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A solution contains 0.270 mol of isopropanol (C3H7OH) dissolved in 3.940 mol of water.?
A) What is the mole fraction of isopropanol?
B) What is the mass percent of isopropanol?
C) What is the molality of isopropanol?
1 Answer
- Roger the MoleLv 77 years ago
A)
(0.270 mol) / (0.270 mol + 3.940 mol) = 0.0641
B)
(0.270 mol C3H7OH) x (60.09535 g C3H7OH/mol) = 16.226 g C3H7OH
(3.940 mol H2O) x (18.01532 g H2O/mol) = 70.9804 g H2O
(16.226 g) / (16.226 g + 70.9804 g) = 0.186 = 18.6% by mass
C)
(0.270 mol) / 70.9804 g H2O x (1000 g H2O) = 3.80 m
