Question

In a 61.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.320. What is the mass of each component?

Answer 1

Ans. # Step 1: Let the mass of methanol and water be X and Y grams respectively.

            Or, X g + Y g = 61.0 g

            So, X + Y = 61.0                  - equation 1

#Step 2:

# Moles of methanol = Mass / Molar mass = X g / 32.04216 g mol^-1 = 0.03121X mol

# Moles of Water = Y g / 18.01528 g mol^-1 = 0.05551Y mol

Now,

mole fraction of methanol = [0.03121X mol / (0.03121X mol + 0.05551Y mol)] = 0.320

            Or, 0.03121X = 0.320 x (0.03121X + 0.05551Y)

            Or, 0.03121X = 0.0099872X + 0.0177632Y

            So, 0.0212228X - 0.0177632Y = 0                      - equation 2

#Step 3: Comparing (Equation 1 x 0.0212228) MINUS equation 2-

            0.0212228X + 0.0212228Y = 1.2945908

            -0.0212228X + 0.0177632Y = 0

            -------------------------------------------------

                                    0.0389860Y = 1.2945908

                        Or, Y = 1.2945908 / 0.0389860 = 33.2066

Hence, mass of water = Y g = 33.21 g

# Now, putting the value of Y in equation 1-

            X = 61.0 – Y = 61.0 – 33.21 = 27.79

Hence, mass of methanol = X g = 27.79 g