Using data from Table 13.3, calculate the freezing and boiling points of each of the following s

Using data from Table 13.3, calculate the freezing and...

Key Concepts

Colligative Properties

Colligative properties are physical properties of solutions that depend on the concentration of solute particles rather than their identity. These include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. In the context of the problem, freezing and boiling points change because the number of solute particles affects the energy required for phase transitions in the solvent.

Molality

Molality is a concentration unit defined as the number of moles of solute per kilogram of solvent. It is particularly useful in colligative property calculations because it remains independent of temperature changes, unlike molarity, due to the use of the solvent's mass instead of volume.

Freezing Point Depression

Freezing point depression is a colligative property where adding a solute to a solvent results in a lower freezing point than that of the pure solvent. This phenomenon occurs because the presence of solute particles disrupts the formation of a solid lattice, requiring a lower temperature to achieve the phase change.

Boiling Point Elevation

Boiling point elevation is another colligative property whereby the addition of a non-volatile solute to a solvent increases the boiling point of the resulting solution compared to the pure solvent. The solute particles reduce the vapor pressure of the solvent, thereby requiring a higher temperature to reach boiling conditions.

van 't Hoff Factor

The van 't Hoff factor represents the number of particles into which a solute dissociates in solution. For electrolytes like NaCl and KBr, accounting for this factor is crucial as they dissociate into ions, effectively increasing the solute particle concentration and amplifying the colligative effects such as boiling point elevation and freezing point depression.

Transcript

00:01 Question number 73 is a freezing point depression, boiling point elevation question.

00:09 In addition, you also need to recognize that the ionic solutes dissociate into their corresponding number of ions.

00:20 This you may have been taught as is the von hoff factor.

00:23 There are a couple ionic compounds that are involved in these calculations.

00:30 You need to use table 13 .3 in order to find the normal freezing and normal boiling points of the solute that is, or the solvent, that is involved in each of these solutions.

00:44 You also need to use this table in order to find the kb and kf values, the boiling point elevation constants and the freezing point depression constants for the specific solvents.

00:59 So let's get started.

01:01 For the first one, we have a 0 .22 molal glycerol solution in ethanol.

01:09 So the boiling point elevation equation is delta tb is going to be equal to the molality of the solute multiplied by the boiling point elevation constant for the solvent ethanol, which we look up to be 1 .22.

01:25 We then get an increase in the boiling point of .27 degrees.

01:31 We go to table 13 .3, and we see that the normal boiling point of the solvent ethanol is 78 .4, so we add on our .27, and we get a new boiling point of this solution at 78 .7 degrees celsius.

01:47 We'll do a similar calculation for the change in the freezing point of ethanol when we have a .22 molal glycerol present.

01:57 We'll take the 0 .22 molal concentration multiplied by the freezing point constant for ethanol which is 1 .99 and we'll get a 0 .44 change in the freezing point.

02:10 However this is a freezing point depression so it will be a negative 0 .44.

02:15 So the new freezing point is going to be the normal freezing point of ethanol negative 114 .6 which is in table 13 .3 minus the change of 0 .44.

02:29 And we get a new freezing point of ethanol with the 0 .22 molal glycerol of negative 115 degrees celsius.

02:38 We then continue this with the three other solutions.

02:41 The next solution contains 0 .224 moles of naphthalene in 2 .45 moles of chloroform.

02:51 They didn't give us the molal concentration in order to use the freezing point.

02:55 Point and boiling point equations, we need to have the solute expressed in a molal concentration.

03:02 So we need to take the mol, to get the molality, we need to take the moles which are provided and divide by the kilograms of the solvent.

03:11 We don't know the kilograms of the solvent, but we do know the moles of the solvent.

03:16 We can convert the moles of the solvent into kilograms by multiplying by the molar mass, and then dividing by a thousand or even easier take the molar mass and convert it to kilograms.

03:35 The molar mass of chloroform is 119 .4 grams per mole, so in kilograms it's going to be 0 .1194 kilograms per mole.

03:46 This will now give us kilograms of the solvent in the denominator divided into the moles of the solute, which is naphthalene.

03:54 This is our molality.

03:57 We'll multiply that by the boiling point elevation constant for the solvent, which is chloroform.

04:06 Its value in table 13 .3 is 3 .63, and we get an increase in the boiling point of 2 .98 degrees celsius.

04:15 So the new boiling point of the chloroform is going to be the normal boiling point of chloroform, 61 .2, plus the change of 2 .98, new boiling point being 6 .4 .2, plus the change of 2 .98.

04:27 64 .2 degrees celsius.

04:31 Then we do the same thing for the freezing point depression.

04:34 We'll calculate the molality the exact same way as we did before and now multiply it by the freezing point depression constant for chloroform, which is 4 .68 in table 13 .3.

04:48 This gives us then a decrease in the temperature of the chloroform 3 .84 degrees.

04:55 So the final freezing point, of the chloroform is going to be its normal freezing point, negative 63 .5 degrees celsius minus the change, which is 3 .84.

05:08 So the new boiling point then of the chloroform solution, sorry, freezing point of the chloroform solution will be negative 67 .3 degrees celsius.

05:18 The next solution is 1 .50 grams of sodium chloride in 0 .250 grams of, or kilograms of water...

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