Solutions and their Properties

[00:00:00]Let's look at solutions and their properties. First, we define a solution.

[00:00:06]A solution is defined as a mixture of two or more substances distributed throughout a single phase.

[00:00:16]Okay. So, a solution is made up of of a solute and a solvent.

[00:00:26]Okay. So solution is equal to solute plus solvent. A solute A solute is a substance substance being dissolved.

[00:00:46]Substance being dissolved. Ag sugar.

[00:00:50]Okay. Then a solvent.

[00:00:54]A solvent is a dissolving dissolving medium aging water. Okay. So sugar plus water is equal to sugar solution.

[00:01:19]Okay. So sugar solution is it forms now a single that's a single phase. Okay. So solute plus water solute plus solvent water forms sugar solution. Okay. So solutions uh we have various we have various types of solutions. So a solution a solution can be can be a gas can be a liquid or a solid. Okay. So let's look at various types of solutions.

[00:02:02]Example, state state of solution, state of solute, then state of solvent.

[00:02:18]For example, air. Air or natural natural gas. State of solution gas.

[00:02:29]State of solute gas. instead of solvent gas. Okay.

[00:02:35]Then we can look at antifreeze.

[00:02:40]Antifreeze. Okay. This is used as coolant in cars car radiators. Okay.

[00:02:48]This one is a liquid. Liquid state of solvent. Uh liquid. Okay.

[00:02:56]Brass.

[00:02:58]Brass is an alloy of zinc and copper.

[00:03:01]Okay. So, it's a solid solid uh solid.

[00:03:08]Okay. Can also look at carbonated water.

[00:03:15]Carbonated water. So, here carbon dioxide dissolved in in water. Carbonated water.

[00:03:25]Okay.

[00:03:27]like carbonated drinks. Okay. State of solution, liquid, state of solute, gas, which is carbon dioxide, state of solvent, that's a liquid.

[00:03:44]Okay. Then sea water.

[00:03:48]Sea water. So we have sea water. We know the salt. Okay. So we know that the solute is salt. Then state of solution liquid state of solvent uh liquid. Okay. Then hydrogen.

[00:04:07]Hydrogen in platinum.

[00:04:11]In platinum. Hydrogen is a gas. Okay.

[00:04:15]This one is a solid. Platinum is a solid.

[00:04:19]Solid. Okay. So these are the various these are the various types of um of solutions. Okay. So you can see clearly here that a solution is not only a liquid. It can be a gas liquid or or solid. Okay. Then solutions are classified are classified as saturated saturated unsaturated unsaturated unsaturated and super saturated.

[00:05:03]Okay. So we start with the uh saturated. Okay. So let's define saturated saturated solution.

[00:05:17]Saturated uh solution. Okay. So a saturated solution uh is formed is formed is formed when no more when no more when no more solute when no more solute will dissolve will dissolve will dissolve in the in the solvent.

[00:05:50]Okay, that's a saturated solution formed where normal solute will dissolve in the in the solvent. Okay.

[00:05:58]The second one unsaturated unsaturated uh uh solution is formed is formed when it can still dissolve is formed when it can still when it can still dissolve.

[00:06:23]dissolve more solute.

[00:06:28]That's unsaturated.

[00:06:31]Okay. Then super saturated super saturated uh solution.

[00:06:41]Okay. is formed when you put is formed is formed when you put when you put when you put solute in the solvent in excess is formed when you put solute in the solvent in the solvent in excess in excess. Okay. So it means the solvent the solvent will not be able to absorb the whole solute. Okay. So in this case uh to make it dissolve you need to increase the the temperature. Okay. So once you hit the solution, we know that most salts most salts are are soluble most salts most salts are soluble in in hot are soluble in hot water.

[00:07:46]In hot water than cold okay than cold. All right. So now uh let's look at um the effect of okay before we we get into the calculations maybe first of all we need to look at uh solution solution content okay so let's look at various ways various ways of expressing solution composition okay or concentration. So let's look at solution composition.

[00:08:39]Solution composition.

[00:08:44]Okay. So we want to look at various ways in which we can express our uh concentration.

[00:08:55]Okay. So the the relative amounts of substances must be specified. Okay. So we encounter terms as dilute. Okay.

[00:09:09]Dilute solution. Okay. So dilute simply means there's relatively uh relatively little solute. Okay, that's the uh dilute. Okay, relatively little solute present.

[00:09:29]Then concentrated.

[00:09:32]Concentrated.

[00:09:35]Concentrated uh means relatively large relatively large large amount amount of solute.

[00:09:50]Okay. So for example when you make a cup of coffee when you make a cup of coffee.

[00:09:58]So suppose this is a B. Okay. So here you add this is coffee coffee. Okay. Coffee coffee.

[00:10:14]So in in cup A you add 1 tbsp of sugar. Okay. So here 1 tbsp 1 tbsp of sugar.

[00:10:31]Then here you add 3 tablespoons 3 table 3 3 tablespoon of sugar.

[00:10:42]Which one do you think is more concentrated? So this one is dilute. A is dilute then B is concentrated. Okay.

[00:10:51]So that's uh the difference. Okay. So the more solute present the more concentrated the solution uh will be. Okay. So let's look at concentration.

[00:11:08]concentration of of solution.

[00:11:13]Concentration of of solution. Okay. So, concentration is a measure is a measure of the amount of solute in a given in a given amount of solvent or solution. Okay. Measure is a measure of the amount.

[00:11:36]Measure of the amount of solute. Measure of the amount of solute in a given in a given solvent in a given amount of solvent or or solution. Okay. Measure of the amount of solute. So we are going to measure the amount of solute present in a solvent or or solution. So measure of the amount of solute in a solvent in a solvent or or solution or solution.

[00:12:11]Okay. So uh other than uh we all know that uh uh marity morality mality m is a common uh is a common measure of concentration that we know. Okay. So marity marity is is equal to moles of solute moles of solute over over liter solution over liter solution that is marity. Okay molar. So the volume moles of solute over volume of of solution.

[00:13:04]So there are also other ways there are also other ways other than monality that we are going to to look at. Okay. So we can measure concentration uh by mass percent mass percent mass percent weight by weight or mass by mass. Okay. So in chemistry uh weight weight and mass can be used interchangeably. Okay. So mass by mass.

[00:13:43]Okay. So in chemist when you say what weight we mean mass the two terms can be used interchangeably. Okay. So mass percent is equal to is equal to mass of solute.

[00:13:58]Mass of solute over mass of solution over mass of solution over mass of solution times 100%.

[00:14:13]That is how we shall be calculating mass percent. Mass percent is equal to mass of solute over mass of solution time 100. Okay. So we also have we also have mality.

[00:14:32]Mality.

[00:14:34]Okay. Mality. So marity is equal to is equal to moles moles of solute.

[00:14:45]Moles of solute over kilogram.

[00:14:51]Kilogram of solvent.

[00:14:55]that is mality. Okay. So the units the units will be more per kg.

[00:15:05]More per kg. Okay. That is mality. So mole per kg more per kg is equal to m which is equal to m small letter m. We all know that with mality it is capital M. So this one is m small letter M. Okay. So we also have more fraction.

[00:15:37]More fraction more fraction which is denoted by the Greek letter Kai. Okay. So mo fraction of component A. More fraction of component A.

[00:15:57]More fraction of component A. So we say K A is equal to mo of A over mo of A plus mo of B.

[00:16:10]That is the mo fraction. Okay.

[00:16:14]Then uh where n is number of of moles? n is number of of moles. Okay. So mo of mo I mean more fraction of solute mo fraction of solute plus mo fraction of solvent is equal to 1. more fraction of solute plus more fraction of solvent is equal to one. Okay. So it means that mo fraction can never be greater than one.

[00:16:56]Okay. That's the total it can it can it can ever achieve one. Okay.

[00:17:05]So that's the the the formula. Okay. So if we want to calculate mole fraction of solvent or solute. So mo fraction of solute is equal to moles of solute. Moles of solute over moles of solute.

[00:17:27]Moles of solute plus moles of of solvent that is mole fraction of solute. Then mole fraction of solvent mole fraction of solvent is equal to moles of solvent.

[00:17:42]Moles of solvent over moles of solvent plus moles of solute plus moles of solute. Okay. Plus mo of solute that is mo fraction. Okay. Then we can also express we can also express the concentration concentration in terms of ppm ppm and ppb.

[00:18:22]Okay. So ppm means parts parts per million.

[00:18:29]Parts per million.

[00:18:32]Then ppb parts per billion.

[00:18:37]parts per billion. Okay. So ppm ppm by weight by weight or or mass by mass or ppm by mass by mass or weight by weight. Okay. So, ppm is equal to is equal to mass of solute.

[00:19:03]Mass of solute mass of solute in g over mass of sample over mass of sample mass of sample in g* 10 to the power 6 that is ppm weight by weight or mass by mass okay then ppb ppb mass by mass is equal to mass of solute mass of solute in grams over mass of sample mass of sample mass of sample in g* 10 ^ 9 okay so if it is weight by volume weight by volume or mass by volume then mass of sample now becomes mass of solution here also mass of solution solution. Okay. So we can also calculate uh ppm by simply dividing micro g micro g micro g per g.

[00:20:19]Okay that is ppm micro g per g it will give you ppm. So ppm is also equal to micro g per per g or or mig mig per kg that is also equal to ppm.

[00:20:41]Then ppb ppb is equal to micro g per micro g per kg micro g per kg or or nanog nano nanog per g. Okay. So this uh these are the other ways in which you can calculate uh calculate ppm and ppb apart from using the the formula. Okay. So that is how we can we can calculate. Okay. So now we can have an example. Let's see how we can we can calculate using various various uh types of concentration uh content. Okay. So we are going to express concentration in various uh various uh specifications. Okay. So a solution example example a solution a solution a solution is prepared a solution is prepared by by mixing by mixing 1.00 g of ethanol 1.00 00 g of ethanol C2 H5O ethanol. Okay. 1 g of ethanol with 100 g with 100 g of of water. 100 g of of water to give a final volume.

[00:22:35]to give a final volume to give a final volume of 101 mill.

[00:22:45]Okay. So, uh calculate calculate a um marity marity.

[00:23:02]Okay. Then B uh more fraction I mean mass percent. So let's start with the mass percent.

[00:23:14]Mass percent. C more fraction.

[00:23:20]More fraction. Okay. More fraction. Um more fraction.

[00:23:27]And mality.

[00:23:30]and D mality mality of ethanol of ethanol. Okay. Solution.

[00:23:43]So we start with with marity the the one we we honor the popular one. Okay. So marity marity is equal to moles of solute moles of solute over liter liter solution moles of solute over liter solution. So in this case ethanol and water you need to identify you need to identify the solute but water water is never a solute okay water is always a solvent. So a solute in this case is ethanol. Okay. So you need to calculate the moles of ethanol. So let's calculate moles of ethanol. Okay. So moles of ethanol.

[00:24:40]Moles of ethanol. We know mo is equal to mass over mass. So the mass of ethanol is 1 g. Okay. 1.00 g. Then molar mass. The molar mass of ethano is 46.07.

[00:24:57]46.07 g per mole. Then you divide 1 / 46.107.

[00:25:07]It is giving us 2.17 * 10 ^ -3 * 10 ^ -2 * 10 ^ -2 moles -2 moles of of ethanol of ethanol. Okay.

[00:25:27]So these are the moles of of ethanol.

[00:25:31]Okay. Then now um marity marity of ethanol is equal to moles of ethanol 2.17 * 10 ^ -2 mo over volume. The volume of the solution the volume of the solution is 101.

[00:25:56]Okay 1. So we convert volume into liters. Okay. So 101 * 103 L. Then now you you divide. Okay. So you're going to find 0.215 moles per liter or simply 0.25 molar. Okay. So that is the the marity.

[00:26:24]Then mass percent.

[00:26:28]Uh so this is our A. Then B let's calculate mass percent.

[00:26:35]Mass percent. So we know mass percent is equal to mass of solute mass of solute over mass of solution over mass of solution time 100. Okay. So in this case mass of solute solute that is 1 g. Okay. Then mass of solution. We know that solution is equal to solute plus solvent. So you add solute.

[00:27:06]Then mass of of solvent which is water 100 g. Okay.

[00:27:15]So you add the two and divide. Okay. So we are going to find 0 0.99 0.990% ethanol.

[00:27:29]Okay. So that is the mass percent. Okay.

[00:27:32]Mass of solute over mass of of solution.

[00:27:38]Then let's calculate the mo fraction the mole fraction of ethano. The mole fraction of of ethanol mole fraction of ethanol. Okay. So mole fraction of ethanol is equal to moles of ethanol. Moles of ethanol over moles of ethanol over moles of ethanol plus moles of of water plus moles of of water. Okay. So we need to calculate moles of of water. Okay. So moles of water moles of water is equal to is equal to so we know that the mass of water the the the mass of water is 100. Okay. So 100 g over 18.02 g per mole. So which is giving us giving us 5.5 5.56 moles of of water. 5.56 moles of water.

[00:28:57]Okay. Then we bring it here. Moles of ethanol we have already calculated which is 2.17 * 10 ^ -2 mo. then over over 2.17 * 10 ^ -2 mo then plus moles of water 5.56 moles okay so you add down then you you divide so when you do the arithmetic you're going to find 0 389 389 this is the more fraction So you can clearly see that the mo fraction is is less than zero. Okay.

[00:29:46]Then finally uh we find the mality mality.

[00:29:55]Let's find the mality m. Okay. So mality is equal to moles of solute. moles of solute over over kg solvent kilograms of solvent. Okay.

[00:30:14]So in this case the moles of solute the moles of solute that is ethanol which is equal to 2.17 * 10 ^ -2 mo. Okay. Then over kilogram of solvent the mass of water is 100. Then we convert it into into kilogram. So 100 * 10 ^ 3 kgs. Okay. Then we divide. So when we divide we're going to find 0 217. Okay. 0 217.

[00:30:56]0.217 more per kg. More per kg. Okay, that is our mality or you can simply say 0.217 m with a small letter M. Okay, so that is how we can we can calculate we can express concentration in various uh concentration uh terms. Okay.

[00:31:31]Of course, there are also other other concentration uh terms that we can use. Okay. So, meantime, we can just look at uh this.

[00:31:42]Okay. So, now let's look at um let's look at the energies.

[00:31:53]The energies The energies of solution The energies of solution formation the energies of solution formation.

[00:32:12]Okay. So formation of liquid solution takes place in three distinct steps.

[00:32:20]Formation of liquid uh solution uh takes place takes place in three distinct distinct steps. Three distinct steps.

[00:32:38]Okay. So we have step one. Okay. So step one is the separation of solute.

[00:32:48]Separation Separation of Solute.

[00:32:53]Separation of solute into individual components into individual individual components.

[00:33:06]Okay. So this is our delta H1.

[00:33:11]Delta H1 Delta H1 uh endothermic. The process is endothermic.

[00:33:20]Endothermic.

[00:33:22]The process is endemic for delta H1. That is the enthalpy of of formation here. That's endothermic.

[00:33:34]Okay. The process is endothermic. It is being absorbed. Then the second step is overcoming overcoming intermolecular forces.

[00:33:48]Overcoming intermolecular forces in the solvent.

[00:33:53]In the solvent.

[00:33:56]Overcoming intermolecular forces in the in the solvent. To make room for solute.

[00:34:03]Okay. to make room to make room for solute to make room for solute. So delta H2 endothermic the process is also endo endothermic for step two. Okay.

[00:34:23]Then step three now is allowing the solute allowing allowing allowing the solute and solvent to inter interact allowing the solvent the solute and solvent to interact to form a solution to form a a solution.

[00:34:55]So that is our delta H3. The process is now exo exothermic. Okay. Heat is being is being uh released. Okay. So the enthalpy of solution, the enthalpy of solution, the enthalpy of solution, the enthalpy of solution, delta H solution. Delta H solution is equal to enthalpy of solution. Delta Solution is equal to delta H1 plus delta H2 plus delta H3.

[00:35:40]Okay. So it's a summation. So we add it's a sum of the individual enthalpies.

[00:35:48]Okay. So that is the enthalpy of solution. You add delta H1, delta H2 and delta H H3.

[00:35:59]Okay. So now if the enthalpy of solution if the enthalpy of solution is negative if the enthalpy of solution is negative.

[00:36:20]If the enthalpy of solution is negative, it means the reaction is exothermic.

[00:36:30]Exothermic. The reaction is exothermic.

[00:36:36]So, and solution forms and solution forms.

[00:36:44]Okay. If enthalpy of solution If enthalpy of solution is positive is positive means it's endothermic no solution no solution forms no solution forms. So solution will only form will only form when the enthalpy of solution is negative.

[00:37:15]Okay. So we can look at uh a table a table uh or examples.

[00:37:28]Sometimes we can we can mix a polar a polar solvent and a nonpolar solvent.

[00:37:38]Okay. And what happens is the solution going to to form. Okay. So mixing mixing okay mixing polar polar. So if we have a polar solute, polar solute plus polar solvent, polar solute plus polar solvent solution.

[00:38:12]Okay. So here solution forms solution forms.

[00:38:20]Okay. So polar solute plus polar solvent solution forms. Okay. So what of nonpolar solute?

[00:38:31]Nonpolar nonpolar solute plus uh polar plus polar solvent plus polar solvent.

[00:38:47]So here no solution forms no solution no solution forms no solution forms okay then what of nonpolar nonpolar nonpolar nonpolar solute nonpolar solute plus nonpolar solvent plus nonpolar solvent.

[00:39:19]Okay. So here solution forms solution forms solution forms. Okay. Then polar solvent polar solute plus nonpolar plus nonpolar solvent.

[00:39:42]So here no solution.

[00:39:45]no solution uh forms. Okay. So always remember like always remember like dissolves dissolves like okay like dissolves like okay so polar polar it forms a solution nonpolar non-polar a solution is is formed. Okay.

[00:40:16]So, we can now look at um let's look at gas solutions. Gas solutions.

[00:40:27]Gas solutions.

[00:40:31]Okay.

[00:40:37]Gas solutions.

[00:40:41]So we are we want to see some some factors. So temperature what are some of the factors affecting affecting the solubility? So we have temperature temperature and and pressure temperature and and pressure. Okay. So temperature affects the solubility of of gases. So the higher the temperature the lesser the solubility of a gas. Okay. So you can see that temperature is inversely proportional to uh solubility. Okay.

[00:41:20]Then pressure pressure has little effect on the solubilities of solids or or liquids. Okay.

[00:41:31]So significantly increase the solubility of a gas. Okay. So solubility pressure the the the pressure affects more of the solubility of a gas than that of a solid and and and liquid.

[00:41:52]Okay. So we are now going to look at pressure and gas solubility.

[00:42:00]Pressure pressure and gas solubility pressure and gas solubility pressure and gas solubility. So the relationship the relationship between gas pressure and concentration of dissolved gas is given by Henry's law. So relationship relationship between gas pressure gas pressure gas pressure and the concentration and the concentration and the concentration and the concentration of the dissolved and the concentration of the dissolved gas dissolved gas is given by Henry Henry's law is given by Henry's Henry's law.

[00:43:05]Okay. So we look at Henry's law. So Henry's law is given by the equation concentration of the gas is equal to K * P gas. Okay. or simply uh C * C is equal to C [snorts] is = K KP C is equal to KP so where C where C is the concentration C is the concentration concentration of the dissolved gas concentration of the dissolved gas C is the concentration of the dissolved gas.

[00:44:00]Okay. Then K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K is Henry's law.

[00:44:06]Henry's law constant.

[00:44:09]Henry's law constant. Henry's law constant or constant characteristic of a particular uh solution. Then P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P is a partial pressure partial pressure. P is a partial pressure.

[00:44:31]P is a partial pressure.

[00:44:35]Partial pressure P is a partial pressure partial pressure of the gaseous solute.

[00:44:46]P is a partial pressure of the gaseous solute.

[00:44:53]P is a partial pressure of the gaseous uh solute above the solution. Gaseous solute above the solution.

[00:45:06]Okay. So that is Henry's Henry's law. Henry's law. Okay. So it states that the the partial pressure the partial pressure of the gas is directly uh proportional to the to the concentration to the concentration. Okay. Pressure direct proportional to the to the concentration. Okay. So we can we can calculate the solubility.

[00:45:42]We can calculate we can calculate the solubility.

[00:45:50]We can calculate the the solubility of a gas at a new temperature using H's law. We can calculate the solubility of a gas of a gas at a new temperature at a new pressure at a new pressure using using Henry's law Henry's law.

[00:46:18]Okay. So the formula is given as new new solubility new solubility new solubility is equal to is equal to solubility solubility time new pressure solubility time new pressure new pressure over OD pressure.

[00:46:54]Okay. [clears throat] So if the pressure changes, we can as well calculating the solubility, new solubility.

[00:47:03]So let's have an example.

[00:47:06]Uh let's see how we can uh calculate calculate the new solubility.

[00:47:17]Okay. Example.

[00:47:27]Example.

[00:47:30]Um, what is the solubility?

[00:47:37]What is the solubility?

[00:47:41]What is the solubility?

[00:47:44]What is the solubility of oxygen of oxygen gas? What is the solubility of oxygen gas at 25°C?

[00:47:57]at 25° C and a partial pressure and a partial pressure and a partial pressure of 150 atmospheres 1,150 atm I mean uh let's use to okay to okay uh to uh if The solubility of oxygen.

[00:48:30]If the solubility If the solubility of oxygen If the solubility of oxygen is 0414 g per 100 m liter ml at at 25° C at 25°C Celsius and 7 and 7 east to okay so this is the question what is the solubility of oxygen gas at 25° C and a pressure of [snorts] of 1502 if the solubility of oxygen is 0414 g per 100 ml 25° and 762.

[00:49:31]Solution uh new solubility is equal to solubility time new pressure new pressure over o pressure.

[00:49:51]Okay. So we substitute the solubility given it's there uh which is equal to 0414 g per 100 per 100 ml then times our new pressure our new pressure our new pressure that is uh 1150 okay uh 1150 1150 to okay over odd pressure odd pressure that is 760 7 to so you can see that the to cancels then you do your arithmetic we are going to find the new solubility 0626 g per 100 m liter. Okay. So that is how we can calculate to find the new solubility.

[00:51:05]So let's have another example on Henry's law. A soft drink.

[00:51:13]A soft drink. A [snorts] soft drink uh is bottled.

[00:51:20]A soft drink is bottled is bottled so that so that at at 25° C at 25° uh Celsius uh contains carbon dioxide contains CO2 contains the carbon dioxide the gas. Gas at a pressure at a pressure of 5.08 atm 5.08 atm. Okay, that's a pressure. Okay.

[00:52:06]Over the liquid. 5.0 atm over the liquid.

[00:52:14]Over the liquid. Assuming Assuming the the partial pressure of carbon dioxide.

[00:52:25]Assuming the partial assuming the partial pressure partial pressure of CO2.

[00:52:35]Assuming the partial pressure of carbon dioxide in the atmosphere.

[00:52:43]in the atmosphere in the atmosphere uh is 4.0 0 * 10 ^ -4 ATM atm Okay. So, calculate calculate the equilibrium calculate the equilibrium equilibrium uh concentrations.

[00:53:14]Calculate the equilibrium concentrations of carbon dioxide of carbon dioxide.

[00:53:23]of carbon dioxide in the soda.

[00:53:30]In the soda in the soda.

[00:53:34]Both before.

[00:53:36]Both before and after.

[00:53:43]both before and after and after the bottle is opened.

[00:53:50]Okay. After the bottle is opened is opened. Okay. Then Henry's law constant Henry's law constant for carbon dioxide is equal to 3.1* 10 to the power -2 more per per liter atmosphere.

[00:54:23]Okay. So this is the question. A soft drink is bottled so that at 25° C contains carbon dioxide gas at a pressure of 5.08 atm over the liquid. Assuming the partial pressure of carbon dioxide in the atmosphere is 4.0 * 10^ -4 atm.

[00:54:46]Calculate the equilibrium concentrations of carbon dioxide in the soda both before and after the bottle is opened.

[00:54:55]So this is Henry's law constant solution uh solution. So uh the first one is before before the bottle before the bottle is opened. Okay. So we know that C is equal to so concentration of carbon dioxide is equal to K * P.

[00:55:21]Okay. Henry's law is equal to K KP. Okay. Concentration of carbon dioxide is equal to K of carbon dioxide.

[00:55:33]His law constant for carbon dioxide time the partial pressure of carbon dioxide.

[00:55:39]So before it is opened it means before it has been exposed to the atmosphere.

[00:55:45]Okay. So we are calculating now the concentration of carbon dioxide in the bottle before it is opened. So K is equal to is = 4.0 * 10 ^ -4 atm.

[00:56:03]Okay I mean uh sorry the K that is atmospheric pressure. Okay. So Henry's law constant which is 3.1 * 10 ^ -2 more per liter atmosphere.

[00:56:24]Okay. Then the partial pressure of carbon dioxide in the bottle is 5.0 5.0 atm. Okay. So you cancel the ATM you cancel then you you multiply you multiply uh you are going to find uh 0.16.

[00:56:46]So the concentration is 0.16 more per liter. Okay. Then we calculate after after bottle after the bottle is opened. after the bottle is opened. Okay. So what we mean that by that after you opened the bottle it means now uh the the pressure the pressure will be equal to the atmospheric pressure since it has not been exposed to the atmospheric uh pressure. Okay. So in the opened bottle the carbon dioxide in the in the in the soda reaches equilibrium with the atmospheric uh pressure. Okay. So what changes is is the pressure. Okay. So C is equal to KP. Okay. So in this case our K remains the same 3.1 * 10 ^ -2 more per liter.

[00:57:57]atmospheres. Okay. Then the atmospheric pressure the atmospheric pressure is 4.0 * 10 ^ -4 * 10 ^ -4 atm okay then you cancel. So we are going to find 1 1.2 okay 1.2 22 * 10 ^ -5 mo per liter. Okay. So you can see uh not the large change not the large change uh in concentration of carbon dioxide.

[00:58:40]Okay. You can see when the bottle is opened you can see how the concentration has has reduced. Okay. So this is why soda gives it goes flat after being opened for uh for a while. Okay. So uh uh these are some of the examples of uh Henry's uh law. Okay. So application of this Henry's law uh it is used for uh deep sea divers also mountain uh climbers. Okay, those who rely on oxygen you know that at high altitude there's less oxygen uh at high altitude. Okay.

[00:59:36]So there are so many applications of of Henry's uh law that you need to to look at. So you can you can do some research on Henry's law applications. You read more okay uh applications of Henry's Henry's law. Okay. You can read more on on that. Okay. So now the next topic that we shall look at is um the the next one that we shall look at in the next lecture uh is the vapor pressure of solutions. Thank you.