Question

The weight fraction of methanol in an aqueous solution is $0.64$. The mole fraction of methanol $X_M$ satisfies:

• A. $X_M < 0.5$
• B. $X_M = 0.5$
• C. $0.5 < X_M < 0.64$
• D. $X_M > 0.64$

Hint:

Choosing a basis of $100\text{ units}$ when dealing with mass or mole percentages simplifies conversion calculations and helps avoid fractional arithmetic errors.

Answer 1

The Correct Option is B

Concept: To find the relationship governing the mole fraction $X_M$ from a given mass fraction, we use the definitions of mole fraction and molecular weight. Let us define the components of the binary mixture:
• Methanol (M): $\text{CH}_3\text{OH}$ with molecular weight $M_M = 12 + (1 \times 3) + 16 + 1 = 32\text{ g/mol}$.
• Water (W): $\text{H}_2\text{O}$ with molecular weight $M_W = (1 \times 2) + 16 = 18\text{ g/mol}$. We are given the weight fraction of methanol as $w_M = 0.64$. Consequently, the remaining mass belongs to water, so the weight fraction of water is $w_W = 1 - 0.64 = 0.36$.

Step 1: Establish a computational basis and determine component masses.
Let us assume a total mass basis of $100\text{ g}$ for the aqueous mixture. Under this basis, the individual masses are directly calculated as: \[ \text{Mass of methanol, } m_M = 100 \times 0.64 = 64\text{ g} \] \[ \text{Mass of water, } m_W = 100 \times 0.36 = 36\text{ g} \]

Step 2: Convert the mass of each component into moles.
Using the formula $\text{Moles } (n) = \frac{\text{Mass } (m)}{\text{Molecular Weight } (M)}$: \[ \text{Moles of methanol, } n_M = \frac{64\text{ g}}{32\text{ g/mol}} = 2.0\text{ moles} \] \[ \text{Moles of water, } n_W = \frac{36\text{ g}}{18\text{ g/mol}} = 2.0\text{ moles} \]

Step 3: Calculate the mole fraction of methanol $X_M$.
The definition of mole fraction for a binary system is: \[ X_M = \frac{n_M}{n_M + n_W} \] Substituting the computed molar values: \[ X_M = \frac{2.0}{2.0 + 2.0} = \frac{2.0}{4.0} = 0.5 \] Thus, the mole fraction exactly equals $0.5$.