Question

Calculate molality,molarity and mole fraction of KI if the density of 20%(mass) aqueous KI is 1.202 g/mL

Answer 1

(a) 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



= 1.506 m

= 1.51 m (approximately)

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

∴Volume of 100 g solution



= 83.19 mL

= 83.19 × 10−3 L

Therefore, molarity of the solution

= 1.45 M

(c) Moles of KI

Moles of water

Therefore, mole fraction of KI



= 0.0263

Answer 2

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