 Okay, so write down the question here a solution of I'll write down the question wait Solve this question all of you a solution of non-volatile solute non-volatile solute in water freezes at freezes at minus 0.80 Degree Celsius The vapor pressure of pure water. So I'll write down in short P0 of H2O the vapor pressure of pure water. It is given 23.51 just a second 23 0.51 MM Hg kf of water is given Which is 1.86 Calculate the vapor pressure The P0 of the solution We have a pressure of solution tell me You need to find out the vapor pressure of solution Tell me fast anyone Okay, try this Anyone got the answer tell me 23.41 23.41 is right. It's very close to the right answer 23.41 is correct Is he what we have to do Here first of all first of all Solution of non-volatile solute is given and it is not mentioned. That's what the solute is electrolyte or non-electrolyte Okay, so if it is not mentioned Then we have to assume that the solute is non-electrolyte Okay, we have to assume that Non-electrolyte solute if it is not given Non-electrolyte solute we have to assume first of all this Okay, now freezing point is given so we'll use this formula of delta tf is equals to i into Kf into m Now when the solute is non-electrolyte, so i is equals to what one Right, but we need to find out the solution pressure solution pressure is what? p0 a minus p a divided by p0 a is equals to i Xb also we can write where i is equals to one again, right? So this p we need to find out Solution pressure this p we need to find out p0 a is given Right p0 a is given what we need we need this xb Mole fraction of b because that is not given in the question mole fraction of b. How do we find out? delta tf is Equals to kf into m delta tf is this so molality first of all we'll find out from here Molality is delta tf 0.80 Divided by kf is 1.86 this molality is around 0.43 Molal solution it is Okay This formula you see if I write this as P0 a another wave I write P0 a minus p a Divided by P0 a Is equals to nb Divided by n a plus nb mole fraction of b Okay, since the solution is dilute all these solutions are very dilute. So n a nb n a is very high in Comparison to nb like this also we can solve or you can solve in other way also that you must have done already So p0 a is what p0 a is 23 point five one So you have to do 23 point five one minus p a which you need to find out divided by 23 point five one and This is nothing but since nb is very small So we can neglect nb in terms of with respect to n a and the ratio is this Right so nb is what nb is This kf unit is what it is Kelvin per molal, right? But one mole this is a unit of this So if nb and a is the moles of water that will be 1000 gram of water we are assuming divided by 18 and Nb will be what Nb will be 0.43 Because you see molality is given you can also find out this in this way molality is what number of moles of solute which is b divided by mass of solvent Right mass of solvent which is water mass of water into 1000 right so mass of water if you are assuming thousand gram Right. This is equals to 0.43 the molality of the solution mass of water Suppose it is 1000 gram that we are assuming here because it is not given or 1 kg So number of moles of B, which is the solute is nothing but 0.43. That is what I have written here Okay, this when you solve nb by n a we are getting as 0.0 0.43 into 18 divided by thousand So when you solve this you'll get pa which is around 23 point 32 mm Hg This is what you need approximation. I have taken with n a plus nb. I am writing it as Nb I am Neglecting with respect to n a all of you understood this Understood. Let me know if you have any doubt Okay, this question you see 0.1 molar aqua solution of mg cl2 of mg cl2 at 300 Kelvin is 4.92 Atmospheric what will be will be the percentage ionization of The salt what will be the percentage ionization of the salt anyone got the answer The first part of the question is 0.1 molar aqueous solution of mg cl2 at 300 Kelvin is 4.92 atmospheric This is again 0.1 molar aqueous solution mg cl2 One thing is missing here the osmotic pressure. This is not given here some mistake here the osmotic pressure of 0.1 molar aqueous solution of mg cl2 at 300 Kelvin is for 4.92 atmospheric The pressure is given see osmotic pressure is given Concentration is given Okay, alpha you need to find out Percentage ionization means alpha you need to find out and to find out alpha we need I So write down the formula for osmotic pressure Calculate I from there and then use the formula of I to get alpha 0.5. Shiram is getting 0.5 close. It's right Yeah, approximately 50% is right Pretty you got that What about Sahana, Laxia? anyone See what we have to do The osmotic pressure pi is equals to we need to write I C R and T I We do not know pi is given in the question which is 4.92 divided by Concentration is 0.1 our value is what 0.0 82 and T is what the temperature is given 300 Kelvin when you solve this I You'll get around 1.99 Okay, or if you take it as 2 then since it is the dissociation case. So I is equals to 1 plus n minus 1 Into alpha n value is what mg cl 2 we have So n value will be mg 2 plus Plus 2 cl minus so n value is 3 here So this will be equals to 1 plus 2 alpha and I is nothing but 2 so when you calculate alpha here, you'll get 0.5 or 50% This is the answer What equipment you want Laxia all of you understood this next question you see a certain substance a certain substance that is a certain substance a tetramerizes Tetramerizes in water in water to the extent of extent of 80 percent a solution of of 2.5 gram of a a in 100 gram of water in 100 gram of water lowers the freezing point lowers the freezing point by 0.3 degree Celsius Find the molar mass of a Do this question anyone got the answer Graph of Delta TB Elevation in boiling point graph. That's what you're asking pretty Okay, so we'll do that pretty we'll discuss that but in the last okay not now If not possible, then we'll discuss that in the glass Anyone got the answer? Okay, I'll do this You see the certain substance a tetramerizes in H2O To the extent of 80 percent means the alpha value is 0.80 Solution of 2.5 gram of a so mass of solute is given That is so I'll write down MB not MA here Mass of solute we have so M of B B solute So mass of MB is 2.5 gram the mass of solvent which is water a is 100 gram and Delta TF is given Which is 0.3 degree Celsius Okay, so what we'll do here first of all we know the Formula of Delta TF is equals to is equals to I Kf into M So first of all we need to find out this I So I is equals to it is a case of association tetramerizes we have so I is equals to One plus one minus One minus one by N Into alpha N value is given in the question which is four so one minus One minus zero point two five and alpha is zero point eight zero So one minus zero point seven five into zero point eight zero and This is nothing but zero point four zero the value you get Okay, I is zero point four zeros. I'll just substitute here Delta TF is zero point three I is zero point four KF for water is one point eight six We know that molality is number of moles Divided by the mass of water is hundred into thousand So the number of moles Will be equals to mass by molar mass mass is two point five Molar mass of B. We need to find out is equals to this Hundred and this will get cancelled and number of moles will be equals to point three Divided by ten into Zero point four into one point eight six and Then when you calculate MB here the mass or the molecular mass MB is equals to Approximately you will get 62 gram That is the answer So I know is getting the right answer Understood this hundred plus two point five. How can you assume that pretty? 60 gram is approximate. Is it clear all of you?