Calculate a) molality b) molarity c) mole fraction of KI (potassium iodide) if the density of 20% (mass/mass) aqueous KI (potassium iodide) is 1.202g/³
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Answered by
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)
Answered by
0
Answer:
1.506,1.44,.0263
Explanation:
20%mass/mass mean 20g solute in 100g solution.
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