Ascorbic acid (vitamin C, C₆H₈O₆) is a water-soluble vitamin. A solution containing 80.5 g of ascorbic acid dissolved in 210 g of water has a density of 1.22 g/mL at 55 °C. Calculate (d) the molarity of ascorbic acid in this solution.

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Hello everyone today. We have the following problem assuming that volumes are additive. Find the concentration of each ion in the solution obtained when 3.8 g of magnesium bromine is dissolved in 200 ml of .0150 moller of sodium bromide. So before we do that we must convert this volume into leaders. And so we're just gonna take our 200 ml and multiplied by the conversion factor. That one leader Is equal to 10 to the negative 3rd middle leaders. This is gonna give us 0.2 leaders until we use that for later. And so next we want to draw the dissolution of both of our solutions are both of our products. So we have our magnesium bromide, we're just going to dissolve into magnesium two plus ions. And then we're gonna have to bromide minus ions. We're gonna do the same thing for our sodium bromide which is going to dissociate into sodium plus ions as well as our bromine minus ions. And so next we're going to find our concentration of magnesium ions. And so how we're gonna do that is we're simply going to take our moles of magnesium and divide that by the solution in leaders. So since we don't have the moles of our magnesium outright, we're gonna have to sell for it. So we're going to take our mass of our magnesium bromide and then we're going to multiply by a couple of conversion factors first we're gonna multiply by the molar mass of magnesium bromide which can be found using the periodic table If we take our mass of two bromides and then one magnesium and at them we get 84.1 g. And lastly, we're going to multiply by our multiple ratio in that in one mole of our magnesium bromide, we only have one magnesium. So we see that we only have one mole of magnesium. We're then going to take that and divide by our volume, which is .2 leaders. And we see that if we cancel out our grams of magnesium bromide and our moles were simply left with moles of magnesium over leaders giving us our polarity of 0. moller for our magnesium. So that's our concentration for magnesium. We're then going to find our concentration for our sodium ions. And so simply to do this, we're going to take however many moles of our how many moles of sodium that we have in this sodium bromide, which is one. We're gonna multiply by its concentration which is point 0150 Moller. In doing this, we're gonna get a concentration for sodium ions as .0150. And so now we have to find our concentration of our bromide from our magnesium bromide. So we're going to put parentheses from our magnesium bromide because we have two of them. And so as before, we're going to take the moles and divide that by leaders. However, since we don't have it out, right, we're gonna have to solve for it. So we have our 3.8 g of our magnesium bromide As before. We're gonna multiply by its molar mass. So one mole of magnesium bromide Is Gonna Equal 1 84.1 g. We're gonna multiply by the multiple ratio in that. In one mole of our magnesium bromide, we have two roaming atoms. So we're gonna say you have two moles of roaming When our units canceled out. We're gonna be left with a concentration of 0.206 molar for our bromide. And then lastly, of course, we have to find our concentration of bromide ions from our sodium bromide. And just as we did before, we're gonna take the number of bromide ions in that and then multiplied by its concentration which is 0.0 1 50 moller Of our sodium bromine to give us 0. dolar of bromine ions. And so since it's said that volumes are additive, what we're gonna do because we're going to add our bromide ions. So we're gonna have our 0.206 Mueller for bromine ions from magnesium bromide and then add our zero 0.1 50 moller bromine ions from our sodium bromine to give us 0.2 to one Moeller of bromine ions and this is going to be total. This is essentially going to give us the following answers. So we're gonna have our 0.103 molar of magnesium ions, we're going to have our 0.1 50 moller of sodium ions, and then we're going to have our 500.221 molar of bromide ions total. And with that we have answered the question overall, I hope this helped, and until next time.

Ascorbic acid (vitamin C, C₆H₈O₆) is a water-soluble vitamin. A solution containing 80.5 g of ascorbic acid dissolved in 210 g of water has a density of 1.22 g/mL at 55 °C. Calculate (d) the molarity of ascorbic acid in this solution.

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Ascorbic acid (vitamin C, C6H8O6) is a water-soluble vitamin. A solution containing 80.5 g of ascorbic acid dissolved in 210 g of water has a density of 1.22 g/mL at 55 °C. Calculate (d) the molarity of ascorbic acid in this solution.

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Ascorbic acid (vitamin C, C₆H₈O₆) is a water-soluble vitamin. A solution containing 80.5 g of ascorbic acid dissolved in 210 g of water has a density of 1.22 g/mL at 55 °C. Calculate (d) the molarity of ascorbic acid in this solution.