The molarity of a solution of sodium acetate (CH3COONa) at 20 C is 7.5 M. Use the graph showing the solubility of sodium acetate as a function of temperature to describe the solution. The solution is . (LO 13.7, 13.8)

(a) Saturated (b) Unsaturated (c) Supersaturated (d) Colloidal

Video transcript

Hi everyone for this problem. It reads true or false based on the graph showing the soluble itty of lead chloride with respect to temperature a .65 molar solution of lead chloride at 20°C is saturated density of lead chloride at 20 degrees Celsius is equal to 5. g per millimeter. So we want to answer whether or not this is true or false. And what we want to answer this in relation to is whether it is saturated. So let's just recall definitions for saturation. So there are three, the first is saturated and when something is saturated, this means the maximum amount of solute is dissolved in the solvent. Then we have super saturated and this is a solution containing more than the maximum amount of solu capable of dissolving at a given temperature and the last that we have is unsaturated, which is a solution containing less than the maximum amount of solute that can be dissolved. And so we want to answer whether or not it's true or false that this is saturated. So that means we're going to need to know what the Um Alaric E. Is. Okay, so let's take a look at our graph based on the graph the soluble itty of lead chloride at 20°C is one g of lead chloride per 100 ml of water. And we see that here. Okay, so one g of lead chloride per 100 ml of water. So here our salute is lead chloride and the solvent is water. So what we're going to want to do to solve this problem is we're going to want to convert the solve ability to polarity so that we can compare. And the number that we're going to compare two is .65 molar. Okay, so this value here. So let's go ahead and do this conversion. So we want to go from soluble itty two modularity. Okay. So what we can do is we can use the density of solute to calculate its volume. Okay. And we know that the volume Based off of the problem. The volume we have, what we're starting with is one g of lead chloride because we have one g of lead chloride per 100 ml of water based on the graph. So our starting point is one g of lead chloride. And we're going to use the density to go from grams. Okay. To volume. So using the density given there is 5.85 grams of lead chloride per one millimeter. Okay, so we see here that grams of lead chloride cancel. And we're left with milliliters which is going to give us the volume. Okay, so our volume is 0. mL. Okay, so this is the volume of our soul. You we need the volume of solution. So we're right here. This is volume of solute. So now we need the volume of the solution and the volume of the solution is going to equal the volume of the sol U plus the volume of the solvent. Okay, so it equals the volume of solution. Excuse me, The volume of the sol ute plus the volume of the solvent. That is going to give us the volume of the solution. So let's go ahead and plug in these values. So the volume of the solvent or the sol ute we just calculated is 0.17094 mL. And the volume of the solvent is 100 mL. This is based off our graph because we based off our graph, it tells us that we have one g of lead chloride per mL of water and water is our solvent. Okay, so the volume of the solution is 100. mL. Alright, so now that we know that we can calculate the similarity. Okay, so the malaria t from here the mill arat E. Is going to equal So we have what we're going to do to calculate the mill arat E. Is, we're going to remember me larry T is grams. The mill arat E is moles per liter. Okay, so we have one g times one mole over 278.1 g. Because we're gonna take our one g and convert that to moles because malaria is moles over leader and this is going to be over Our volume. But we need to convert this to leaders. So we're going to have 100.17094 which is our volume of the solution. And this is middle leaders. And we want to convert this from middle leaders to leaders. So we're going to multiply and it's going to be multiplied by one leader over 1000 mL. Okay, so once we do this, Armel arat e is 0. molar. So this is the value we're going to compare to the malaria t given in the problem. Okay, so we have 0.035897 Moller that we just calculated this value is less than 0.65 molar. So that means the solution contains more than the maximum amount of salt capable of dissolving at a given temperature. So this means it is super saturated out of the different saturation we talked about above. So because it is super saturated. The answer to this problem is false. Because the problem says based on the graph it is saturated and that is false, it is not saturated, it is actually super saturated. So that is going to be the answer for this problem. And that is it for this problem. I hope this was helpful