For a solution containing 25% ethanol, 25% acetone, 25% acetic acid, and 25% water, match the mole fractions of the substances with their values.

A-s, B-r, C-q, D-p

The correct answer is option 1. A-s, B-r, C-q, D-p.

Molar mass of Ethanol \((C_2H_5OH)\) = 46 g/mol

Molar mass of Acetone \((CH_3COCH_3)= 58 g/mol

Molar mass of Acetic acid (CH₃COOH)= 60 g/mol

Molar mass of Water \((H_2O)\)= 18 g/mol

Moles of Ethanol\(= \frac{25 \text{ g}}{46 \text{ g/mol}} \approx 0.543 \text{ mol} \)

Moles of Acetone \(= \frac{25 \text{ g}}{58 \text{ g/mol}} \approx 0.431 \text{ mol} \)

Moles of Acetic acid \(= \frac{25 \text{ g}}{60 \text{ g/mol}} \approx 0.417 \text{ mol} \)

Moles of Water \(= \frac{25 \text{ g}}{18 \text{ g/mol}} \approx 1.389 \text{ mol} \)

Total moles \(= 0.543 + 0.431 + 0.417 + 1.389 = 2.780\, \ mol\)

Mole fraction of ethanol \(= \frac{0.543}{2.780} \approx 0.195 \)

Mole fraction of acetone \(= \frac{0.431}{2.780} \approx 0.155 \)

Mole fraction of acetic acid \(= \frac{0.417}{2.780} \approx 0.150 \)

Mole fraction of water \(= \frac{1.389}{2.780} \approx 0.500 \)

Given these calculations, the correct matching is: A-s, B-r, C-q, D-p

This means:

The mole fraction of ethanol (A) corresponds to 0.195 (s).

The mole fraction of acetone (B) corresponds to 0.155 (r).

The mole fraction of acetic acid (C) corresponds to 0.150 (q).

The mole fraction of water (D) corresponds to 0.500 (p).