Question

A solution contains 0.470 mol of isopropanol (C3H7OH) dissolved in 3.320 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?

Answer 1

A. The mole fraction of isopropanol in the solution is 0.124

B. The mass percent of isopropanol is 12.4%

C. The molality of isopropanol is 7.86 M

A. How to determine the mole fraction of isopropanol

• Mole of isopropanol = 0.470 mole
• Mole of water = 3.320 mole
• Total mole = 0.470 + 3.32 = 3.79 mole

• Mole fraction of isopropanol =?

Mole fraction = mole / total mole

Mole fraction of isopropanol = 0.470 / 3.79

Mole fraction of isopropanol = 0.124

B. How to determine the percentage of isopropanol

• Mole of isopropanol = 0.470 mole
• Total mole = 3.79 mole

• Percentage of isopropanol =?

Percentage = (mole / total mole) × 100

Percentage of isopropanol = (0.470 / 3.79) × 100

Percentage of isopropanol = 12.4%

C. How to determine the molality

• Mole of isopropanol = 0.470 mole
• Mole of water = 3.320 mole
• Mass of water = 3.320 × 18 = 59.76 g = 59.76 / 1000 = 0.05976 Kg

• Molality of isopropanol =?

Molality = mole / Kg of water

Molality of isopropanol = 0.47 / 0.05976

Molality of isopropanol = 7.86 M

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