Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options given.

For a solution containing 25% ethanol, 25% acetone, 25% acetic acid and 25% water

(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

The correct answer is option 1. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

Given the solution contains:

25 g ethanol (\(C_2H_5OH\))

25 g acetone (\(CH_3COCH_3\))

25 g acetic acid (\(CH_3COOH\))

25 g water (\(H_2O\))

First, we calculate the number of moles for each component:

Ethanol (\(C_2H_5OH\)):

Molar mass of ethanol = 46 g/mol

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

Acetone (\(CH_3COCH_3\)):

Molar mass of acetone = 58 g/mol

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

Acetic acid (\(CH_3COOH\)):

Molar mass of acetic acid = 60 g/mol

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

Water (\(H_2O\)):

Molar mass of water = 18 g/mol

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

Next, we calculate the total number of moles in the solution:

Total moles = \(0.543 + 0.431 + 0.417 + 1.389 = 2.78 \text{ mol}\)

Now, we calculate the mole fraction for each component:

Mole fraction of ethanol, \(X_{ethanol} = \frac{0.543}{2.78} = 0.195\)

Mole fraction of acetone, \(X_{acetone} = \frac{0.431}{2.78} = 0.155\)

Mole fraction of acetic acid, \(X_{acetic acid} = \frac{0.417}{2.78} = 0.150\)

Mole fraction of water, \(X_{water} = \frac{1.389}{2.78} = 0.500\)

Therefore, the correct option is 1. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)