00:01
Hello students, so the question is to calculate the freezing point of the solution which has the exosmotic pressure of 10 .9 atmosphere.
00:11
So, in step 1 by the help of the exosmotic pressure formula we will be calculating.
00:20
So, the step 1 is to calculate is to calculate molarity since 5 which is the exosmotic pressure is equal to i m which is molarity into rt.
00:35
So, here i is the vent of factor which is 1 for the glucose.
00:44
Since here the glucose is not dissociable in the water so the vent of factor is equal to 1 and m is the molarity.
00:54
So, here m is the molarity which we need to calculate so it is molarity, r is the universal gas constant and it is equal to 0 .0821 atmosphere liter atmosphere per mole per kelvin and t is the temperature which is given and it is 298 kelvin.
01:23
So, by putting the values over here in the formula of exosmotic pressure we can calculate for m which is molarity and it is equal to 10 .9 atmosphere which is the value of exosmotic pressure divided by the vent of factor which is 1 into the r which is the gas constant and it is 0 .0821 liter atmosphere per mole per kelvin multiplied by the temperature which is 298 kelvin.
01:57
So, by calculating it we will get the molarity and it is equal to 0 .445 mole per liter.
02:11
So, this is the molarity.
02:13
So, from molarity we got to know that 1 liter of solution consists of 0 .445 moles of the glucose solute.
02:30
Now, in step 2 we will be calculating is to calculate molality.
02:38
So, the molality which is denoted by small m and it is equal to the number of solute which is denoted by the number of moles which is denoted by n divided by the weight of solvent in kgs.
03:00
So, here we know from the step 1 we know that 1 liter of solution consists of 0 .445 moles.
03:09
So, here the number of moles which is n is equal to 0 .445 moles.
03:17
Now, to calculate for the weight of the solvent it will be so here the weight of the solvent is equal to the density into volume since the density which is noted by rho is equal to mass upon volume.
03:34
So, from here we will be taking the volume which is v is equal to 1 liter and density which is given and it is 1 .031 kg it will be 1 .031 gram per ml.
03:52
So, from here we can calculate the weight of the solvent.
03:58
So, by putting these values we will get the weight of the solvent which will be 1 .031 kg.
04:05
So, this is the weight of the solvent...