Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. c. 5.5% NaNO₃ by mass (in water)

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Hello, everyone. Today we have the following problem. Calculate the freezing point and boiling point of each aqueous solution assuming complete association of the solute. So we have 5.5% of our sodium nit nitrate by mass and the solvent is water. So to do this, we need morality which is the moles of our salute divided by our kilograms of our solvent. So we want to assume 100 g of solution to make this a bit easier. And so what we will do is we have our or to find our moles of our salute, we have our 5.5 g of our sodium nitrate. You multiplied by its smaller mass. If we take three oxygens and have their mass added to the mass of one sodium and one oxygen, we get a molar mass of 85 g per mole. Now solving for this, we get 0.064706 moles. And then of course, we have to find our mass of our water in kilograms because that is our solvent and we take 100 g and we subtract that by the 5.5 g of our salute to get 94.5 g. However, we must convert this to kilograms as our units. So we have 1 kg is equal to 10 races of the power of 3 g. And this gives us 0.0945 kg. We can then solve for our morality. So our moles of solute which was 0.064706 moles can be divided by the kilograms of solvent which was 0.0945 kg. And we get a modality of 0.68472 bals. Now, first in calculating our freezing point depression, we use the formula that it is equal to I which is the VHA factor multiplied by the constant BKF constant and they're multiplied by the morality. So what is Van Hoff factor? Well, because this is sodium nitrate, this will dissolve when it's in an aqueous solution into sodium and nitrate ions. So you have two ions. So our Van Hoff factor is two, they multiply that two by the constant or the freezing point, constant of water, which is 1.86 degrees Celsius divided by Malas, they multiply by the malas that we have, which was 0.68472. And then if we solve, we get a freezing point depression of 2.55. However, we must take into account that the freezing point. So the freezing point is equal to the freezing point of the solvent plus the freezing point depression. So it's zero degrees because that is the freezing point of water. And we subtract that by the freezing point of our salute, which is 2.55 °C. And this gives us an answer of negative 2.55 °C. Now, moving on to our boiling point elevation that we saw for, it's pretty similar in that we have the van factor multiplied by the malas, but this time to use the content for boiling point. So our van ho factor was two. Now the boiling point elevation or the boiling point for water, the constant for water is zero 0.5 12 degrees Celsius divided by Mola. And then the Molas of course, is 0.68 472, solving four hour boiling point elevation. You would have an answer of 2.55. But because this is elevation and because the boiling point of the solvent has to be added to the boiling point of elevation, we get this plus 100 °C for an answer of 100 0.70 °C. And our original equation should have equaled just 0.7. So this should have been 0.7 °C. And if we're at your choices, we see your choice c best reflects this overall. I hope this helped. And until next time.

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. c. 5.5% NaNO₃ by mass (in water)

Verified Solution

Key Concepts

Colligative Properties

Dissociation of Ionic Compounds

Freezing Point Depression and Boiling Point Elevation Formulas

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. c. 5.5% NaNO3 by mass (in water)

Verified Solution

Key Concepts

Colligative Properties

Dissociation of Ionic Compounds

Freezing Point Depression and Boiling Point Elevation Formulas

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. c. 5.5% NaNO₃ by mass (in water)