K b and K f depend only on the SOLVENT. Below are some common values. Use these
ID: 893747 • 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 T as defined above will be a negative number for freezing point depression. Therefore, Kf must also be given as a negative number.
1) The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT .
How many grams of DDT, C14H9Cl5 (354.5 g/mol), must be dissolved in 249.0 grams of benzene to reduce the freezing point by 0.450°C ?
g DDT.
2)The boiling point of ethanol CH3CH2OH is 78.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in ethanol is estrogen (estradiol).
If 10.07 grams of estrogen, C18H24O2 (272.4 g/mol), are dissolved in 157.2 grams of ethanol ...
The molality of the solution is m.
The boiling point of the solution is °C.
3)The boiling point of chloroform CHCl3 is 61.70°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in chloroform is estrogen (estradiol).
How many grams of estrogen, C18H24O2 (272.4 g/mol), must be dissolved in 215.0 grams of chloroform to raise the boiling point by 0.400 °C ?
g estrogen.
Thank you.
T = T(solution) - T(pure solvent) * m = (# moles solute / Kg solvent) Kb = boiling point elevation constant. Kf = freezing point depression constant.Explanation / Answer
1) -deltaTf = Kf x m
where,
Kf = -5.12 oC/m
delta Tf = depression in freexing point = -0.450 oC
m = molality of solution = moles/kg of solvent = g/molar mass of solute x kg of solvent
Feed values,
-0.45 = -5.12 x m
m = 0.088 m
0.088 = g/354.5 x 0.249
grams of DDT required to lower the freezing point of benzene = 7.77 g
2) Using the formula,
T[solution] - T[pure] = m x Kb
where,
T[solution] = unknown
T[pure] = 78.50 oC
Kb = 1.22 oC/m
m = molality = 10.07 g/272.4 g/mol x 0.1572 kg = 0.235 m
Feed values,
T[solution] = 0.235 x 1.22 + 78.50 = 78.79 oC
3) delta Tb = Kb x m
where,
Kb = 3.67 oC/m
delta Tb = 0.4 oC
m = g of estrogen/272.4 x 0.215
feed values,
m = 0.4/3.67 = 0.109 = g/272.4 x 0.215
grams of estrogen = 6.38 g
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