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 .

Answer 1

Molar Mass of Kl = 166 g/mol

Density = 1.202 g/ml

Mass/Mass % = 20%

Mass of solution = 100g

Mass of solute = 20g

Mass of solvent = 80g = 0.08 kg

Density = Mass of solution / Volume of solution

1.202 = 100 / V

V = 100/1.202

V = 83.19 ml

V = 0.083 L


Moles of Kl = Given mass/Molar Mass = 20/166 = 0.12
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a) Molality = Moles of Kl /Mass of solvent (Kg)

= 0.12 / 0.08

= 1.5 m
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b) Molarity = Moles of solute / Volume of solution

= 0.12 / 0.083

= 1.44 M
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c) Mole Fraction = Mole of solute / Moles of solute + Moles of solvent

= 0.12 / ( 0.12 + 4.44 )

= 0.12 / 4.56

= 0.026
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Thanks !

Answer 2

Hey !!

20% ( mass/mass ) aqueous KI solution means that

Mass of KI = 20 g

Mass of solution = 100 g

      Mass of solvent ( water ) = 100 - 20 = 80 g

(1) Calculation of molality

         Molar mass of KI = 39 + 127 = 166 g mol⁻¹

            nKI = 20 g/ 166 g mol⁻¹

                     = 0.120

Molality of solution = wKI / wH₂O × 1000

           = 0.120 / 80 × 1000

           = 1.5 m

(2) Calculation of molarity

        Density of solution = 1.202 g mL⁻¹

        Volume of solution = 100 g / 1.202 g mL⁻¹

                                     = 83.2 mL

                                     = 0.0832 L

Molarity of solution = nKI / Volume of the solution in L

                     = 0.120 mol / 0.0832 L

                      = 1.44 M

Alternatively,

              Molarity = mass / mass% × d × 10 / Molar mass of KI

= 20 × 1.202 × 10 / 166

= 240.4 / 166

= 1.45 M

(3) Calculation of mole fraction of KI

          nKI = 0.120 mol

     nH₂O = Mass of water / Molar mass of water

            = 80 g / 18 g mol⁻¹

           = 4.44 mol

xKI = nKI / nH₂O + nKI

             = 0.120 mol / 4.44 mol + 0.12 mol

              = 0.120 mol / 4.560 mol

              = 0.0263

Hope it helps you !!