of KI `= (0.12)/(0.12+4.44)=0.0263`. |
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