The Vapour Pressure of Solutions

[00:00:01]we can now look at the vapor pressure of of solutions. Okay. So we shall now consider more on the nonvalat solutes and also ionic ionic solutes. We want to see how this affects the vapor pressure of a solvent.

[00:00:26]So a nonvolatile solute lowers lowers the the vapor pressure of a solvent because the dissolved uh volatile solute uh decreases the number of solvent molecules per unit volume and lower the escaping uh tendance of the solvent uh molecules. Okay. So uh detailed uh studies of vapor pressures detailed studies of vapor pressures of solutions.

[00:01:12]Vapor pressures of solutions uh containing nonvolatile containing nonvolatile um nonvolatile solutes uh were were carried out by uh Franco Franco M route from 1918 to 1901.

[00:01:50]Okay.

[00:01:53]uh he was a French a French uh chemist or physicist. Okay. uh Francois uh so he's the one who did some detailed studies on the uh vapor pressures of of solvents containing a nonvolatile uh solution B solution is equal to K solvent * P solvent.

[00:02:40]Okay. So this is Rout's law where uh P solution P solution is observed observed vapor pressure observed uh vapor pressure of solution observed vapor pressure vapor pressure of solution.

[00:03:08]Okay. Then ka solvent this one ka solvent ka solvent is a more fraction more fraction of solvent more fraction of solvent. Okay. Then p not solvent p not solvent is the vapor pressure.

[00:03:37]vapor pressure of pure solvent.

[00:03:44]Okay, so that's rout's law. Okay, so Rout's law is a linear equation.

[00:03:55]Rout's law is a linear equation.

[00:04:03]Rout's law is a linear linear equation of the form y = mx + b. Okay. So let's compare p is = k solvent * p solvent.

[00:04:25]Okay. So in this equation in this equation y is = p not solution then x x is= k solvent.

[00:04:40]Okay then m is= p not uh solvent. So it means a plot a plot a plot of P solution. This is this is Kai solvent more fraction of the solvent gives a straight line gives a straight line.

[00:05:07]It gives a straight line straight line of slope. slope equal to P not uh solvent.

[00:05:16]Okay. So we have something like this.

[00:05:23]Okay. So we have vapor pressure of the solution then plotted against more fraction.

[00:05:32]Okay. So the slope of the line the slope of the line uh that's vapor pressure of the pure of the pure uh solvent. Okay, that's a vapor the vapor pressure.

[00:05:51]So the the lowering the the lowering the the lowering of vapor pressure depends on the number of solute particles present.

[00:06:07]So lowering lowering depends depends on the number on the number of solute solute particles solute particles present.

[00:06:24]Okay. So we can we can do an example.

[00:06:30]Let's see how we can we can use um we can use routes law example calculate calculate the expected calculate the expected expected vapor pressure.

[00:07:01]Calculate the expected vapor pressure at 25°C.

[00:07:08]at 25° C for a solution.

[00:07:14]for a solution prepared.

[00:07:18]Prepared by dissolving.

[00:07:22]Prepared by dissolving.

[00:07:26]By dissolving uh by dissolving 158 g 158 g of of sucrossse.

[00:07:44]158 g of sucrossse.

[00:07:47]Sucrossse that is sugar table sugar of sucrose.

[00:07:53]Molar mass of sucrossse.

[00:07:56]Molar mass molar mass of sucrossse is equal to is equal to 343 343 343.3 g g per mole. That's the molar mass of of sucros in in six 40 43.5 ml of water of water of water. Okay. At 25°C, at 25°C, the the density of water, the density of water is 0.99 9971 g per m per mill. And the vapor pressure.

[00:09:02]And the vapor pressure.

[00:09:06]And the vapor pressure of pure water of pure water is 23.76 to. Okay. So this is the question.

[00:09:21]Calculate the expected vapor pressure at 25° C for a solution prepared by dissolving 158 g sucrossse mar 343.3 g per mole in 643.5 ml of water at 25°C.

[00:09:41]The density of water is 0.9971 g per ml and vapor pressure of pure water is 23.676 676 uh to solution.

[00:09:53]Okay. So what we need here um uh first of all we need to determine the mo fraction of of water which is the solvent. Okay. So we need we need first to calculate the moles of sucrossse. Okay. So we can first calculate moles of sucrossse.

[00:10:19]We start with moles of sucrossse. Mass of sucrossse that is 158 158 g over molar mass of sucrossse 343.3 g per mole. So we we divide. Okay. Uh uh we are going to find 0 0.4 4616 moles moles of sucrose.

[00:10:50]Okay. So we have the moles of sucrossse.

[00:10:53]So what we need now is to calculate the mass of water. But from the question here we've been given volume of water.

[00:11:01]So we can use density to calculate the mass of water. Okay. Since we have density and volume then we can calculate uh the mass. Okay. So we know that density is equal to mass over over volume. So mass is equal to density time uh times volume. Okay. The density given is 0.99 0.9971 g per ml time volume 643.5 ml. Okay. So cancel the mill. Then we have our mass our mass is equal to 641.6 641.6 g of water. Okay. So now we can calculate the moles of of water since we know the mass. Okay. So moles of water moles of water is equal to mass of water which is 6 41.6 g over molar mass 18.02 g per per mole 18.02 g per mole. So which is equal to uh we now divide.

[00:12:24]Okay. So when we divide we shall get 35.63. 363 35.63 moles of of water. Okay. 35.63 moles of of water. So that is our uh that is the number of of moles. Okay. So now we can calculate we can calculate the the the mo fraction. Okay. So, more fraction of water. More fraction of of water.

[00:13:02]Moore fraction of water is equal to is equal to moles of water.

[00:13:09]Moles of water over moles of water plus moles of sucrose plus moles of sucrossse. Okay. Then now we substitute moles of water 35 63 moles over over 35.63 moles plus moles of sucrossse moles of sucrossse 0.4 0.4616 moles. Okay. So we can add uh down then we divide. Okay. So our mole fraction is equal to is equal to 0.9 0.9 98 0.9873.

[00:14:10]Okay. So that is the mo fraction the mo fraction of of water. Okay. more fraction of water. So we have all the information we need to calculate the the the the partial pressure I mean the the vapor pressure the vapor pressure of the uh solution. Okay. So we are now going to apply routes routes law. Okay.

[00:14:47]So vapor pressure of of the solution vapor pressure of the solution is equal to is equal to mole fraction of water time vapor pressure of pure water. So a mo fraction of water 0.98 0.9873 times the pure water vapor pressure of pure water that is 23.76 to then we multiply then we find 23 23 7 23.46 46 top. Okay. So you can now compare you can now compare uh we see that the vapor pressure of water has been has been lowered in the presence of the nonvolatile.

[00:15:51]You can see it has been lowered from 23.76 to 2346.

[00:16:00]Okay. So the vapor pressure the vapor pressure has been has been has been lowered from 23.76 to 23.46 in the pure in the pure state. Okay. So in the pure state is 23 76. In the presence of the volatile nonvolatile solute it becomes 23 23.46 in the in the solution. Okay. So the vapor pressure has been lowered by 0 030 to that is when we subtract 23.76 minus minus this. Okay. So we can see clearly here that the nonvolatile uh solutes lower lowers the vapor pressure of the uh solution. Okay. So a sol sucrossse it's a molecular it's a molecular uh solid okay it does not dissociate in uh in solution. So now let's look at the situation uh whereby the solute is an ionic ionic compound. Okay. So let's see how we can calculating calculating vapor pressure calculating vapor pressure calculating vapor pressure containing containing an ionic ionic compound.

[00:17:53]Okay. Calculating vapor pressure containing an ionic uh compound. Okay.

[00:18:01]So, predict predict the vapor pressure predict the vapor pressure of a solution prepared of a solution prepared prepared by by mixing of a solution prepared by mixing by mixing uh 35 g 35.0 g of sodium sulfate 35.0 0 g of sodium sulfate. Molar mass of sodium sulfate is equal to 142 142 g per mole. 142 g per mole with 175 with 175 g of water. 135 g of of water, okay, of water at 25°C.

[00:19:15]at 25° uh Celsius. Okay. So, the vapor pressure of pure water at 25 the vapor pressure of pure water at 25° C is 23 23.76 to okay. So, we do the same.

[00:19:40]First of all, we find we find the moles of water. Then we also find the moles of the solute. Okay. So, we can straight away find moles of water. Okay. So, moles of water is equal to the gram of water is 175.

[00:20:01]175 over 18.02 g per mole. So we divide 175 over that.

[00:20:12]It's giving us 9 9.72 9.72 moles of of water. 9.72 moles of water. Then we calculate moles of sodium sodium sulfate. moles of sodium sulfate. Uh uh moles of sodium sulfate mass of sodium that is 35 g. Then molar mass is 142 142 g per mole g per mole which is equal to 35 over 1 13 1 / 142. It's giving us 0.24246 246 moles of sodium sulfate. Okay. So now here we are dealing with an ionic ionic compound which dissociates in in water. So when sodium when sodium sulfate is dissolved in water, sodium sulfate dissolved in water, it will dissociate into sodium ions and sulfate and the sulfate ions. Okay. So you can see here we have two sodium then two there. So we are balancing the equation. So total we have three. We have a one here, a two there plus a one. So three. Okay. So now uh we are now going to So the total number of particles now the total number of particles present will be three 3 * time the the moles of sodium sulfate produced. Okay. So we are now going to say moles of solute.

[00:22:12]Moles of solute is equal to 3 * 0 246 moles. Okay. Which will now give us which will now give us 0.73 0.738 8 moles moles of solute. Okay. So these are the mo that we shall now use in our routes uh routes law. Okay. So uh we apply now routes law. Okay. So P solution is equal to kai water time p water. Okay. So in this case our uh we have not yet found the the the mo fraction. Okay. So now we need more fraction of of water. More fraction of water. Okay. Before we we substitute in our routes law. Okay. So the mole fraction of water uh we take note of the the moles. Okay. So we are going to use moles of water. Okay. Moles of water and moles of the solute. Okay. So can create some some space here.

[00:23:38]Okay.

[00:23:40]Okay. So we have what we uh so mole fraction of of water is equal to moles of of water over moles of water plus moles of of solute plus moles of solute.

[00:23:57]Okay. So, moles of water 9.72 moles over 9.72 moles plus 0 738 moles. Okay. So which is equal to uh so our mo fraction our mo fraction is 0.92 okay 0.9 29 this is the mo fraction then we use our equation okay mo fraction of water 0.929* vapor pressure of pure quart 23.76 to so we we now multiply. So when multiply we find 22 point.1.

[00:24:55]So you can see here how greatly the the vapor pressure of pure water has been lowered from 23 to 22 uh.1. Okay. So that is how we can calculate. So you need uh to identify if you are dealing with molecular molecular solutes or ionic ionic solutes. Okay. So in most cases for ionic solutes you don't forget to dissociate the the salt which will give you the total number of moles of solute.

[00:25:40]Okay. So for ionic for ionic solutes they have to be dissociated in water. Then for molecular such as sucrose, glucose. So these they don't they don't form ions in in in solution. Okay. So I I which is okay. We are going to come to that. Okay. So for molecular it means I is equal to one.

[00:26:10]Okay. Then for ionic for ionic it will equal to the number of the dissociated ions. So we shall come to I which is a vant vanov factor. Okay. So we can also look at non ideal non ideal uh solutions non ideal solutions. Okay. So the non ideal non ideal solutions non ideal okay so the vapor pressure of non ideal solution uh deviates deviates from loud's law so vapor pressure of non ideal solution deviates deviates from rout's law.

[00:27:17]Okay. It deviates from loud loud's law because uh because solute because solute solvent interaction solute solvent intermolecular intermolecular intermolecular attractions intermolecular uh attractions intermolecular attractions are either weak are either weak. Okay. So if solvent if solute intermolecular attractions are weak then we have a positive deviation.

[00:28:07]Okay. So we have a positive positive deviation that is when the solute solvent intermolecular attractions are weak then when they are stronger or strong when they are strong then we have a negative a negative uh deviation okay [clears throat] a negative deviation from the forces is in the pure in the pure uh liquid. Okay. So, let's look at positive positive deviation.

[00:28:49]Positive deviation.

[00:28:52]Positive uh deviation. Okay. So when mixing the two liquids, when mixing the two liquids, when mixing the two liquids, uh, so if A, if AB, when mixing the two liquids, if AB intermolecular forces are weaker, if AB intermolecular forces are weaker, are weaker? If AB intermolecular forces are weaker than the pure AA and BB forces, okay? Than pure A a A and BB forces.

[00:29:49]than a AA and BB forces. Molecules escape more easily. Okay, molecules escape more easily.

[00:30:02]Molecules escape more easily.

[00:30:11]Okay. So it means now it means uh the pressure the pressure the pressure is greater than is greater than the pressure the vapor pressure of A plus the vapor pressure of of B.

[00:30:37]Okay. So to find the total pressure to find the total pressure if you are mixing two volatile volatile liquids you can simply add you can simp simply add the the the the vapor pressure of the two solutions. This is what uh we have done uh here. Okay. So total pressure P total is equal to the vapor pressure of of solution A plus vapor pressure of solution solution B. So we know that the vapor pressure of the solution is equal to mole fraction time vapor pressure of the pure pure uh solvent. Okay. So that is um uh positive uh deviation. So an example of this an example of this uh so this one is higher you can see pressure is greater than that. So it's higher it's higher than ideal higher than ideal. Okay. So the vapor pressure is greater than the vapor pressure of the two uh two solutions. So example of this is when ethano ethanol is mixed with acetone. Example ethano plus acetone ethanol plus acetone will give a positive uh deviation. and also acetone.

[00:32:19]Acetone acetone plus carbon carbon dulfide.

[00:32:25]Okay. Then the enthalpy of mixing is greater than uh zero. Enthalpy of mixing enthalpy of mixing is greater than zero.

[00:32:39]It means the reaction is is endothermic.

[00:32:43]is endothermic reaction is endothermic and the change in volume and change in volume is greater than zero as well.

[00:32:54]Okay, means that the volume the volume expands expands. Okay. So we can now look at um uh negative negative deviation negative uh deviation negative uh deviation.

[00:33:17]Okay. So when a when a when a when a intermolecular attractions when AB intermolecular attractions are stronger are stronger are stronger than AA a and bb. Okay. They are stronger than pure are stronger than pure AA and BB interactions interactions.

[00:33:57]Okay. The escaping tendance of molecules is reduced is reduced. escaping tendency of molecules is reduced is reduced.

[00:34:14]Okay. So, AB AB intermolecular forces are stronger than pure A and and BB.

[00:34:24]Okay. So in in such a case uh it means now the vapor pressure P the vapor pressure P the vapor pressure P is less than K A plus K B P B okay the vapor pressure is less than uh the sum of the two uh solutions Okay.

[00:34:55]So this is lower than ideal. Lower than ideal.

[00:35:01]Lower than ideal. Okay. So the the enthalpy of mixing the enthalpy of of mixing is less than is less than zero.

[00:35:16]And the reaction is exo exothermic.

[00:35:20]reaction is exo exothermic.

[00:35:25]Okay. Then examples of this uh examples uh chloroform chloroform plus we know chloroform is polar. Okay.

[00:35:41]Uh chloroform chloroform plus acetone.

[00:35:47]Chloroform plus acetone gives a negative uh deviation due to strong hydrogen uh bonding and also water. Water plus nitric nitric acid water plus nitric acid okay also gives the negative uh deviation. Okay. Then let's look at key formulas for deviations.

[00:36:21]Key formulas for deviation.

[00:36:26]Key formulas for deviations. Okay. Key formulas for deviation. So for an ideal solution, the total pressure for an ideal solution for an ideal solution the total pressure the total pressure which is denoted PT the total pressure uh PT PT relies directly on rout's law.

[00:36:57]Okay. So PT PT is equal to is equal to K A P A plus K B P B. Okay, that is how we can find the total pressure. I explained earlier on that you add the two pressures of of solution. Okay. So uh so when you have A and B two V volatile uh uh uh liquids the the the total pressure is the sum is the sum of their vapor pressures. Okay. So we know that where uh this one k a and k b are more fractions are more fractions of a more fractions of a and b respectively. Okay, there are more fractions of A and B. Then P not P A P A and P not B. These are vapor pressures of pure A and and B. Okay. So in any ideal solutions the total pressure graph causes causes positive uh deviation or negative uh deviation. So we can show that on on the graphs. Okay. So for an ideal solution the graph would look like this.

[00:38:42]Okay. So here we have vapor pressure.

[00:38:47]Vapor pressure. Okay. So when you you plot uh a graph. So let's plot a graph.

[00:38:59][clears throat] We can now plot a graph of vapor pressure versus more fraction. Okay. So more fraction.

[00:39:11]So here we have more fraction on the x axis more fraction. Okay. So we can have k kai a and also k b. Okay. So this one. So here it can be k a is equal to 1. Okay. Then here is = z. Then b here is = z. then b here is equal to to one. Okay. So for an ideal solution the graph looks like this. Then down.

[00:39:50]Okay. So this is for an ideal solution.

[00:39:55]Okay. So this is a vapor pressure of solution. Vapor pressure of of solution.

[00:40:03]Then here vapor pressure. Okay. So, uh then when this is for an ideal ideal solution.

[00:40:14]Okay. So, now let's see the deviations.

[00:40:21]Let's see the deviations.

[00:40:23]Uh so we we first plotting uh for ideal solution. The graph is like this. Okay.

[00:40:34]This is for ideal Okay. Then for weak solute uh weak solute weak solute.

[00:40:46]So this is for weak solute.

[00:40:49]Weak solute weak solute solvent.

[00:40:55]Weak solute solvent interaction. Weak solute solvent interaction.

[00:41:04]Okay. that is positive deviation.

[00:41:09]So for positive deviation it will look like this. Okay like that that is positive uh uh deviation then okay so you can see it's deviating from the ideal. So this is vapor pressure of of solution.

[00:41:35]So that is positive deviation. It's it's caving uh outwards. Okay. Then down here. So this is vapor pressure. Then more fraction of A and B. So let's look at negative negative uh deviation.

[00:41:53]So with negative deviation uh first we do for ideal. Okay. So for negative it will go down down like that.

[00:42:07]Okay. Then down here. Okay. Then down here.

[00:42:15]Okay. So for negative it caves it caves downward. So this is vapor pressure. vap pressure of of solution. Then this is a more fraction. Okay. So this is how the the graphs look like. So here it means this is now a strong strong solute strong solute solvent solvent interaction. So you need to take note. So for weak solute solvent interaction that is positive then for uh strong solute solvent interaction that is uh negative uh deviation. Okay.

[00:43:10]So example uh we can We can do an example.

[00:43:22]Uh we have these two hex zen hex zen.

[00:43:30]Uh so for each of the following solutions would you expect the the solution to be ideal? Okay. To be ideal show a positive or a negative or a negative deviation. So hexane and and chloroform.

[00:43:49]Hexane and chloroform and chloroform.

[00:43:56]Hexane and chloroform.

[00:44:00]Okay. So what do we expect uh hexane and and chloroform? Do we expect a positive or or or negative?

[00:44:13]Okay. So we know that hexane is a nonpolar and chloroform is polar. So this one is non nonpolar then this one is polar. So it's positive [snorts] positive uh deviation. Okay. Then ethyl alcohol.

[00:44:37]Etho alcohol plus plus water.

[00:44:43]A alcohol plus water. Okay. So both are polar. So here we So both these two both are are polar. Both are polar.

[00:44:57]Okay. So we expect a negative a negative uh deviation a negative deviation.

[00:45:06]Okay.

[00:45:08]So that is how we can uh we can Okay.

[00:45:13]Then I think another one uh another example hexen hexen and octane hexen and octane hexane and octane.

[00:45:28]Okay. So both are nonpolar. with similar molar masses. So this is ideal. This is ideal. Okay. This is ideal. All right.

[00:45:41]So uh we can we can look at an example.

[00:45:51]uh we see how we can uh calculate calculate the total pressure of two volatile of two volatile uh uh liquids. Okay. So let's uh uh calculate uh an example uh calculate a solution A solution is prepared by mixing.

[00:46:27]A solution is prepared prepared by mixing by mixing by mixing 5.81 g of acetone of acetone 5.81 81 g of of acetone. The mar mass of acetone C3 H8 O okay of acetone molar mass [snorts] molar mass of acetone is equal to is equal to 58.1 58.1 g per mole grams per mole and and 11.9 g of chloroform form 11.9 g of chloroform of chloroform. Okay. Molar mass of chloroform molar mass molar mass molar mass of chloroform is equal to one 119.4 119.4 before g per mo g per mole. 119 g per mole. Okay. Then at 35°C, at 35°C the the at 35°C this solution this solution this solution has a total pressure of 2602.

[00:48:21]has a total pressure total pressure of 2 C to that is a total pressure. Is this an ideal solution?

[00:48:35]Is this an ideal Is this an ideal solution? Is this an ideal uh solution?

[00:48:47]Okay. Then the vapor pressure of pure acetone.

[00:48:52]The vapor pressure of pure acetone.

[00:48:57]The vapor pressure of pure acetone and pure chloroform and pure chloroform and pure chloroform.

[00:49:14]uh at 35°C.

[00:49:17]At 35°C, uh uh 345 to for acetone and and 290 293 to for respectively that is for for chloroform. Okay. So we want to to decide or to find out if this uh combination is ideal. Okay. So what to do first? We need we need to to calculate to calculate the the total pressure and and compare and compare to the total pressure that is uh here 260.

[00:50:13]Okay. So we know that we know that total pressure P total will equal to more fraction of A. So a in this case is acetone more mole fraction of acetone time vapor pressure of pure acetone plus mo fraction of chloroform time vapor pressure of pure chloroform.

[00:50:41]Okay. So now we can we can calculate the number of moles for for each. Okay. So moles of acetone moles of acetone is equal to So what is the mass of acetone? The mass of is given 5.81 g. Okay. So 5.81 g over molar mass. The molar mass of acetone is 58.1 58.1 g per mole. So which is equal to uh we add we divide we are going to get 0.10 moles of acetone then moles of chloroform moles of chloroform moles of chloroform the mass of acetone of chloroform is 11.9 11.9 g then molar mass of acetone 119 119.4 24 119.4 g per mole. Okay. So when you divide we still find the same number of of moles.

[00:51:56]We find the same number of of moles.

[00:52:01]Okay. So we have the same number of moles of acetone and moles of of chloroform are the same are the same. So if they are the same it means the mole fraction the mole fraction for both is the same. Okay since they have the same number of moles. So what we do is mole fraction of acetone is equal to moles of acetone over moles of acetone plus moles of of chloroform. So we have 0.1 over 0.1 + 0.1.

[00:52:40]Okay, you can see that it is uh so here it will now go to 0 0.5 0.5.

[00:52:51]Okay, since it's now half 0.1 + 0.2 then 0.1 over that 0.5.

[00:53:00]So which is also so this mo fraction which is also equal to the mo fraction of of chloroform. Then now we substitute in the in the equation. Okay. So our equation is this one. Okay. So the mole fraction 0.54A.

[00:53:26]Then the the partial pressure of acetone the partial pressure of acetone is equal to u it's given. Okay. The partial pressure of acetone which is 4 43 * 345 345 to then plus uh 0.5 0.5 partial pressure of of chloroform which is 290 293 okay 293 3 to so [snorts] we multiply and add. So when we multiply and add we are getting 319 319 to 319 to then now we compare we compare you can see the total pressure uh that was found uh experimental.

[00:54:29]So experimental uh expected expected is 3 319 but experimental experimental is expected I mean expected is what we have been given. This is uh experimental.

[00:54:51]Okay. So uh uh what expected expected uh expected pressure uh so this is expected expected expected 319 okay 319 that is expected then experimental experimental is that one given experimental that is two sist.

[00:55:28]Okay. So uh we can say twoist is less than uh 3 319. So the solution the solution is not ideal. Okay. The solution the solution is not ideal. is not ideal.

[00:55:51]It's not ideal.

[00:55:53]Is not ideal. It shows a negative deviation.

[00:55:59]It shows a negative uh deviation.

[00:56:04]Okay. Since twoist is less than 3 319. Okay. It shows a negative deviation from L's law indicating that acetone and chloroform molecules attract each other more strongly than like molecules than like molecules than like molecules reducing the escaping tendons.

[00:56:40]Okay. So there's a uh there's a a a strong uh attraction between uh these two. Okay. Than the pure uh uh molecules. Okay. So that's a negative. It shows a negative uh deviation deviation from loud law. Okay.

[00:57:19]So solution not ideal shows negative negative deviation.

[00:57:30]Okay. And you know how the graph looks like for negative uh deviation.

[00:57:37]Okay. So the the experimental vapor pressure is much lower than the value predicted by rout.

[00:57:49]Hence our conclusion.

[00:57:52]Okay. So we can end here. Then the next the next remaining uh part is on collleative uh properties.

[00:58:05]Thank you.