# 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). As density and %

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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).
As density and % by mass is given, so find the mass of solute and solvent (as x % solution contains x g solute in (100 - x) g solvent).
Find volume of the solution, by using,,
Volume =("Mass")/("Density")
Recall the formulae of molality, molarity and mole fraction, to calculate them.
Molality =("Mass of solute/ molar mass of solute")/("Mass of solventin kg")

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a) Molality
Weight of KI in 100 g of water = 20 g
Weight of water in the solution =100-20=80" g"=0.08" kg"
Molar mass of KI =39+127=166" g mol"^(-1)
Molality of the solution (m) =("Number of moles of KI")/("Mass of water in kg")
=((20"g")//(166" g mol"^(-1)))/((0.08" kg"))=1.506" mol kg"^(-1)=1.506" m"
b) Molarity ltbbrgt Weight of the solution = 100 g
Denisty of the solution =1.202" mL"^(-1)
Volume of the solution =("Weight of solution")/("Density")=((100" g"))/((1.202" g mL"^(-1)))
=83.16" mL"=0.083" L"
Molarity of the solution (M) =("Number of gram moles of KI")/("Volume of solution in litres")
=((20"g")//(166" g mol"^(-1)))/((0.083" L"))=1.45" mol L-1"=1.45" M"
c) Mole fraction of KI
(Number of moles of KI)n_("KI")=("Mass of KI")/("Molar mass of KI")
=((20" g"))/((166" g mol"^(-1)))=0.12" mol"
(Number of moles of water) n_("H"_(2)"O")=("Mass of water")/("Molar mass of water")
=((80" g"))/((18" g mol"^(-1)))=4.44" mol"
(Mole fraction of KI) x_("KI")=(n_("KI"))/(n_("KI")+n_("H"_(2)"O"))=((0.12" mol"))/((0.12+4.44)" mol")
=(0.12)/(4.56)=0.0263