School of Chemistry
University of KwaZulu-Natal, Westville Campus
Chemical Reactivity – CHEM120
Tutorial 10: 21st & 22nd October 2010
1)
Phase 2
Phase 1
Phase 3
a) For the phase diagram above, identify whether the substance is a gas, liquid or a solid in the indicated regions (phase 1, phase 2, and phase 3) •
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Phase 1 = Solid Phase 2 = Liquid Phase 3 = Gas b) At 80 °C, what happens to this substance when the pressure is raised from 200 Torr to 800 torr? •
Gas to liquid (condenses) c) At what temperature does this substance vaporize when the pressure is 300 Torr? •
~ 70 °C d) Is this CO2, H2O, CH4 or petrol? •
2)
Water (the solid Æ liquid line (B – D) leans slightly to the left, typical of water) 1.10 g of an unknown compound reduces the freezing point of 75.22 g of benzene (C6H6, Kf = 5.12 °C m‐1) from 5.53 to 4.92 °C. What is the molar mass of the compound? Equation needed: ∆
Re – arrange to solve for molality (m) ∆
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Recall the definition of molality: From this we can find the moles of solute Benzene in Kilograms: .
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Moles of Solute = 0.07522 x 0.119140625 Recall basic calculation of moles: Rearrange for Molar Mass (M) .
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Answer = 122.7 g mol (1.22 x 10 g mol ) 3)
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Calculate the vapour pressure at 25 °C of a solution containing 165 g of the non – volatile solute, glucose, C6H12O6, in 685 g H2O. The vapour pressure of water at 25 °C is 23.8 mmHg. Equation needed: Psoln = xsolventP°solvent Moles Glucose = 0.91586681 Moles Water = 38.02344687 .
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PH2O = 0.976479637 x 23.8 mmHg = 23.24021535 Answer = 23.2 mmHg 4)
A cellophane bag, which acts as a membrane permeable only to water, contains a 2 M sugar solution. The bag is immersed in a 1 M sugar solution. What will happen? (A) The bag will soon contain more solution that will be more concentrated than 2 M. (B) The bag will soon contain more solution that will be less concentrated than 2 M. (C) The bag will lose sugar and the solution in it will become less concentrated. (D) The bag will lose water and the solution in it will become more concentrated. 5)
What is the mole fraction of water in 200. g of 95% (by mass) ethanol, C2H5OH? (A) 0.050 (B) 0.12 (C) Mass Solution Mass Ethanol Mass Water Molar Mass C2H5OH Molar Mass H2O Moles C2H5OH Moles H2O 200 190 10 46.0688 18.0152 4.124266 0.555087 g
g g
Total n = χH2O 4.679353 mol 0.118625 g mol‐1 g mol‐1 mol
mol
0.56 (D) 0.88 95% = mass ETOH/200 g x 100% 0.95 = mass ETOH/200 g (0.95 x 200 g) = mass ETOH Mass Water = 200 ‐ 190