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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Let the mass of solution in water = 100 g The mass of KI = 20 g ` therefore ` Mass of solvent (water) = 100 - 20 = 80 g = 0.080 kg (a) Calculation of molality Molar mass of KI = 39 + 127 = 166 g `"mol"^(-1)` Number of moles of KI `= 20/166 = 0.120` Molality of solution = Number of moles of KI / Mass of solvent in kg ` = (1.120)/(0.080) = 1.5 m` (b) Calculation of molarity Density of solution = 1.202 g` mL^(-1)` Volume of solution ` = (100g)/(1.202 g mL^(-1) ) = 83.2 mL = 0.0832 L` (volume = mass/density) Molarity of solution = Moles of solute / olume of solution in L ` = (0.120)/(0.0832) = 1.44 M` (c) Calculation of mole fraction of KI Number of moles of KI = 0.120 Number of moles of water = Mass of water / Molar mass of water ` = 80/18 = 4.44` Mole fraction of KI = `(0.120)/(0.120 + 4.44) = (0.120)/(4.560) = 0.0263`
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