In question 1 11. It talks about how external pressure must be applied more than the osmotic pressure in order to allow solvent from a hi concentrated solution on one side of a semi permeable membrane to send its water molecules through the membrane to the other side. That might be pure water. This osmotic pressure obeys the laws similar to the ideal gas law where P. B equals Nrt. It then shows you that if we replace P with osmotic pressure that we get an over V. Which is related to more clarity. So we have osmotic pressure being equal to polarity, multiplied by RT. With this equation, it asks you to determine the osmotic pressure at 25 degrees Celsius of a 250.20 molar sucrose solution. Well, sucrose only separates, well, it doesn't separate into anything one unit of sucrose creates uh one mole of sucrose. Well, one molecule of sucrose doesn't separate like ionic compounds do. So it's polarity is it's osmolarity. So when using the equation, osmotic pressure will be equal to the straight up polarity of the sucrose solution. 0.20 Multiplied by are multiplied by T. The r value is 0.8 to 06 Leader atmospheres per kelvin mole. So we'll end up getting pressure in units of atmospheres when using this equation. T however, needs to have units of kelvin because our our value has units of kelvin innit leader atmospheres per kelvin mole. Therefore will convert the 25 degrees Celsius. That is provided into kelvin temperature by adding to 73 and we get an osmotic pressure for the solution of 0.49 atmospheres. For part B. It talks about seawater containing 34 g of salts. For every leader of solution that wants you to assume that the salute consists entirely of sodium chloride and to calculate the osmotic pressure of seawater 25 degrees Celsius. We'll do the same thing as we did before. Osmotic pressure will be equal to the polarity multiplied by are multiplied by T. But this needs to be the osmolarity referring to the total polarity of all the salute ions in solution. So if we have 3.4 g of sodium chloride per liter, we can convert the gram sodium chloride into molds sodium chloride by dividing by the molar mass sodium chloride. So we will then have the grams will cancel will have units of mold sodium chloride per liter sodium chloride. That will be the more clarity sodium chloride. But the osmolarity of the solution because sodium chloride separates completely into sodium ions and chloride and ions will be two times that. So this whole thing right here will be the osmolarity or the end value. Well then multiply that by R. Value. And then the temperature again 25 degrees Celsius plus 2 73 to get our kelvin temperature and we get 285 atmospheres for part C. It states that the average osmotic pressure of blood is 7.7 atmospheres at 25 degrees Celsius. So what concentration of glucose will be isotonic with blood? Well, we'll use the same osmotic pressure equation but now we'll be solving for the osmolarity of a glucose solution. Knowing the pressure, if the pressure is 7.7 atmospheres, that'll be set equal to the osmolarity of the glucose solution. Multiplied by the R value. Same as always 0.8 to 06 Multiplied by the kelvin temperature. Which will be the 25 plus 2 73 or 2 98 kelvin rearrangement of this equation gives us an osmolarity of 315 Mueller for glucose because glucose is not separate into ions. It's osmolarity. Is it straight up polarity now for part D. Why says I'm is an enzyme that breaks bacterial cell walls. The solution that contains 0.150 g of the enzyme in 212 mL of solution has an osmotic pressure a 2120.953 tour at 25 degrees Celsius. And for this question, it asks you to calculate the molar mass of the list design. Well, if we go back up here, you'll notice where we put in the molar mass, we put in the molar mass right here. So if we know the pressure and we don't know the molar mass, then this is where our unknown variable will be found in our calculation. It'll look something like this. We'll take the pressure and tour and converted into atmospheres by dividing by 7 60. This will then be osmotic pressure and atmospheres. We'll set that equal to the mass that we have multiplied by one over the molar mass as we did up here. One over the molar mass. Well then divide that whole thing by the leaders of the solution. The two 10 mL can be converted into leaders by dividing by 1000. So now we have this right here. If we knew molar mass would give us the polarity of the license. I'm solution because the licensee does not separate into ions which was not necessarily obvious in the question. But if it's not a common ionic compound, you can assume that the more clarity is the osmolarity and it does not separate this whole thing right here then is the polarity. We will multiply that by the R. Value and then by the T. Value. And all we have now is an algebraic equation for which we can solve for the molar mass. So I'll save you some space here and not take you through all the algebra. But the molar mass ends up becoming 13,900 g per mole. Then for part E. Osmotic pressure of acquis solution of a certain protein was measured in order to determine the protein smaller mass. The solution contained 3.50 mg of dissolved protein in five mL. The osmotic pressure at 25 degrees Celsius was 1.542 Or what is the molar mass. So this will be another calculation very similar to what we did up here. We'll take the tour converted into atmospheres by dividing by 7 16. We'll set that equal to the mass, multiplied by one over the unknown molar mass in order to get moles. Well then divide the moles by the volume. In order to get more clarity. five mL can be converted into leaders by dividing by 1000. Mhm. This whole thing here then is the polarity. Well multiply that by there are value 0.8 to 06 Multiply that by the temperature. And then again, all we have left is an algebraic expression for which we will solve the molar mass. So without going through all the algebra. Hopefully, you know, enough to get through the algebra, you just need to isolate this essentially what we're doing is we're gonna multiply both sides by mm mm is going to come over here and then we'll divide both sides by this number right here. When we do that, we have mm equal to all of this over here. Except that part with this being divided on both sides. And we get 8910 g per mole for answer