By D. A. Burgess (auth.)
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Extra info for Work Out Chemistry ‘A’ Level
Example text
C) Give estimates for the molar freezing-point depression constants for (i) water, (ii) benzene. (WJEC) The partial pressure of nitrogen in the atmosphere is approximately 4/5 atmospheric pressure. This is because 1. About 4/5 of the molecules in the atmosphere are nitrogen molecules. 2. About 4/5 of the mass of the atmosphere is nitrogen. 3. The mean velocity of nitrogen molecules is about 4/5 that of all the molecules present in the atmosphere. 1 E None of these 33 S g of a non-volatile compound of relative molecular mass 80 is dissolved in 70 g of a solvent having a relative molecular mass of 160 and a vapour pressure of 200 mmHg at 280 K_ Which of the following is the expected vapour pressure, in mmHg, of the resulting solution at 280 K?
If the time for S02 to diffuse was 80 s, the time for the same volume of CH4 to diffuse is 40 s. 10 Further Questions 211 30 g of methanol and 80 g of ethanol form an ideal solution at 20°C. What is meant by the term ideal solution? 99 kPa respectively, calculate (i) the partial vapour pressure of each component vapour and (ii) the total vapour pressure above the solution. 12 Chlorobenzene and bromobenzene are completely miscible. 013 x 105 Pa (Pa = N m- 2 ). 04 x 104 Pa, calculate the mole fraction of chlorobenzene (i) in the liquid mixture, (ii) in the vapour.
Does this make sense? Would you expect the vapour to be depleted in the less volatile component? (b) Two immiscible liquids provided they are thoroughly agitated will behave as independent liquids; each will exert its normal vapour pressure at room temperature. The total vapour pressure above the mixture is therefore the sum of the VPs of the pure liquids. A+B VP A B Temperature In the diagram, A is the VP curve for one liquid and B is that for the other. The total VP is shown by the curve A + B.