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Question
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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Solution
Given: Let mass of solution in water = 100 g, then mass of KI = 20 g
∴ Mass of solvent (water) = 100 − 20 = 80 g = 0.080 kg
(a) Molar mass of KI = 39 + 127 = 166 g mol−1
∴ Moles of KI = `(20 "g")/(166 "g mol"^(-1))`
= 0.120 mol
Molality of solution = `"Number of moles of KI"/"Mass of solvent in kg"`
= `(0.120 "mol")/(0.080 "kg")`
= 1.5 mol kg−1
(b) Density of the solution = 1.202 g mL−1
∴ Volume of solution = `"Mass"/"Density"`
= `(100 "g")/(1.202 "g mL"^(-1))`
= 83.2 mL
= 0.0832 L
Molarity of the solution = `"Number of moles of solute"/"Volume of solution in L"`
= `(0.120 "mol")/(0.0832 "L")`
= 1.44 M
(c) No. of moles of KI = 0.120 mol
No. of moles of water = `"Mass of water"/"Molar mass of water"`
= `(80 "g")/(18 "g mol"^(-1))`
= 4.44 mol
Mole fraction of KI = `"Number of moles of KI"/"Total number of moles in solution"`
= `0.12/(0.120 + 4.44)`
= `0.120/4.560`
= 0.0263
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