When 3.89 g of a nonelectrolyte solule is dissolved water (o make 245 mL ol' solution at 22 %C.the solution exerts a osmotie pressur of 933 tOrr: What Ilhe mol...

A 188 mg sample of a nonelectrolyte isolated from throat lozenges is dissolved in enough water to make $10.0 \mathrm{mL}$ of solution at $25^{\circ} \mathrm{C} .$ The osmotic pressure of the resulting solution is 4.89 atm. Calculate the molar mass of the compound.

Sucrose, which is table sugar, Just normal sugar. So which is C 12 h 22 0 11. And we first need to find it smaller mass capital them. So the way I like toe fall from all the mass is first by listing out each Adam each different kind of element and then taking its Stuckey metric amount. So basically, the number of each Adam I'm multiplying that by its atomic mass Can't This list is in grams per mole again. All right, that I'll top there. Just gonna use those numbers. And we basically get 342 grams per mole of stickers. So it's simple. We always do was then add everything else. Okay, so now that we have that, we've also been given the mass of sucrose to be 17 0.1 cramps and we're looking for its morality, probably in water. Okay, So morality, then is moles over mass and usually normally moles of solute. And now most of solvent or massive solvent over massive solvent. Okay, so let's just scroll down. Sure. So first we need to find the most off salute. So if you do that, we just know that moles is equivalent to masked by molar mass. We have the mass appear 17.1 g and we have the Mueller Mass down there. 3 42 of them. If you just divide those two together we get 0.499 Most soiled. Yeah. Yeah, I believe we were also given the mass of the water. Yes, which is the solvent. What? So then that mass was zero point 275 g or kilograms? Kilos of water. So now we just take that The molds that we just found of solvent and divide that by mass off water. I'm sorry, I can't spell That's an l morality. Okay, so if I can scroll down came in. So just to reiterate it Morality Moles of salute over mass in kilos of solvent not a solution solvent. So then that gives us zero point one 18 17 morality units. Most per kilo came. Let's just quickly summarize what we did. We first. We know that morality is most of Seoul ute over solvent. So we took the massive salute, divided it by the molar mass of solute to find its malls and then divided models over the mass of the water into

Well, first, um, calculate the number of moles. 35 grams of the, uh known electrolytes were not told exactly what it is, but we're told that the molar mass is 58 grams for more so we can calculate numbers of moles, which is 580.6 serum. Oh, we can then calculate the morality, which is moles of Saul Utes over kilograms of water. We're told that we have 600 grams of water, which is 6000.6 kilograms of H 20 and then that leaves us with the morality of just one. We can now calculates the boiling point elevation. Since it's a non electrolyte, we do not need the Vantaa factor. The constant for water is 0.512 degrees Celsius for Melo will ality and the boiling point elevation is 0.0.512 degrees Celsius. The come bowling point of pure water. Uh, at uh during this experiment, we're told his 99.7 to 5 degrees Celsius plus the elevation of 50.512 degrees Celsius, would leave us with a boiling point elevation of 100.237 degrees Celsius.

Well, first, um, calculate the number of moles. 35 grams of the, uh known electrolytes were not told exactly what it is, but we're told that the molar mass is 58 grams for more so we can calculate numbers of moles, which is 580.6 serum. Oh, we can then calculate the morality, which is moles of Saul Utes over kilograms of water. We're told that we have 600 grams of water, which is 6000.6 kilograms of H 20 and then that leaves us with the morality of just one. We can now calculate the boiling point elevation. Since it's a non electrolyte, we do not need the Vantaa factor. The constant for water is 0.512 degree Celsius for Melo morality and the boiling point elevation is 0.0.512 degrees Celsius. The come bowling point of pure water Uh, at uh are during this experiment, we're told his 99.7 to 5 degrees Celsius, plus the elevation of 50.512 degrees Celsius, would leave us with a boiling point elevation of 100.237 degrees Celsius.

Okay, so we've got to salute here. Right. X. And sugar. So let's start by defining X as the mass of X. So then we'll have .10 -X. Will equal the mass of the sugar because the total was pointing one g. I'm also given a freezing point depression. So let's go ahead and use our equation to find the morality here Changed by 5° K. F for water is 186. So our morality comes out to be .269. And morality is moles per kilogram of solvent. Well, we had one g of solvent, so that's going to be .001 kilograms. So that will give us a total of 269. So I'm sending -4 malls of solitude. So the malls of the sugar plus the moles of the X. Have to equal our total moles. So our total moles is 2.69. So I'm sending negative four. So that's going to equal X grams of X. And let's change grams to malls. We're told that has a molar mass of 4, 10 plus .1 -X. And that's the grams of our sugar. So let's go ahead and change that moles. And that's 3 42. Okay, so our total moles here equal the moles of X plus the moles of sugar. So I'm gonna simplify these. Okay, we're gonna do 1/4 10 1/3 42. So basically we're just trying to solve for X now 2, 4, 4 times 10 to -3 times X. Plus We're going to have to 9 to I'm still in the negative four -2.92. I'm saying the -3 X. So that will give us 2.3. I'm sending negative five 4.8 times. Send a -4 X. No, we'll go ahead and solve for X. And that's going to come out to be .047. And we define that as the mass of X. So our percentage of X in our Solution is going to be the .047 g Of X, divided by the total mass, which was .1 grams. And if we multiply by 100, we'll get that. It is 47 of X.