3) A solution is prepared by dissolving 2 mLof methanol in 100.0 mL of water. Th

Question

3) A solution is prepared by dissolving 2 mLof methanol in 100.0 mL of water. The final volume of the solution is 118 mL. The densities of methanol and water are 0.782 g/mL and 1.00 g/mL, respectively. For this solution, calculate molarity, molality, mass percent and mole fraction. Be careful with "solution" vs. solvent' for this problem. (Hint: the mass of the solution is the mass of the solute the mass of the solvent.) (12 pts) Molarity Molality: Mass Mole fraction: Hint! Start by figuring out: volume of Solvent Mass of Solvent Moles of solvent: volume of Solute: Mass of Solute Moles of Solute Volume of Solution Mass of Solution:

Explanation / Answer

Mass of methanol = 0.782 x 20.2 = 15.796g

Mass of water = 1 x 100 = 100g

Molarity = moles/volume (L)

                = 0.493 x 1000/100

                = 4.93 M

Molality = 0.493/0.1

                 = 4.93m

Mole fraction = 0.493/0.493+5.55

                          = 0.0815

Mass percent = mass fraction x 100

Mass fraction = mass of methanol / mass of methanol + mass of water

                          = 15.79/ 115.79

                          = 0.1363

Mass percent = 0.1363 x 100 = 13.63 %

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