n isotonic solution must be approximately 0.30 osmol/L. How much KCl is needed to prepare 175 mL of an isotonic solution?

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Welcome back everyone. The concentration of an isotonic solution is approximately 0.750 osmoles per liter, determine the mass in grams of sodium bromide required to prepare an isotonic solution with a volume of 250 mL. So beginning with the fact that sodium bromide consists of the metal sodium and the nonmetal bromide. This is then an ionic compound which based on the solubility rules for bromides is actually going to be a soluble bromide salt and will fully dissociate into the sodium plus one cat ion and one mole of the bromide B R one minus anion. So it yields two ion products. And we're going to begin by determining the molar or molar concentration of sodium bromide in the solution. And so we would determine that by first beginning with the given factor from the prompt, which is that we have 0.750 osmoles. So that is os osmoles os M O L per liter, which we should recall that from our textbooks, one osmo per liter is equivalent to one mole of ions per liter. So we can interpret that as 0. moles of ions per liter. And we will then multiply to go from moles of ions in the denominator to then moles of sodium bromide in the numerator. Recall that for molar, our units should be moles per liter. And so we can interpret from the dissociation of sodium bromide that we've got two moles of ions per mole of sodium bromide in the numerator. So canceling out moles of ions, we are left with moles of sodium bromide per liter for our final units of molar. And that will result in the molar of sodium bromide as 0.375 moles per liter. So now with the molar figured out, we're then going to recall the formula from our textbooks where the moles of our substance can be found from taking the molar multiplied by its volume. And this is just an interpretation of our term molar which again, we stated is moles of our substance per liter. And so from our prompt, we are given the volume of the solution with sodium bromide as 250 mL. And so to solve four moles of sodium bromide, we would take its molar which we determined above as 40. moles per liter. And we will multiply by its volume in the solution as a whole being mL which we will multiply to convert. From then Millis in the denominator to leaders in the numerator where recall that our prefix Milly tells us we have an equivalent of 10 to the negative third of our base unit leader. So canceling out millis we are left with leaders for volume, which we can cancel out with liters from the 100.375 moles per liter term. And that leaves us with moles as our final unit. And this is going to simplify two moles of sodium bromide being equal to a value of 0.9375 moles of sodium bromide. So now with moles of sodium bromide determined, we can finally get to the mass of sodium bromide that is required to prepare the solution in which we would say that we've got the rearrangement so that we've, we can say that the mass of our substance is going to be found by setting that equal to the moles of our substance multiplied by its molecular weight. And that can come from our textbooks and notes and so plugging in our nodes, we've got the moles of sodium bromide which we determined above as 0.9375 moles of N A B R, which is multiplied by its molecular weight from the periodic table, which we would refer as 102.894 g of sodium bromide per mole. And so canceling out units of moles of sodium bromide, we are left with grams as our final unit for mass. And we will then have the mass of sodium bromide required for the solution equal to a value of about 9. g of sodium bromide that is required. And this is going to be our final answer to complete this example. I hope that this made sense and let us know if you have any questions.

n isotonic solution must be approximately 0.30 osmol/L. How much KCl is needed to prepare 175 mL of an isotonic solution?

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Key Concepts

Osmolality

Isotonic Solution

Molarity and Mass Calculation