125ch13c(1) - 27 13.34) Note the denominator in mass % is...

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Unformatted text preview: 27 13.34) Note the denominator in mass % is the total mass of solution Total mass = mass solute + mass solvent Technically, the total mass in part b is: 1000 g H2O + 0.0079 g Sr2+ = 1000.0079 g soln & ppm = (0.0079 g Sr2+)/(1000.0079 g soln) x 106 = 7.8999 = 7.9 ppm Since the amount of Sr2+ is so minute compared to the amount of H2O there was no problem in ignoring it in the denominator in part b above. 28 13.35) A solution contains 14.6 g CH3OH in 184 g H2O. a) Find mol fraction of CH3OH, XCH3OH mol CH3OH XCH3OH = ----------------------total mol solution (No unit) 1 mol CH3OH ? mol CH3OH = 14.6 g CH3OH x --------------------- = 0.4556 mol CH3OH 32.04 g CH3OH 1 mol H2O ? mol H2O = 184 g H2O x ------------------ = 10.12 mol H2O 18.016 g H2O 0.4556 mol CH3OH XCH3OH = --------------------------------- = 0.04271 = 0.0427 (0.4556 + 10.12) mol soln b) Find mass % CH3OH (also called wt % or wt/wt %) mass CH3OH mass % CH3OH = ------------------------ x 100% total mass solution 14.6 g CH3OH mass % CH3OH = ------------------------------------- x 100% (14.6 g CH3OH + 184 g H2O) = 7.3514 % = 7.35 % CH3OH c) Find molality of CH3OH, m mol CH3OH m = ----------------kg H2O 0.4556 mol CH3OH m = -------------------------- = 2.4765 = 2.48 m CH3OH 0.184 kg H2O 29 13.37) molarity, mol solute M = --------------L solution a) mol Mg(NO3)2 0.540 g Mg(NO3)2 1 mol Mg(NO3)2 103 mL soln ? ------------------- = ------------------------ x ------------------------- x ---------------1 L soln L soln 250.0 mL soln 148.31 g Mg(NO3)2 = 0.01456 M = 0.0146 M Mg(NO3)2 30 31 13.44) 32 13.46) 33 13.46) (cont.) 34 13.48) 35 13.48) (cont.) 36 13.48) (cont.) ...
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This note was uploaded on 02/19/2012 for the course CHEM 125 taught by Professor Zellmer during the Winter '08 term at Ohio State.

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