What mass of salt (NaCl) should you add to 1.00 L of water in an ice cream maker to make a solution that freezes at -10.0 °C? Assume complete dissociation of the NaCl and density of 1.00 g>mL for water.

Video transcript

Welcome back everyone. Our next problem says, what mass of salt and ac should you add to 1.00 L of water in an ice cream maker to make a solution that freezes at negative 10.0 °C assume complete association of the N AC L and density of 1.0 g per milliliters for water. Our answer choices are a 58 g. B 160 g, C 540 g or D 1.4 kg. So in this problem, we're talking about the phenomenon of freezing point depression that when you add a solu that's dissolved in a solution, you lower the freezing point. So in this case, salt and we have an equation to calculate it, which is at delta T subscript F. So the change in the freezing temperature is equal to lowercase I K subscript F and lowercase M. And in this equation little I or lowercase I is the Van Taho factor. That's a measure of how your solute dissolves uh in, in the solution. So the number of particles from the salute when it dissolves and then our K subscript F is equal to the freezing point, depression constant, and this is something we can look up for our salute. So in this case, we want the KF of water which is going to equal 1.86 degrees Celsius kilograms over moles. And then the lower case M in this case is the morality of the salute or morality of the solution. I should say, which is defined as the moles of salute per kilogram of solution. And then finally, we should add in when looking at lowercase I, we look at that number of particles from the solute. When it dissolves, we're told that sodium quad completely dissolves. So in this case, our little I factor is two. So it will equal two, four sodium chloride. So we know that as well. So we've got our little I, we've got our KF, we can calculate our delta TF because we're given, we obviously know the freezing normal freezing point of water. We're given the freezing point that we're aiming for. So what we're focusing on calculating here is our lower case M the morality of this solution because that's going to allow us to calculate the number of moles of salt we need, which can lead to the mass of salt that we need. So that's what we're aiming for. So we want to solve our equation four M. So we'll have M equals delta TF, subscript F divided by lowercase IKF. So our delta TF. Well, that's pretty straightforward. We know we want to lower the freezing point by 10 °C since we want our solution to freeze at negative 10, and we know the freezing point of water is zero. We're just going to write it out formally just to make sure we're getting our signs correct. So it would be the freezing point. So TF under standard conditions minus the TF that we're looking for. So that will equal 0.0 °C minus negative 10.0 °C. So we end up with a delta TF of 10.0 °C. So again, useful to write that out just to make sure we have a sign correct for what we're using. So let's plug in all our numbers. So we have M equals are delta TF. So 10.0 °C will be on the top divided by our lowercase I which is two. And then our KF which again, we know the freezing point, depression, constant of water is 1.86 °C kilograms divided by moles. And that is going to equal, let's look at how our units cancel out. Only unit on the top is degrees Celsius. So that cancels out and note that our other units are, and then we have kilograms in the denominator and then we have them divided by moles. So that's going to put our final units as moles per kilogram, which makes sense because we are, we're solving for the morality of a solution. So when we do this math here we end up with 2.6882 sort of rounded off little MN AC L because of that moles of salt per kilogram of our salute or a solvent, excuse me, bows of solu per kilogram of solvent. So per kilogram of water. Well, we're given how many liters of water we have. So we want to convert those liters to kilograms. So we have 1.00 L of water. We'll use a conversion factor multiplied by 1000 mL divided by 1 L and then our density of water. So the way to convert between volume and mass is given to us as 1.00 g per milliliter. So we want milliliters to cancel out. So we'll say 1.0 g on the top divided by 1 mL. So now we have grams and we need to go all the way to kilograms. So one more conversion vector. So multiplied by 1 kg divided by 1000 g. So looking at our units liters cancel, milliliters, cancel grams, cancel and we're left with kilograms. We have one multiplied by 1000 divided by one divided by 1000. We end up very simply with 1.0 kg of water. So now we know how many moles of salt per kilogram of water. We know that we have a kilogram of water, scroll up a little bit here to the groom, our calculation. So since we know we have 2.6882 moles of salt moles of N ac L per kilogram of water. And we know we have 1 kg of water. We take that 2.6882 moles multiply it by the molar mass of sodium or of sodium chloride. Excuse me. So multiply that 5 58.44 g per mole and that will equal 1.57 or 157 excuse me, 0.10 g of sodium chloride. We have to look back at our significant figures here. And when we look up at the numbers, we're given the original problem. You can see that we have three significant figures. So we'll go ahead and erase this and have it in significant figures. 160 g sodium chloride. So when we look at our answer choices, we have choice B corresponding to that 160 g. So we need to add 160 g of salt to a liter of water to make a solution that freezes at negative 10 °C. See you in the next video.