please explain the reason behind the answer The vapor pressure of CCI3F at 300 K

Question

please explain the reason behind the answer

The vapor pressure of CCI3F at 300 K is 856 torr. If 12.0g of CCI3F is enclosed in a 1.2 - L container, will any liquid be present? Given that the heat of fusion of water is -6.02 kj/mol. that the heat capacity of H2O(l) is 75.2 J/mol middot K and that the heat capacity of H2O(s) is 37.7 J/mol middot K, calculate the heat of fusion of water at -15 degree C Two liquids. A and B. have vapor pressures at a given temperature of 24 mmHg and 36 mmHg, respectively. We prepare solutions of A and B at a given temperature and measure the total pressures above the solutions. We obtain the following data: Predict the total pressure above a solution of 2 mol A and 4 mol B. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

11.122 We need to assume that we are at 300K (warm room temp).We want to know how many moles of CCl3F we have and then use the ideal gas laws to see how much volume this much gas would occupy. If it would occupy more than the volume of our container, 1.2 liters, then we will have some liquid. Molar mass of C~12, Cl ~ 34.5, F ~14, so ~120, so we have about a tenth of a mole. At STP a mole would take up 22.4 l, so we have around a twentieth of the volume a mole would occupy and we have about a tenth of a mole, so we have about twice as much CClF as would vaporize. If we were closer, we would multiply by some ratios, for example 856/760 says we have more vaporized than we would expect, and 300/273 says we would need a bigger volume to hold a mole. These two adjustments tend to cancel each other, so with our 2x factor, it is safe to say we have some liquid.

11.146 We are below the freezing point by a lot, which makes this easy to freeze. We can imagine taking this water up to zero and then freezing it. To take liquid water from -15 to 0 takes 15*75.2 J for each mole heated. To melt frozen water at zero takes 6020 J/mole. Since we are doing the opposite of both, we get 1125J + 6020J = 7.145 KJ, so -7.14 kJ (not positive about htis one)

11.148 from the given answers, we see that we have a weighted average of the two vapor pressures. 2 and 4 is the same ratio as 1 and 2, so we expect 32 mm Hg. The amount of excess liquid in the container has a very small effect on the vapor pressure.

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