1. Calculate the freezing and boiling points of a solution prepared by dissolving 15.5g of AI(NO3)3 in 200.0g of water. (Molar mass of AI(NO3)3 is 212.996 g/mol).
2. A solution is prepared by dissolving 120 grams of NaCI in 450 grams of water. Find the freezing and boiling point of this solution. (Molar mass of NaCI is 58.44 g/mol).
1. Molality = Mole/kg = Mass/(molar mass × mass in kg)
= 15.5/(212.996 ×0.2) = 0.36
Tsoln = TWater + k.m.i
= 100 + 0.51(0.36)(4)
= 100.73°C (boiling point)
Freezing point
Tsoln = Tice + k.m.i
= 0 +(-1.86)(0.36)(4)
= -2.68°C
2. Molality = Mole/kg = Mass of solute /(molar mass × mass of solvent in kg)
120/(58.44×0.45) = 4.56
Tsoln = TWater + k.m.i
= 100°C + 0.51(4.56)(2)
= 104.65°C (boiling point)
Freezing point
Tsoln = Tice + k.m.i
= 0° + (-1.86)(4.56)(2)
= -16.96°C
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