Chemistry 20L Exam Study Questions Winter 2004 Answer Key Open lab notebook, lecture guides and hand-written notes. No lab manuals or texts. Bring a calculator. 1. (a) percent error = (0.0220.54 + 0.0030.254 + 0.053.21) X 100 = 2.8% ~ 3% (b) Absolute error = (30.078 - 20.174 + 9.813) ± (0.003 + 0.001 + 0.005) = 19.717 ±0.009 % error = 0.00819.717⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x 100 = 0.045% ~ 0.05% (c) absolute error = [(0.0020.642+ 0.0042.413) X 1.549] + 0.002 = 0.007 + 0.002 = 0.009 % error = (0.009(0.642)(2.413)− 0.501) X 100 = 0.0091.048 X 100 = 0.86% ~ 0.9% (d) 12.635 ± 0.005(5.967± 0.003) + (0.478 ± 0.004) = 12.635±0.0056.445±0.007 % error = (0.00512.635 + 0.0076.445) X 100 = 0.148% ~ 0.1% 2. (a) 1.625 X 2.8% = ±0.045 ~ ±0.05 (b) ± 0.009 (c) ± 0.009 (d) 12.6356.445x0.148% =±0.003 or 12.6356.445x0.1%=±0.002 3. average = 20.905; average deviation = 0.035; relative average deviation = 0.17% 4. Molarity = moles solute/liter solution Normality = equivalents solute/liter solution Weight percent (w/v) = grams solute/100mL solvent Weight percent (w/w) = grams solute/100g solvent PPM = parts per million pH = -log [H+] 5. (a) molarity is the appropriate unit when discussing chemical reactions since stoichiometry is based on mole relations. Molarity is dependent on temperature because of volume changes with temperature. PPM is useful when small concentrations are under investigations and the frame of reference is always at this level (b) weight percent is useful when the substances are solids and the preparation is a “recipe” only and does not convey any sense of a reaction that stoichiometry.(c) like weight percent, volume percent is a “recipe” instruction and is useful for concentrations when the solute and solvent are liquids. 6. (a) Molecular weight of ethanol = 45 g/mol Volume of 1.00 x 10-3 mol = 0.051 mL PPM = 0.045g/1000g = 45g/106g = 45 PPM (b) wt % = 0.0045g/100g = 0.0045% (c) vol % = 0.051 mL/1000 mL = 0.0051 mL/100 mL = 0.0051% 7. 2.00 x 10-4 M 4.00 x 10-4 N 0.00196% (w/v) 19.6 PPM 8. The molecular weight of sodium hydroxide NaOH, is 40.08g. The final concentration is 0.4063g40.00g/molX1000mL/L100.00mLX10.00ml50.00mL= 0.2032M 9. The question asks for copper not the salt. It is the concentration of the heavy metal that is usually the important issue.) Weight of copper in the sample =weight of sample x AW of Cu /MW of CuSO4.5H2O) = 0.1037g x 63.55 g/mol249.7 g/mol = 2.639 10-2 g concentration of Cu in diluted solution = 2.639 x 10-2g0.100L⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 0.005L0.100L⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1.320 x 10-3 g/L Since this is a dilute aqueous solution 1.00 mL ~1.00g PPM Cu ≈1.320 x 10-2g1000g⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 1000g1000g⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ =13.20g106⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 13.20 ≈ 13.2PPM 10. 0.56% Note that the names of the pieces of equipment (e.g. 100-mL volumetric flask, 5-mL pipet) do not indicate the precision of the equipment. In part (b) of this question you calculate the uncertainty in the concentration. The absolute error in (a) is ±0.06 ppm. Rounding down to 13.2 is very conservative, and loses some of the significance of the measurements; reporting 13.20 implies a higher precision than is actually known if you assume ±1 in the last digit. The unambiguous way to report the answer is 13.20 ±0.56%. 11. Absorbance is a measure of the light absorbed by a solution. Specifically it is the negative log of the fraction of light transmitted through the solution. Because it is a logarithm, it is a dimensionless quantity. b is the symbol for the thickness of the solution through which the light being absorbed passes. b is often called the ‘cell length” or “path length” and is reported in centimeters. ε is the molar extinction coefficient. It is the amount of light absorbed by 1 cm of a 1M solution of the compound. 12. 44.5%13. (a) 2390/M (c) 2.09 x 10-4 14. (a) A 60% octane solution will begin to boil at 90oC. The composition of the vapor in equilibrium with the solution is 70% hexane. 0 0.2 0.4 0.6 0.8 1.0MOLE FRACTION OF HEXANEOCliquidvaportie line boiling curve orvaporization curvecondensation curve vapor and liquid co-exist in thisregion T E M P E R A T U R E6080100120140 (b) Gas chromatogram of original mixture. injection hexane octane time Paper direction (c) Gas chromatogram of first distillate.15. (a) moles of OH- = moles of H2PO4- = (0.500M x 0.01941L) = 9.705 x 10-3 (moles of H2PO4- )/3 = moles of STPP = 3.235 x 10-3 weight of STPP in sample = moles STTP x MW = 3.235 x 10-3 mol x 371 g/mol = 1.200g % STPP in sample = 1.200g/2.000g x 100 = 60.0% (b) Inherent error = sum of % errors in buret reading, concentration of base and weighing of original sample = 0.00022.0000⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ +0.0010.500⎛ ⎝ ⎜ ⎞ ⎠ ⎟ +0.0319.41⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎤ ⎦ ⎥ x 100 = 0.36% 16. (a) The end point in a titration is when an indicator changes color signaling (one hopes) that the equivalents of acid and base that have been mixed are equivalent. The equivalence point in a titration occurs when the equivalents of acid and base that have been mixed are identical whether or not there is any indication experimentally that this is the case. (b) Volumetric glassware is calibrated by the manufacturer to give a precision of 0.1% based on proper use. Non-volumetric glassware may have volume indications to give an estimate of the volume contained or delivered to a precision of 1 – 5%. (c) “Weak” coffee is a common description of the concentration of the coffee. A weak acid refers to a substance that does not dissociate completely in solution. The acid that is put into the solution can have any concentration. (d) Absorbance is a measure of the amount of light interacting with the molecules in a solution and not passing through the solution. It is related to the transmittance by the equation A = -log T. The transmittance, T, is the fraction of incident light that passes through a solution. (e) Accuracy refers to the closeness of a set of results to the true value for that result. The precision refers to the deviation or variation of a set of results from the mean. 17. See the lab technique procedures 2, 5, 4, and 9 (pages 2 – 9) and the procedure in Assignment 8 on pages 84 and 85 in the lab
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