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FinalExamSolutions0 - CHM 2045 Final Exam SOLUTIONS Spring...

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CHM 2045 Final Exam SOLUTIONS Spring 2009 Form Code A Instructions: On your scantron enter and bubble in your name , UF ID number , and Form Code (see above). Careful! Bubbling errors will not be negotiated . 1. A sample of Mn metal reacts completely with an excess of hydrochloric acid: Mn(s) + 2HCl(aq) → MnCl 2( aq) + H 2 (g) The hydrogen gas produced is collected over water at 25.0ºC. The volume of the gas is 7.80 L, and the pressure is 0.980 atm. Calculate the mass (g) of manganese metal consumed in the reaction. (Vapor pressure of water at 25ºC = 23.8 mmHg.) (1) 4.6 g Mn (2) 16.6 g Mn (3) 19.8 g Mn (4) 21.7 g Mn (5) 24.6 g Mn Pressure of H 2 gas = 0.980 – (23.8/760) = 0.949 atm n of H 2 = PV/RT = (0.949)(7.80) / (0.0821)(298) = 0.303 mol H 2 (0.303 mol H 2 )(1 mol Mn / 1 mol H 2 ) = 0.303 mol Mn (0.303 mol Mn)(54.94 g/mol Mn) = 16.6 g Mn 2. The boiling point of a glucose solution (d = 1.16 g/mL) is 100.21 ºC. Calculate the osmotic pressure of that solution at 298K. (glucose = 180.2 g/mol ; k fp for water = 1.86 o C/m ; k bp for water = 0.520 o C/m) (1) 9.4 atm (2) 10.5 atm (3) 157 atm (4) 2.99 atm (5) 31.6 atm ΔT b = 0.21ºC = (0.520)( m ) so m = 0.40 mol glucose per kg solvent Assume 1 kg of solvent and (0.40 mol)(180.2 g/mol) = 72 g glucose --> 1072 g total mass of solution (1072 g) / (1.16 g/mL) = 924 mL = 0.924 L solution 0.40 mol glucose per 0.924 L solution = 0.43 mol/L = M π = MRT = (0.43)(0.0821)(298) = 10.5 atm 3. What is the strongest intermolecular force that exists between SO 2 and I 2 ? (1) dipole-dipole (2) ion-dipole (3) London dispersion (4) dipole-induced dipole (5) ion-induced dipole SO 2 has a net dipole moment; I 2 does not The attraction between the two molecules is therefore a dipole-induced dipole 4. The solubility of NH 4 Cl at 20°C = 37g per 100 g H 2 O. If a solution is prepared by dissolving 95 g of NH 4 Cl in 200.0 g H 2 O at 60°C, what mass of NH 4 Cl will recrystallize when the solution is cooled to 20°C? (1) 21 g (2) 58 g (3) 37 g (4) 74 g (5) 48 g 37g/100g = 74g/200g solubility 95g – 74g = 21 g 5. Copper crystallizes in a face-centered cubic unit cell, and copper has an atomic radius of 128 pm. What is the density of copper? (1) 1.35 g/cm 3 (2) 2.71 g/cm 3 (3) 4.98 g/cm 3 (4) 7.66 g/cm 3 (5) 8.90 g/cm 3 FCC --> 4 atoms per unit cell, so: (4 atoms/cell)(63.55 g/mol)(1 mol / 6.022x10 23 atoms) = 4.22x10 -22 g/cell a = 4r/√2 = 362 pm = 3.62x10 -8 cm a 3 = (3.62x10 -8 cm) 3 = 4.74x10 -23 cm 3 = volume of cell d = mass/volume = (4.22x10 -22 g) / (4.74x10 -23 cm 3 ) = 8.90 g/cm 3 6. Arrange NH 3 , PH 3 , and CH 4 in order from lowest to highest boiling point. (1) PH
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This note was uploaded on 09/25/2011 for the course CHM 2046 taught by Professor Veige/martin during the Spring '07 term at University of Florida.

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FinalExamSolutions0 - CHM 2045 Final Exam SOLUTIONS Spring...

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