Question

Mole fraction of aqueous solution of $CH_3COOH$ is 0.1. Select incorrect statement -

• A. %(w/w) = 27
• B. Molality = 6.17 m
• C. $\frac{n_{H_2O}}{n_{CH_3COOH}}=9$
• D. $\frac{W_{H_2O}}{W_{CH_3COOH}}=27$

Answer 1

Answer: (D)

D

Answer 2

The problem asks us to identify the incorrect statement regarding an aqueous solution of $CH_3COOH$ where its mole fraction is 0.1.

Let $X_{CH_3COOH}$ be the mole fraction of $CH_3COOH$ and $X_{H_2O}$ be the mole fraction of $H_2O$. Given: $X_{CH_3COOH} = 0.1$. Since it's an aqueous solution, the only other component is water. The sum of mole fractions in a solution is 1. $X_{CH_3COOH} + X_{H_2O} = 1$ $0.1 + X_{H_2O} = 1$ $X_{H_2O} = 0.9$

We know that mole fraction is defined as $X_i = \frac{n_i}{n_{total}}$. So, $\frac{n_{CH_3COOH}}{n_{CH_3COOH} + n_{H_2O}} = 0.1$ and $\frac{n_{H_2O}}{n_{CH_3COOH} + n_{H_2O}} = 0.9$.

Let's check each option:

Option (C): $\frac{n_{H_2O}}{n_{CH_3COOH}}=9$ Divide the mole fraction of $H_2O$ by the mole fraction of $CH_3COOH$: $\frac{X_{H_2O}}{X_{CH_3COOH}} = \frac{n_{H_2O} / (n_{CH_3COOH} + n_{H_2O})}{n_{CH_3COOH} / (n_{CH_3COOH} + n_{H_2O})} = \frac{n_{H_2O}}{n_{CH_3COOH}}$ $\frac{n_{H_2O}}{n_{CH_3COOH}} = \frac{0.9}{0.1} = 9$. So, statement (C) is correct.

To evaluate other options, let's assume a basis for calculation. Let $n_{CH_3COOH} = 1$ mole. From $\frac{n_{H_2O}}{n_{CH_3COOH}} = 9$, we get $n_{H_2O} = 9 \times 1 = 9$ moles.

Now, calculate the masses of $CH_3COOH$ and $H_2O$. Molar mass of $CH_3COOH$: C=12, H=1, O=16 $M_{CH_3COOH} = (2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \text{ g/mol}$. Molar mass of $H_2O$: H=1, O=16 $M_{H_2O} = (2 \times 1) + 16 = 18 \text{ g/mol}$.

Mass of $CH_3COOH (W_{CH_3COOH}) = n_{CH_3COOH} \times M_{CH_3COOH} = 1 \text{ mol} \times 60 \text{ g/mol} = 60 \text{ g}$. Mass of $H_2O (W_{H_2O}) = n_{H_2O} \times M_{H_2O} = 9 \text{ mol} \times 18 \text{ g/mol} = 162 \text{ g}$.

Option (A): % (w/w) = 27 Percentage by weight (% w/w) of $CH_3COOH = \frac{\text{Mass of } CH_3COOH}{\text{Total mass of solution}} \times 100$ Total mass of solution = $W_{CH_3COOH} + W_{H_2O} = 60 \text{ g} + 162 \text{ g} = 222 \text{ g}$. % (w/w) = $\frac{60 \text{ g}}{222 \text{ g}} \times 100 = \frac{6000}{222} \approx 27.027 \%$. Rounding to the nearest integer, it is 27. So, statement (A) is correct.

Option (B): Molality = 6.17 m Molality (m) is defined as moles of solute per kilogram of solvent. Solute: $CH_3COOH$, Solvent: $H_2O$. Moles of solute ($n_{CH_3COOH}$) = 1 mole. Mass of solvent ($W_{H_2O}$) = 162 g = 0.162 kg. Molality (m) = $\frac{n_{CH_3COOH}}{W_{H_2O} \text{ (in kg)}} = \frac{1 \text{ mol}}{0.162 \text{ kg}} \approx 6.1728 \text{ m}$. Rounding to two decimal places, it is 6.17 m. So, statement (B) is correct.

Option (D): $\frac{W_{H_2O}}{W_{CH_3COOH}}=27$ Using the calculated masses: $\frac{W_{H_2O}}{W_{CH_3COOH}} = \frac{162 \text{ g}}{60 \text{ g}} = \frac{162}{60} = \frac{27}{10} = 2.7$. The statement claims the ratio is 27. So, statement (D) is incorrect.

The question asks to select the incorrect statement.

The final answer is $\boxed{D}$

Explanation of the solution:

1. From the mole fraction of $CH_3COOH$ (0.1), determine the mole fraction of $H_2O$ (0.9).
2. Calculate the ratio of moles of $H_2O$ to $CH_3COOH$: $\frac{n_{H_2O}}{n_{CH_3COOH}} = \frac{0.9}{0.1} = 9$. This confirms option (C) is correct.
3. Assume 1 mole of $CH_3COOH$. Then, moles of $H_2O$ are 9 moles.
4. Calculate the masses: $W_{CH_3COOH} = 1 \text{ mol} \times 60 \text{ g/mol} = 60 \text{ g}$. $W_{H_2O} = 9 \text{ mol} \times 18 \text{ g/mol} = 162 \text{ g}$.
5. Calculate % (w/w) of $CH_3COOH$: $\frac{60}{60+162} \times 100 = \frac{60}{222} \times 100 \approx 27.03\%$. This confirms option (A) is correct.
6. Calculate molality: $m = \frac{1 \text{ mol}}{0.162 \text{ kg}} \approx 6.17 \text{ m}$. This confirms option (B) is correct.
7. Calculate the ratio of masses: $\frac{W_{H_2O}}{W_{CH_3COOH}} = \frac{162}{60} = 2.7$. This contradicts option (D) which states the ratio is 27. Therefore, statement (D) is incorrect.