Problem 2:
Total mass of solution = 1200.0 g NaCl + 7300.0 g H2O = 8500.0g Total volume of solution = 8500.0 g NaCl solution x (1 mL/1.089
g NaCl solution) = 7805 mL or 7.805 L Molarity = 1200.0 g NaCl x (1 mol NaCl/58.5 g NaCl) x (1/7.805
L) = 2.63 mol/L NaCl or 2.63 M NaCl Molality = 1200.0 g NaCl x (1 mol NaCl/58.5 g NaCl) x (1
/ 7.3000 kg) = 2.80 mol NaCl/kg or 2.80 m NaCl Mass percent = 1200.0 g NaCl x (1/8500.0 g solution) x 100
= 14.118 % NaCl Mole fraction NaCl
Mol NaCl = 1200.0 g NaCl x (1 mol NaCl/58.5 g NaCl) = 20.5
mol NaCl
Mol H2O = 7300.0 g H2O x (1 mol H2O/18.0 g H2O) = 406 mol
H2O
Total mol = 20.5 NaCl + 406 mol H2O = 426.5 mol solution Mol fraction NaCl = 20.5 mol NaCl / 426.5 mol solution =
0.0481 mol fraction NaCl
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