Raoult freezing point depressionSolutions

1) The molality is moles of solute per kilogram of solvent. Each solution has 1.00 g solute per 100. g (0.100 kg) solvent. Find the number of moles in 1.00 g solute by dividing by the molar mass; for example, for methyl iodide

moles of solute = = 7.04x10-3 mol.

So the molality is that number of moles divided by the mass of benzene in kg (namely 0.100 kg); for methyl iodide, the molality is 7.04x10-2 mol/kg.

Now solve the freezing-point depression equation for Kf:

Kf = ∆T/m.

So for methyl iodide

Kf = 0.335°C/(7.04x10-2 mol/kg) = 4.76°C kg/mol.

The table under exercise 2 contains molalities and freezing point depression constants for each substance.

2) Freezing point depression constants computed for each solute are tabulated below.

Solute / molecular weight / ∆T(°C) / molality(mol/kg) / Kf=∆T/m
methyl iodide / 142 / 0.335 / 0.070 / 4.76
chloroform / 119.5 / 0.428 / 0.084 / 5.11
carbon tetrachloride / 154 / 0.333 / 0.065 / 5.13
carbon disulfide / 76 / 0.654 / 0.132 / 4.97
ethyl iodide / 156 / 0.331 / 0.064 / 5.16
ethyl bromide / 109 / 0.461 / 0.092 / 5.02
hexane / 86 / 0.597 / 0.116 / 5.13
ethylene chloride / 99 / 0.491 / 0.101 / 4.86
turpentine (-pinene) / 136 / 0.366 / 0.074 / 4.98
nibrobenzene / 123 / 0.39 / 0.081 / 4.80
naphthalene / 128 / 0.391 / 0.078 / 5.00
anthracene / 178 / 0.287 / 0.056 / 5.11
methyl nitrate / 77 / 0.64 / 0.130 / 4.93
dimethyl oxalate / 118 / 0.417 / 0.085 / 4.92
methyl salicylate / 152 / 0.339 / 0.066 / 5.15
diethyl ether / 74 / 0.671 / 0.135 / 4.97
diethyl sulfide / 90 / 0.576 / 0.111 / 5.18
ethyl nitrile / 55 / 0.938 / 0.182 / 5.16
ethyl formate / 74 / 0.666 / 0.135 / 4.93
ethyl valerate / 130 / 0.384 / 0.077 / 4.99
allyl thiocyanate / 99 / 0.519 / 0.101 / 5.14
nitroglycerine / 227 / 0.22 / 0.044 / 4.99
trigylceride of butyric acid / 302 / 0.161 / 0.033 / 4.86
trigylceride of oleic acid / 884 / 0.056 / 0.011 / 4.95
acetaldehyde / 44 / 1.107 / 0.227 / 4.87
chloral / 147.5 / 0.342 / 0.068 / 5.04
benzaldehyde / 106 / 0.473 / 0.094 / 5.01
camphor / 152 / 0.338 / 0.066 / 5.14
acetone / 58 / 0.85 / 0.172 / 4.93
dibutyl ketone / 142 / 0.359 / 0.070 / 5.10
mean / 5.01
standard deviation / 0.117

To download worked spreadsheet, go to:

Rather cursory inspection shows that all of the freezing point depression constants are pretty close to 5.0° kg/mol. How close? Almost all of the values are within 0.2 of this value. Statistical quantities such as the mean and standard deviation are commonly used to describe data sets. The mean, or arithmetic average, can usually be thought of as a representative or typical outcome. The mean value of Kf is 5.01° kg/mol. The standard deviation is a measure of the spread of the data. Roughly 2/3 of the points in a data set can be expected to fall within one standard deviation of the mean. The standard deviation here is 0.12, or 2.4% of the mean. Only about 2% of the points in a data set can be expected to fall more than two standard deviations away from the mean. The only point in this set to do so is that of methyl iodide (shown in bold red), and it is not that far away.

In sum, a standard deviation 2.4% of the mean is not bad for this kind of experiment; Raoult is correct in concluding that the data do not suggest that Kf is anything other than the same for all the solutes. Furthermore, the methyl iodide point is not so egregious an outlier as to warrant discarding.