Mhm. When we calculate the freezing point depression, you use the formula that the change in freezing point is equal to the freezing point. Constant comes to morality times the degree of ionization. So if we're dealing with solutions of water, the freezing point constant is 1.86 degrees Celsius per mobile. Um, which means if we have a solution, that is 0.1 Molo, uh, Yuria. Then you're going to plug in, Let's do it and read the freezing point Constant, which is in black and then in radio plug in the concentration. Let me put the units. This is degrees Celsius per melo, not small. And then we have times the degree of organization. Which area is a non electrolyte? It won't split into ions, so the degree of organization is one. So basically you're multiplying 1.86 times 0.1 Okay, which will give you 0.186 degrees Celsius. That's the freezing point. Depression. Now, water freezes at zero degrees Celsius. So you just have to subtract, right? Mhm. Yeah. And you will get negative 0.186 degrees Celsius as your new freezing point. Mhm. If you have ammonium nitrate, which is N h four n 03 Then it will split into NH four plus an N 03 minus. So I is going to be, too. It forms two ions. Let me put here that this was your idea. So you use the same constant because you have the same solvent. We're still using the same morality. Except we're multiplying by two. Yeah. Oh. So what's going to give us about 0.38? We can get the exact number on our calculator. 0.186 to 0.372 Okay. Yeah. And when we subtract it from zero, we're gonna get negative. 0.372 degrees Celsius is going to be the freezing point. Just Yeah. Mm. Next, we have hcl So for H c l. I equals two because it forms h plus and cl minus. So it's going to be the exact same numbers as in part B. Which means that the freezing point will be the same. 0.372 degrees Celsius. Right? I'll erase part A to make room. Mhm here for Sandy? Yeah. I mean, for d e N f Okay, so let's do part D. Your commission. Mm. See, a C l two. Okay, so that's gonna be one calcium. And to chlorine. I equals three. Yeah. So our delta T f is 1.86 times 0.10 Molo, times three. Yeah. Okay. And that comes out to 0.558 So when you subtract that from zero, you get that. The new freezing point is negative. 0.5 58 degrees Celsius. Yeah. Okay. Mhm. No, I'll erase this on the bottom just to make more room. No. Mhm. Yeah. Mm. Yeah. Mhm. Yes. M g s 04 Thank you. I is equal to chew. So it's the same thing as, um, Parts B and C. And your answer is going to be sure. Yeah. Negative. 0.372 degrees Celsius. Mhm for part F, we have C two. Yeah, h 50 h, which is ethanol. And that doesn't lionize. So I equals one, just like for your area, in part a. Okay, right. And then that would make our answer the same as in part a negative. 0.18 six degrees Celsius. Okay. And Lastly, we have acetic acid. Okay. Same thing. Yeah. Mhm. Yeah, as in mhm parts, Uh, B c N e acetic acid lionizes into h plus an acetate. So I equals two. And that would make our answer negative 0.372 degrees Celsius, which shows that collaborative properties like freezing point depression depends only on the amount of solid present. And that's why all the numbers are the same for the same degree of ionization, regardless of the type of solid.