K b and K f depend only on the SOLVENT. Below are some common values. Use these
ID: 893760 • Letter: K
Question
Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow.
0.512
-1.86
1.22
-1.99
3.67
2.53
-5.12
2.02
*Please note that Tf as defined above will be a negative number. Therefore, Kf must also be given as a negative number.
You may also see Tf defined as the magnitude of the freezing point depression, a positive number. In this case, Kf will also be a positive number. Regardless of the sign used for Kf, remember that the freezing point of the solution is always depressed.
1) The freezing point of water H2O is 0.00°C at 1 atmosphere.
How many grams of chromium(III) acetate (229.2 g/mol), must be dissolved in 249.0 grams of water to reduce the freezing point by 0.450°C ?
g chromium(III) acetate.
2) The boiling point of water/H2O is 100.00 °C at 1 atmosphere.
If 10.07 grams of iron(II) bromide, (215.7 g/mol), are dissolved in 157.2 grams of water ...
The molality of the solution is m.
The boiling point of the solution is °C.
3) The boiling point of water is 100.0°C at 1 atmosphere. (Kb(water) = 0.512°C/m)
How many grams of aluminum chloride (133.3 g/mol), must be dissolved in 215.0 grams of water to raise the boiling point by 0.400°C ?
g aluminum chloride.
Explanation / Answer
1. Given the depression in freezing point, dTf = 0.450 °C
Kf for water = 1.86 °C / m
Now dTf = Kf x m
=> m = dTf / Kf = 0.450°C / 1.86 °C / m = 0.242 m
Hence 0.242 mole of chromium(III) acetate, Cr(CH3COO)3 is present in 1000 g of solvent(water)
However the mass of water taken = 249.0 g.
Hence the moles of chromium(III) acetate, Cr(CH3COO)3 that would be present in 249.0 g water
= 0.242 mol x(249 g / 1000 g) = 0.0603 mol
Molecular mass of chromium(III) acetate = 229.2 g/mol
Hence the mass of chromium(III) acetate equivaleent to 0.0603 mol chromium(III) acetate
= (0.0603 mol chromium(III) acetate ) x (229.2 g / 1 mol chromium(III) acetate) = 13.8 g (answer)
2. Moles of Iron(iii)bromide, FeBr3 taken = mass / molecular mass = 10.07 g / 215.7 g/mol = 0.04669 mol
Mass of water taken = 157.2 g
Hence molality of the solution can be calculated as
m = moles of solute present in 1000 g of water = 0.04669 mol x (1000 g / 157.2 g) = 0.2970 m (answer)
Kb for water = 0.512 °C / m
Hence Elevation in boiling point, dTb = Kb x m = (0.512 °C / m) x 0.2970 m = 0.1521 °C
Hence boiling point of the solution = (100 + 0.1521) °C = 100.1521 °C (answer)
3.
Given the elevation in boiling point, dTb = 0.400 °C
Kb for water = 0.512 °C / m
Now dTb = Kb x m
=> m = dTb / Kb = 0.400°C / 0.512 °C / m = 0.781 m
Hence 0.781 mole of aluminum chloride, AlCl3 is present in 1000 g of solvent(water)
However the mass of water taken = 215.0 g.
Hence the moles of aluminum chloride, AlCl3 that would be present in 215.0 g water
= 0.781 mol x(215.0 g / 1000 g) = 0.168 mol
Molecular mass of aluminum chloride, AlCl3 = 133.3 g/mol
Hence the mass of aluminum chloride, AlCl3 equivaleent to 0.168 mol aluminum chloride, AlCl3
= (0.168 mol aluminum chloride, AlCl3) x (133.3 g / 1 mol aluminum chloride, AlCl3) = 22.4 g (answer)
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