# Calculate the (a) molality, (b) molartiy, and (c) mole fraction of KI if the density of 20% ( mass // mass ) aqueous KI is 1.202 g m

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Calculate the (a) molality, (b) molartiy, and (c) mole fraction of KI if the density of 20% ( mass // mass ) aqueous KI is 1.202 g m L^(-1).

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Calculatetion of molality of solution
Weight of water in the solution= 20 g
Weight of water in the solution=100-20=80g=0.08 kg
Molar mass of KL=39+127=166g mol^(-1).
"Molality of solution (m)"=("No of gram moles of KL")/("Mass of water in kg")=((20g)//(166g mol^(-1)))/((0.08 kg))
Calucalation of molarity of solution
"Weight of solution"=100g , "Density of solution" = 1.202g mL^(-1).
"Volume of solution"=("Weight of solution")/("Density")=((100g))/((1.202g mL^(-1)))=83.19 mL=0.083L
"Molarity of solution (m)"=("No. of gram moles of KL")/("Voluume of solution in litres")=((20g)//(166g mol^(-1)))/((0.083L))
Calculation of mole fraction of KL
n_(KL)=("Mass of KL")/("Molar mass of KL")=((20g))/((166g mol^(-1)))=0.12 mol
n_(H_(2)O)=("Mass of water")/("Molar mass of water")=((80g))/((18g mol^(-1)))=4.44 mol
x_(KL)=n_(KL)/(N_(KL)+n_(H_2O))=((0.12 mol))/((0.12+4.44)mol)=(0.12)/(4.56)=0.0263.`

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