00:01
Hello, everyone.
00:02
This is ricky, and today we're working on problem number 60 from chapter 11.
00:06
And we are going to be solving for the malarity, the molar mass of insulin, given the osmotic pressure.
00:17
So here we have the formula for osmotic pressure.
00:21
Let's rearrange it to solve for malarity.
00:25
So malarity is equal to osmotic pressure over the gas constant times temperature.
00:32
And remember, molarity is equal to moles over liters of solution.
00:43
So we can actually rearrange this further by solving for moles.
00:55
So we get moles of insulin is equal to the osmotic pressure times the amount, the liters of or the volume of the solution over the rate of the gas constant times temperature.
01:18
And so molar mass is equal to one gram or is equal to grams per mole.
01:31
And so if this is the moles of insulin used, we can plug this into here and then solve directly for a molar mass by using the grams of insulin used in the solution.
01:52
So let's write out our variables.
01:56
Moor mass equals unknown.
02:01
Insulin is equal to 7 .0 grams...