Using data from Table 13.3, calculate the freezing and boiling po...

Question

Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (b) 20.0 g of decane, C10H22, in 50.0 g CHCl3;

Accepted Answer

Hello everyone. Today, we are being asked to determine the freezing and boiling points of a solution comprised of g of octane dissolved in five moles of cyclo hexane. So we want to find the Malawi overall. But to do that we need two components. We need the moles of octane and we also need the kilograms of octane. So solving for moles, we can simply take our 15 g of octane that we have and we can multiply that by the molar mass we can say that one mole of octane Is equal to 114.23 g. Our units are going to cancel out and we are going to be left with 0.1313 moles of octane Which is CH- 818. And then we need to find kg of our cycle hexen. And when do do that. We can take our five moles that we have that. We can multiply that by our molar mass which is 84.16 g per mole. And then we also need to multiply this by our one g per 10 to the third grams. In order to get rid of that grams and give us kilograms once again, our units are going to cancel out here and we are going to be left with 0. to one kg of our cyclo hexane. So we'll just say cyclo hexane right here and it's all for morality. We simply take our molds of our soul salute, which is 0.1313 moles. And we divide by our kilograms of solvent which is 0.4 to one kg. And when do that. We get 0.3119 morality. And so now we can solve our freezing points and are boiling points. So our freezing point is going to be our delta T. F. Which is our freezing point depression. And we're going to equal that to our constant times our morality. And so we do that. We get 20.8 degrees Celsius for morality, multiplied by commonality that we saw before, which is 0.3119 M and we end up with 6.49 degrees Celsius. We take that temperature or we take our freezing point of our octane And we say that that is going to be 6.47°C. And we multiply that by the temperature we just sold for which is 6.49 and we get negative 0.10.0175°C for our freezing point. Something for boiling point is going to be a very similar manner. So we're going to label this as the boiling point here, we're going to say our delta T. B. Or boiling point elevation is going to be our KB or a constant times modality. We're going to plug in the boiling point constant for Benzene that's going to be 2.92°C of morality times 0.3119 Morality giving us 0.91°C. As before. We're going to take that TV, we're going to take the boiling point of that compound. We're gonna say that's 80.74°C and added to 0.19°C to give us 81.7°C. I hope this helped, and until next time.

Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (b) 20.0 g of decane, C10H22, in 50.0 g CHCl3;

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