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#Chem 162-2008 homework 1st &amp; 2nd week

#Chem 162-2008 homework 1st &amp; 2nd week - Hill...

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Chem 162-2008 Homework 1 st nd Week 21. Describe how you would prepare 2.30 kg of an aqueous solution that is 4.85% NaNO 3 by mass. 4.85 g NaNO 3 /100g solution → gNaNO 3 /2300g solution Numerator conversion: Done. Denominator conversion: Done 4.85g NaNO 3 /100g solution = 0.0485gNaNO 3 /gsolution = 111.55gNaNO 3 /2300gsolution Procedure: Mix 111.55g NaNO3 and (2300.00 g – 111.55g =) 2188.45 g water. 27. Cholesterol readings are described on page 486. What mass in grams of cholesterol is contained in an adult with a cholesterol number of 234? (The average human adult has about 5.0 L of blood.) I don’t see a discussion of cholesterol on page 486, but a cholesterol number of 234 means 234 mg cholesterol /dL of blood Convert (234 mg cholesterol/dL blood) to (g cholesterol/5.0 L blood) Numerator: Convert mg to g 234 mg x (1g/1000 mg) = 0.234 g cholesterol Denominator: Convert dL to L 1 dL x 0.1L/dL = 0.1L 0.234 g cholesterol/0.1L = 2.34 g cholesterol/L = 11.7 g cholesterol/5.0 L blood 29. Express the following aqueous concentrations in the unit indicated. (a) 5μg trichloroethylene/L water, as ppb trichloroethylene (b) 0.0025 g KI/L water, as ppm of KI (c) 37 ppm SO 4 2- as molarity of sulfate ion [SO 4 2- ] (a) (5μg trichloroethylene/L water) → (g trichloroethylene/10 9 g water) Numerator: 5 μg x (1g/10 6 μg) = 5 x 10 -6 g Denominator: Assume that in such a dilute aqueous solution the density is 1g/mL 1L x (1000mL/L) x (1g/mL) = 1000g 5x10 -6 g trichloroethylene/1000 g water = 5.0 g trichloroethylene/10 9 g water = 5 ppb trichloroethylene in water. (b) 0.0025 g KI/L water, as ppm of KI 0.0025 g KI/L water → gKI/10 6 g H 2 O Numerator: 0.0025 g KI → g KI Done Denominator: L water → g water L x (1000 mL/L) x (1 g/mL) = 1000 g Therefore, 0.0025 g KI/1000 g H 2 O = 2.5 g KI/1000000 g H 2 O = 2.5 ppm KI (c) 37 ppm SO 4 2- as molarity of sulfate ion [SO 4 2- ] 37 g SO 4 2- /10 6 g solution → mol SO 4 2- /L solution 1

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Numerator: 37 g SO 4 2- /96gmol -1 = 0.3854 mol SO 4 2- Denominator: 10 6 g aq solution x (1 mL/g) x (1L/1000 mL) = 10 3 L aq solution 0.3854 mol SO 4 2- /10 3 L solution = 3.854 x 10 -4 mol/L = 3.854 x 10 -4 M 33. Calculate the molality of a solution that has 1.02 kg sucrose, C 12 H 22 O 11 , dissolved in 554 g water. 1.02 kg sucrose/554 g water → mol sucrose/kg solvent Numerator: 1.02 kg sucrose x (1000 g/kg) x (1/298.34gmol -1 ) = 3.419 mol sucrose Denominator: 554 g solvent x (1 kg/1000 g) = 0.554 kg solvent Numerator and denominator together: 3.419 mol sucrose/0.554 kg solvent = 10.2 mol sucrose/kg solvent = 6.17 m I disagree with the Solutions Manual’s answer. We are working with sucrose, not glucose. 35.
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#Chem 162-2008 homework 1st &amp; 2nd week - Hill...

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