# Chem II

Find the a) % solution, b) molality and c) mole fraction of the acid 16.0 M H2SO4 d = 1.634 g/ml for the solution

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1. density = 1.634 g/mL.
mass of 1000 mL = 1.634 g/mL x 1000 mL = 1634 grams.
M = mols/L = 16.0 mols/liter of solution; therefore, we have 16.0 mols in 1634 g solution.
16.0 mols H2SO4 = 16.0 mols x 98 g/mol = 1568 g H2SO4.
%H2SO4 = (1568 g H2SO4/1634 g soln)*100 = xx % H2SO4.

mole fraction H2SO4= mols H2SO4/(mols H2SO4 + mols H2O).
You have mols H2SO4 from above. Grams H2O = 1634-1568 = yy g H2O and that divided by 18 = mols H2O.

molality = mols/kg solvent. You have mols H2SO4 and you have g H2O (from the mole fraction calculation). Change g H2O to kg and you have it.

Post your work if you get stuck. Check my thinking.

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