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.

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.

Solution

(a)

Mass % is given so that take total mass = 100 g

Mass of  KI = 20 % of 100  = 20 g

Mass of water  =  total – mass of KI  = 100 – 20 = 80 g of water

Mass of solvent (water) in kg = 80 /1000 = 0.08 Kg

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

Use the formula

 

Number of moles of KI = 20 / 166  = 0.1205 mol

Use the formula 

 

Molality of KI = 0.1205 / 0.08 = 1.51 m

 (b)

Given that the density of the solution = 1.202 g mL−1

Volume of 100 g solution  = 100 / 1.202 = 83.2 mL

Divide by 1000 to convert in liter we get = 83.2/1000 =0.0832 liter

Use the formula 

Plug the values we get

Molarity of the solution = 0.1205/0.0832 =1.45 M

(c)

Molar mass of water(H₂O) = 1 × 2 + 16  = 18

Number of moles of water = 80/18 = 4.4444

Plug the values we get

Mole fraction of KI  = 0.1205’/(0.1205+ 4.4444) = 0.026