Suppose `5 g` of acetic acid is dissolved in one litre of ethanol. Assume no reaction in between them. Calculate molality of resulting solution if density of ethanol is `0.789g/(mL)`.

Options

A. `0.1056`
B. `0.056`
C. `0.156`
D. `0.16`

Answer

To calculate the molality of the solution formed by dissolving 5 g of acetic acid in 1 liter of ethanol, we can follow these steps:

### Step 1: Calculate the number of moles of acetic acid (CH₃COOH)
The molar mass of acetic acid (CH₃COOH) is approximately 60 g/mol.

\[ \text{Moles of CH₃COOH} = \frac{\text{mass of CH₃COOH}}{\text{molar mass of CH₃COOH}} = \frac{5 \text{ g}}{60 \text{ g/mol}} = \frac{1}{12} \text{ mol} \approx 0.0833 \text{ mol} \]

### Step 2: Calculate the mass of ethanol
Given the density of ethanol is 0.789 g/mL, and we have 1 liter (1000 mL) of ethanol, we can calculate the mass of ethanol using the formula:

\[ \text{Mass of ethanol} = \text{Density} \times \text{Volume} = 0.789 \text{ g/mL} \times 1000 \text{ mL} = 789 \text{ g} \]

### Step 3: Convert the mass of ethanol to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of ethanol from grams to kilograms:

\[ \text{Mass of ethanol in kg} = \frac{789 \text{ g}}{1000} = 0.789 \text{ kg} \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:

\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.0833 \text{ mol}}{0.789 \text{ kg}} \approx 0.1056 \text{ mol/kg} \]

### Final Answer
The molality of the resulting solution is approximately **0.1056 mol/kg**.

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