# 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.

By Shivank Goel|Updated : August 10th, 2022

(a) We know that

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

20% (mass/mass) aqueous KI refers to 20g of Kl is present in 100g of solution

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

Molality of solution = moles of Kl/ mass of water in Kg

Substituting the values

= 20/166/ 0.08

= 1.506 m

= 1.51 m (approx)

(b) Given

The density of solution = 1.202 g mL⁻1

Volume of 100 g solution = mass/density

Substituting the values

= 100/1.202

= 83.19 ml

= 83.19 x 10⁻3 L

Molarity of solution = 20/166 mole/ 83.19 x 10⁻3 L

= 1.45 M

(c) Moles of Kl = 20/166 = 0.12 mol

Moles of water = 80/18 = 4.44 mol

Mole fraction of Kl = moles of Kl/ moles of Kl + moles of water

Substituting the values

= 0.12/ (0.12 + 4.44)

= 0.0263

Therefore, the molality is 1.51 m, molarity is 1.45 M, and mole fraction of KI = 0.0263.

Summary:

## (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL⁻1.

(a) molality is 1.51 m

(b) molarity is 1.45 M and

(c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL⁻1 is 0.0263.

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