Match the entries of column I with appropriate entries of column II a…

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

Column I	Column II
(a) Mole fraction of ethanol	(iv) 0.195
(b) Mole fraction of acetone	(iii) 0.155
(c) Mole fraction of acetic acid	(ii) 0.150
(d) Mole fraction of water	(i) 0.500

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)