calculate (a) molality (b) molarilty & (c) mole fraction of KI if the density of 20% (mass/mass)aqueous KI solution is 1.202 g/ml

Asked by Aditya | 28th Dec, 2012, 05:27: PM

Expert Answer:

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

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 - 20) g of water = 80 g of water

Therefore, molality of the solution

 = moles of KI/mass of water in Kg  = (20/166)/ 0.08

= 1.506 m

= 1.51 m (approximately)

(b) It is given that the density of the solution = 1.202 g mL?1

mass/ density =Volume of 100 g solution

 100g/ 1.202 gml-1 

= 83.19 mL

= 83.19 × 10?3 L

Therefore, molarity of the solution = (20/166) / 83.19 × 10?3 L

= 1.45 M

(c) Moles of KI

Moles of water = 80/18 = 4.44

Therefore, mole fraction of KI = moles of KI/ moles of KI + moles of water = 0.12/0.12 + 4.44

= 0.0263

 

= 0.0263

Answered by  | 28th Dec, 2012, 06:36: PM

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