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.
(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 = Moles of KI / Mass of water in kg
= 20/166 / 0.08 m
= 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 = Mass / Density
= 100g / 1.202g mL-1
= 83.19 mL
= 83.19 × 10 - 3 L
Therefore, molarity of the solution = 20/166 mol / 83.19 × 10 - 3 L
= 1.45 M
(c) Moles of KI = 20/166 = 0.12 mol
Moles of water = 80/18 = 4.44 mol
Therefore, mole fraction of KI = Moles of KI / (Moles of KI + Moles of water)
= 0.12 / (0.12+4.44)
= 0.0263
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 5: Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (m....
(iv) 
Comments
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Thanks