SolProbAns

# SolProbAns - Revised Spring'99 Answers to the Solution...

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Revised Spring ’99 Answers to the Solution Chemistry Supplementary Sheet 1. A solution of sulfuric acid prepared by dissolving 40.0 g of H 2 SO 4 in 60.0 g of H 2 O has a density of 1.303 g/mL. Calculate: a) the percent H 2 SO 4 by weight b) the mole fraction of H 2 SO 4 c) the molarity of the solution d) the molality of the solution e) the normality of the solution In problems of this type, where you need to convert one set of concentration units to other concentration units, one needs to determine the moles and mass of each component present and the volume of the solution. Then using the appropriate concentration unit definitions, the desired concentration unit can be determined. Note: the density given in the problem is that of the entire solution not the density of the solvent or the solute in the system. In this problem the grams of solute, H 2 SO 4 , and the number of grams of solvent, H 2 O , are given. Thus, one first needs to calculate the total mass of solution, the total volume of solution, and the moles of each component present. # g sol’n = # g H 2 SO 4 + # g H 2 O = 40 g H 2 SO 4 + 60 g H 2 O = 100 g sol’n 1 mL sol’n 1 L # L sol’n = 100 g sol’n x ---------------------- x ------------- 0.0767 L sol'n 1.303 g sol’n 1000 mL 1 mole H 2 SO 4 # mole H 2 SO 4 = 40 g H 2 SO 4 x ----------------------- = 0.41 mole H 2 SO 4 98 g H 2 SO 4 1 mole H 2 O # moles H 2 O = 60 g H 2 O x -------------------- = 3.33 moles H 2 O 18 g H 2 O total # moles of sol’n = # moles H 2 SO 4 + # moles H 2 O = 0.41 moles + 3.33 moles = 3.74 moles sol’n # g H 2 SO 4 40.0 g H 2 SO 4 a) % H 2 SO 4 = --------------------- X 100 % = ------------------------- X 100 = 40.0 % # g sol’n 100 g sol’n # moles H 2 SO 4 0.41 moles H 2 SO 4 b) X H 2 SO 4 = ------------------------- = ------------------------------ = 0.11 # moles sol’n 3.74 moles sol’n # moles H 2 SO 4 0.41 moles H 2 SO 4 c) Molarity (M) = -------------------- = -------------------------------- = 5.35 M # L sol’n 0.0767 L sol’n # moles H 2 SO 4 0.41 moles H 2 SO 4 1000 g d) Molality (m) = -------------------- = ------------------------------- X ---------------- = 6.83 m # Kg H 2 O 60 g H 2 O 1 Kg Since H 2 SO 4 contains 2 hydrogen atoms that can react with base, there are 2 equivalents of H per mole of H 2 SO 4 . # equiv. 2 equiv. e) Normality (N) = M x ------------------ = 5.35 M x ---------------- = 10.70 N

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Solution Chemistry Supplementary Sheet Answers Page 2 1 mole 1 mole 2. A solution prepared by dissolving 15.6 g of benzene, C 6 H 6 , in enough CCl 4 to give 450 mL of solution has a density of 1.15 g/mL. Calculate: a) the molarity of the solution b) the percent C 6 H 6 by weight c) the mole fraction of C 6 H 6 d) the molality of the solution This problem is similar to the previous problem, except that the total volume of solution is given rather than the amount of solvent. In this problem, C 6 H 6 is the solute , since it is dissolved in the solvent , CCl 4 . The density of the solution
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## This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.

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SolProbAns - Revised Spring'99 Answers to the Solution...

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