A solution of methanol, CH3OH , in water is prepared by mixing together 128 g of methanol and 108 g — Solutions Chemistry Question

STEPS:

1. Understand the concept of mole fraction: The mole fraction of a component in a solution ($\chi$) is defined as the number of moles of that specific component divided by the total number of moles of all components in the mixture. For this solution:
$$\chi_{\text{methanol}} = \frac{n_{\text{methanol}}}{n_{\text{methanol}} + n_{\text{water}}}$$
2. Determine the molar masses of the substances: Use the periodic table to find the molar masses of methanol ($\text{CH}_3\text{OH}$) and water ($\text{H}_2\text{O}$):
* Methanol ($\text{CH}_3\text{OH}$): $12.0\text{ g/mol (C)} + 4.0\text{ g/mol (H)} + 16.0\text{ g/mol (O)} = 32.0\text{ g/mol}$.
* Water ($\text{H}_2\text{O}$): $2.0\text{ g/mol (H)} + 16.0\text{ g/mol (O)} = 18.0\text{ g/mol}$.
3. Convert the given masses to moles:
* Moles of methanol:
$$n_{\text{methanol}} = \frac{128\text{ g}}{32.0\text{ g/mol}} = 4.0\text{ moles}$$
* Moles of water:
$$n_{\text{water}} = \frac{108\text{ g}}{18.0\text{ g/mol}} = 6.0\text{ moles}$$
4. Calculate the total number of moles in the solution:
$$n_{\text{total}} = 4.0\text{ moles (methanol)} + 6.0\text{ moles (water)} = 10.0\text{ moles}$$
5. Calculate the mole fraction of methanol:
$$\chi_{\text{methanol}} = \frac{4.0\text{ moles}}{10.0\text{ moles}} = \mathbf{0.40}$$
This matches Option D.

*

WHY_OTHERS_WRONG:

• A is incorrect: 0.80 is much too high. A student might arrive at this value through a severe math error, such as dividing the moles of water ($6.0$) by the moles of methanol ($4.0$) and subtracting from 1, or by incorrectly approximating the molar mass of methanol as being much smaller than its true value.
• B is incorrect: 0.60 represents the mole fraction of water in the solution ($\frac{6.0\text{ moles water}}{10.0\text{ total moles}} = 0.60$). This is a common error where the student solves for the solvent's mole fraction instead of the solute's mole fraction.
• C is incorrect: 0.50 would be the mole fraction if the solution had an equal number of moles of water and methanol (a 1:1 mole ratio). A student might select this if they mistakenly assumed that having similar mass values ($128\text{ g}$ and $108\text{ g}$) meant they also had equal molar amounts.
• E is incorrect: 0.20 would be the mole fraction of methanol if the total moles in the system were doubled to $20.0$ (or if the calculated moles of methanol were halved), representing a division or stoichiometry calculation error.