A solution of an unknown salt in methanol is found to have...

A solution of an unknown salt in methanol is found to have an osmotic pressure of 4.75 atm at 300.0 K. All that is known about the salt is that has i=3. If the solution volume is 25.00 ml and it contains 0.335 g of the unknown salt, what is the molar mass of the unknown? Use R = 0.08206 L\cdot atm/mol\cdot K. 43.8 g/mol 119. g/mol 208. g/mol 69.5 g/mol

Transcript

00:01 So in the question, we need to find out the three different different part.

00:04 Along with that, we are going to be a first finding out a molality.

00:08 Malarity we need to find out in the very first.

00:10 So according to the given data, molarity is given 1 .0 to 1 molar.

00:15 Now we are going to be assuming over here.

00:19 Let's assume the mass of solvent.

00:24 The mass of solvent we are going to be assuming over here is 1 kg.

00:29 And the mass of i can write it the mass of any seal is provided us is 1 .021 so the mass of i can write it so the mass of 1 .021 mole of nacl is equal to 1 .021 multiplied by 5 8 .44 so our answer when we are going to do the calculations we are going to get 59 .67 gram and the total mass of solution which is provided in the the question is i can write it over here total mass of solution is equal to 59 .67 multiplied by 10 risk to minus 3 plus 1 kg so solving this we are going to get it 1 .059 k khae or we can write it over here one zero five nine then this much gram so volume of solution i can write it after that we are going to be writing in the next step the volume of solution we are proceeding we are finding now the volume of solution so putting up over here 1059 gram multiplied by ml divided by 1 .037 gram doing the calculations we are going to get it 1 .01 .21 .m.

02:08 Or we can write it instead of that in the liter 1 .021 liter it's going to become our like answer of the volume of solution.

02:19 Next i can write it that the thing which we need to find out that will be the what.

02:25 Mularity is equal to how much is going to become 1 .021 divided by 1 .0 divided by 1 .0 2 1 .0.

02:38 So it will become a 1 molar.

02:41 So this is our required solution for our part 1 is completed.

02:47 Now moving ahead, i would like to represent the part 2 in which we need to find out whether the thing or we can say whether the solution is a heterogeneous or homogeneous or saturated.

02:58 So i can write it the thing which is provided at a hundred degrees salubility we are provided over here is equal to 250 gram divided by thousand emil sorry 100 ml we are going to get it over here that is 2 .5 gram divided by m l and if we are going to be taking i can write if we take 500 gram in 150 ml then then the concentration is going to become like this way 500 gram divided by ml divided by 150.

03:40 So our answer which we are going to get it is 3 .33 gram divided by emm.

03:46 So that is what we have calculated.

03:48 We have calculated that this much gram we have been taken in the 150 ml then this will be our answer.

03:55 So moving ahead, i can say that this is more than the given cell.

04:01 So at what at a hundred degrees celsius so our solution will be the i can write it a solution would be what it's going to become a saturated it will become a saturated solution or a solution it can be saturated one or it can be the solutions would be i can write it over here a solution may would be a hadrojoujoujou hydrogenous because some amount of the salute won't be dissolved.

04:42 So this is our required solution for the second part.

04:47 Moving ahead with a third part, i can write it over here because this is actually divided into the four different different section.

04:54 So this will be our third part...

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