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HW1_23

# HW1_23 - 12 moles HCl X = 438 g HCl 1 mole HCl#g H 2 O = g...

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Homework due 1/23 Calculate the molality, percent composition by weight, and mole fraction of HCl in an aqueous 12 M HCl solution that has a density of 1.20 g/mL. You are given one set of concentration units and expected to determine the concentration of the solution in different sets of concentration units in this problem. In this case, the solution was given as a 12 M HCl solution. Since this is an aqueous solution, the solvent is water. Since no specific amounts are given one must assume some amount of solution present and do all of the calculations relative to that amount. I have found it most useful to assume one of the unit in the denominator of the concentration given. In this case, if one assumes 1 L (1000 mL) of solution then there are 12 moles of HCl present . The rest of the calculations based on this assumption follow. 1.2 g sol’n # g solution = 1000 mL sol’n X -------------------- = 1200 g solution 1 mL sol’n 36.5 g HCl # g HCl =
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Unformatted text preview: 12 moles HCl X --------------------- = 438 g HCl 1 mole HCl #g H 2 O = # g sol’n - # g HCl = 1200 g sol’n - 438 g HCl = 762 g H 2 O = 0.762 Kg H 2 O 1 mole H 2 O # mole H 2 O = 762 g H 2 O x -------------------- = 42.33 moles H 2 O 18 g H 2 O total # moles of all components in the solution = # moles HCl + # moles H 2 O = = 12 moles HCl + 42.33 moles H 2 O = 54.33 moles total # moles HCl 12 moles HCl molality (m) = ------------------------- = ----------------------------- = 15.75 m # Kg H 2 O 0.762 Kg H 2 O # moles HCl 12 moles HCl mole fraction =X HCl = --------------------------------------- = ------------------------------ = 0.221 # moles HCl + # moles H2O 54.33 moles total # g HCl 438 g HCl % composition = ---------------------- X 100 % = ---------------------- X 100 % = 36.5 % # g sol’n 1200 g sol’n Other approaches are possible. These will work as long as all of the data is kept consistent form the initial assumption....
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