part A The normal boiling point of water, H2O is 100.00 °C and its Kbp value is
ID: 985632 • Letter: P
Question
part A The normal boiling point of water, H2O is 100.00 °C and its Kbp value is 0.512 °C/m. Assuming complete dissociation of the electrolyte, if 11.22 grams of sodium nitrate (NaNO3, 85.00 g/mol) are dissolved in 236.2 grams of water, what is the boiling point of the solution?
part b The normal freezing point of water, H2O is 0.00 °C and its Kfp value is 1.86 °C/m.
Assuming complete dissociation of the electrolyte, if 10.25 grams of potassium iodide (KI, 166.0 g/mol) are dissolved in 178.3 grams of water what is the freezing point of the solution?
Explanation / Answer
Answer – Part A) We are given, normal boiling point of water, H2O = 100.00 °C
Kb = 0.512 °C/m, mass of NaNO3 = 11.22 g , mass of water = 236.2 g
Molar mass of NaNO3 = 85.00 g/mol
First we need to calculate the moles of NaNO3
Moles of NaNO3 = 11.22 g / 85.0 g.mol-1
= 0.132 moles
Molality of NaNO3 = 0.132 moles / 0.2362 kg
= 0.559 m
The Van’t off factor i for NaNO3 = 2
We know, Tb = i*Kb*m
= 2*0.512 °C/m*0.559 m
= 0.572oC
boiling point of the solution = pure solvent boiling point +Tb
= 100 + 0.572
= 100.6oC
Part B) Given, mass of KI = 10.25 g, mass of water = 178.3 g
Kf = 1.86 °C/m, Molar mass of KI = 166.00 g/mol
First we need to calculate the moles of KI
Moles of KI = 10.25 g / 166.0 g.mol-1
= 0.0617 moles
Molality of KI = 0.0617 moles / 0.1783 kg
= 0.346 m
Now we know,
Tf = i*Kf*m
Tf = 2*1.86oC/m * 0.346 m
= 1.29 oC
So, Freezing point of solution = Pure solvent freezing point - Tf
= 0.00oC – 1.291oC
= -1.29 oC
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