00:01
So what is given to us is you have 115 grams of c -11 c -12 h -22 -o -11 in 350 grams of water and you have to find the elevation and boiling point.
00:22
So for the first part, let's try to derive the formula.
00:26
So we know delta t is equal to i.
00:35
B kb where i is band of coefficient b is molality and kb is ebulio scope constant and delta t is change in temperature okay so delta t is equal to delta t b solution minus this is not delta this is a calculating a difference now t b solvent so this is going to be the boiling point right and b which is molality is equal to number of moles of c12 h 22 o11 divided by mass of water okay this is important so to calculate number of moles of c12 h 22 o11 what you need is given mass divided by molar mass o11 okay so this is what you require to calculate this um solution what you have to do is we'll substitute these equations now so, n, c12 h -22 -011 is given mass over molar mass of water, sorry, molar mass of c12 h -22 -o -11.
02:59
So, molecular mass of c -12 h -22 -o -11 is 12 multiplied by 12 .211, 22, 21, 22, multiplied by 12, by 1 .01 plus 11 multiplied by 16.
03:22
So this gives us a total of 342 .34 grams per mole.
03:34
So, nc12 h -22 .011 is equal to given mass...