A solution contains 25% ethanol, 25% acetone, 25% acetic acid and 25% water by mass. Match the mole fraction of each substance (List-I) with its value (List-II). (Molar masses in g mol−1: ethanol 46, acetone 58, acetic acid 60, water 18)
| List-I | List-II |
| (A) Mole fraction of ethanol | (p) 0.500 |
| (B) Mole fraction of acetone | (q) 0.150 |
| (C) Mole fraction of acetic acid | (r) 0.155 |
| (D) Mole fraction of water | (s) 0.195 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-s, B-r, C-q, D-p
The mole fraction of a component is its number of moles divided by the total number of moles of all components:
\( \displaystyle x_i=\frac{n_i}{n_{\text{total}}} \qquad n=\frac{\text{mass}}{M} \)
Take 100 g of solution, so each component weighs 25 g.
Ethanol: \( n=\frac{25\,\mathrm{g}}{46\,\mathrm{g\,mol^{-1}}}=0.5435\,\mathrm{mol} \)
Acetone: \( n=\frac{25\,\mathrm{g}}{58\,\mathrm{g\,mol^{-1}}}=0.4310\,\mathrm{mol} \)
Acetic acid: \( n=\frac{25\,\mathrm{g}}{60\,\mathrm{g\,mol^{-1}}}=0.4167\,\mathrm{mol} \)
Water: \( n=\frac{25\,\mathrm{g}}{18\,\mathrm{g\,mol^{-1}}}=1.3889\,\mathrm{mol} \)
Total: \( n_{\text{total}}=0.5435+0.4310+0.4167+1.3889=2.7801\,\mathrm{mol} \)
(A) Mole fraction of ethanol — This matches (s) 0.195. \( x=\frac{0.5435}{2.7801}=0.195 \)
(B) Mole fraction of acetone — This matches (r) 0.155. \( x=\frac{0.4310}{2.7801}=0.155 \)
(C) Mole fraction of acetic acid — This matches (q) 0.150. \( x=\frac{0.4167}{2.7801}=0.150 \)
(D) Mole fraction of water — This matches (p) 0.500. \( x=\frac{1.3889}{2.7801}=0.500 \)
Check: 0.195 + 0.155 + 0.150 + 0.500 = 1.000. Water has the smallest molar mass, so the same 25 g of it holds the most moles and the largest mole fraction.