Which of the following solutions will give rise to a greater osmo...

Question

Which of the following solutions will give rise to a greater osmotic pressure at equilibrium: 5.00 g of NaCl in 350.0 mL water or 35.0 g of glucose in 400.0 mL water? For NaCl, MW = 58.5 amu; for glucose, MW = 180 amu.

Accepted Answer

Welcome back, everyone. Choose the solution that will result in the greatest osmotic pressure at equilibrium. We have choice A which states 42 g of calcium bromide in 450 mL of water with a molecular weight of 199.89 g per mole. Choice B which states 65 g of sucrose and 550 millis of water with a molecular weight of 342.30 g per mole. Choice C which states 57 g of potassium iodide in ml of water with a molecular weight of 166 g per mole. And choice d which states that all of the solutions will result in the same osmotic pressure. So the key to getting to our answer is going to be understanding that the increase in osmolarity for our solution will correspond to an increase in osmotic pressure at equilibrium for the solution. So in simple terms, the higher the osmolarity we would have therefore a higher amount of dissolved particles in the solution resulting in a higher osmotic pressure at equilibrium. Next, let's also recall that to calculate osmolarity, we would find this by taking the concentration of our solute. So c represents that variable. This is our concentration of our solute in units of moles per liter, which also describes polarity. And so this variable C is then multiplied by I which represents our van Hoff factor. And recall that our van Hoff factor describes our average number of particles which can be either ions or molecules formed per formula unit of solute when dissolved in a solution. So for shorthand, we'll just say it describes particles. So to get to the correct answer, we're going to need to calculate the osmolarity of each answer. Choice. Let's begin with choice. A Where we're given a solution of calcium bromide CABR two. Now recognize that this is an ionic compound. And so this will based on our solubility rules be aqueous and dissolve fully into the following ions. We reform our calcium two plus Kion and our bromide anion. And notice that we have two moles of bromide on the reactant side. So we'll place a coefficient of two in front of our bromide anion on the product side. Now, we can see that from this fold association of our calcium bromide we have on our product side, the formation of three dissolved ions. So three moles of dissolved ions and we can interpret this as therefore, three particles, meaning our van Hoff factor is equal to three. And so to calculate our osmolarity of this solution. We would take the given mass of our calcium bromide as 42 g. And we want to convert to get to units of moles per liter. For our term c again, our concentration of our solute. So multiplying by our conversion factor to go from grams of calcium bromide in the denominator to moles of calcium bromide in the numerator, we're going to recall our molecular weight given in choice a as 199.89 g in the denominator for one mole in the numerator of calcium bromide and canceling out grams of calcium bromide. We have moles of calcium bromide and we're gonna take this entire first calculation and place it around brackets so that we can divide to introduce our unit of liters so that we have the volume as 450 mL given in choice A for the solution of calcium bromide, which we will multiply in the denominator by a conversion factor to go from milliliters of calcium bromide to liters of calcium bromide solution in the numerator where we recall that our prefix milli tells us that we have an equivalent of 10 to the negative third of our base unit leader in the numerator. So canceling out milliliters of calcium bromide, we're left with liters of calcium bromide for our denominator. We'll also place this in brackets and the entirety of our quotient here in brackets represents our terms C and now we're just going to multiply it by our osmolarity, which we determine to be three, specifically three particles. So in our calculators, we will simplify and actually, we'll show this step by step. So we would simplify. So that in our next step, we have for our numerator, a value of 0. moles of calcium bromide. And then in our denominator, we would get 0.45 L of calcium bromide. This quotient in parentheses is still multiplied by our three particles or Van Hoff factor. And in our next line, we would simplify so that we have the product between the results of our quotient being 0.46693 times three particles, Which gives us our osmolarity of our calcium bromide solution equal to about 1.4. And sorry, before I continue for our previous step, our units are moles per liter times our particles. And our osmolarity will be in units of osmoles. And sorry, Ozal should have a capital O. So os moles per liter, which again are equivalent to moles per liter times particles or we can say os molar. So we need to compare the rest of our choices to this calculation in choice A for the osmolarity. Moving on to choice B, we're considering 65 g of sucrose. And we want to recall that sucrose has the following chemical formula C 12 H 22 0 11 and recognized that each atom in this chemical formula is a nonmetal. So this is a molecular compound and therefore, this is a non electrolyte because notice that sucrose is actually a sugar and therefore, it does not dissociate into ions when dissolved in water because the molecule of sugar will remain intact in solution without breaking up into smaller particles. And so each sugar molecule in solution of water will just contribute as one particle. And due to all of these regions, we would say that the van Hoff factor is equal to one or one particle. So to calculate the osmolarity for this Suro solution, we would begin with the mass of 65 g of sucrose C 12 H 22 11. And we're going to multiply by a conversion factor to go from grams of sucrose in the denominator to mold of sucrose in the numerator. Recall our molecular weight given in choice B for sucrose solution as or just for sucrose, the molecule as 342.30 g of sucrose per mole of sucrose, one mole of sucrose in the numerator, canceling out grams of sucrose. We have mold of sucrose in our numerator. We're going to place this entire calculation in brackets and divide by a second set of brackets in the denominator where we're going to introduce our unit of volume. So beginning with the volume of our solution in choice B being 550 mL. We want to multiply by a conversion factor to go from milliliters in the denominator to liters in the numerator. And again, our prefix Milly tells us that we have an equivalent of 10 to the negative third power of our base unit leader in the numerator canceling out milliliters. We are left with leaders in our denominator and now multiplying our quotient in brackets by our van Hoff factor which we determine to be one particle and simplifying in our next line, we have our numerator being the result of 0.189, 89 moles of gros Divided by the result of our denominator which is 0.55 L. So then this is quotient is now placed in parentheses and multiplied by a van Hoff factor of one particle. And in our next line, this simplifies to the product between 0.345255 moles per liter, Multiplied by one particle. And from this product, we get our osmolarity of our solution as about 0. osmos per liter or Oz Mueller. And notice that we have a lower osmolarity which will correspond to a lower osmotic pressure at equilibrium. And so we would rule out choice B let's move on to choice C which considers potassium iodide solution. Recognized that potassium iodide is an ionic compound and this is not one, this is not an exception of our soluble iodide. So this is a soluble iodide and therefore will fully dissociate into the following ion as our potassium plus one cat and our I died one minus anion. I notice that we form two products of dissolved dissolved ions. So we have two moles of dissolved ions. Therefore, we have two particles in solution. And our van Hoff factor is equal to two. And so calculating for our osmolarity Of the solution, we have the mass of the solution as 54 g of potassium or sorry, the mass of our solute as 54 g of potassium iodide multiplying by our conversion factor to go from grams of potassium iodide in the denominator to moles of potassium iodide. In the numerator, we recall that our given molar molecular weight of potassium iodide is 166.0 g of potassium iodide equivalent to one mole in the numerator, canceling out grams of potassium iodide. We have moles of potassium iodide left. We're going to place this calculation in brackets and divide it by our denominator which will also be in brackets where we're given our volume of water that this potassium iodide is dissolved in as 350 mL of solvent. So multiplying by our conversion factor to go from milliliters, Two leaders in the numerator recall that our prefix milly tells us that we have an equivalent of 10 to the negative third of our base unit leader in the numerator canceling out. Mill leaders were left with leaders in the denominator and now multiplying our quotient in brackets by our van Hoff factor of two particles. We're going to continue to simplify our calculations. So just moving down the screen for more room, we have the results for our numerator being a value of 0.3, 253 moles of potassium iodide. And in our denominator, we have the result as 0.35 liters of water solvent, we're still multiplying this quotient in parentheses now by two particles and this will simplify as a product between 0.92943 moles per liter multiplied by two particles in which we have our final osmolarity for our potassium iodide solution as about 1.85 osmoles. And this should be a capital O again. So osmoles per liter or os molar and between choice A and C for choice A, we have 1.4 osmoles per liter. And so looking at our answer choices, we can rule out choice A because the osmolarity is lower than choice C and we've already ruled out choice B. We can see that we do not have the same osmotic pressure for each solution. So we can rule out choice D. And this leaves us with choice C as the only solution that has the greatest osmotic pressure at equilibrium due to it having the highest osmolarity value as 1.85 os molar. So C is our final answer. I hope this made sense. And let us know if you have any questions.

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Which of the following solutions will give rise to a greater osmotic pressure at equilibrium: 5.00 g of NaCl in 350.0 mL water or 35.0 g of glucose in 400.0 mL water? For NaCl, MW = 58.5 amu; for glucose, MW = 180 amu.

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