Question:
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.

Answer:

(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

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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Question:
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.

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Question: 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.

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Question:
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.