Question

\(5 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{COOH}\) is dissolved in one litre of ethanol. Suppose there is no reaction between them. If the density of ethanol is \(0.789 \mathrm{~g} / \mathrm{mL}\) then the molality of resulting solution is: (a) \(0.0256\) (b) \(0.1056\) (c) \(1.1288\) (d) \(0.2076\)

Short answer

The molality of the solution is approximately 0.1056, making option (b) correct.

Step-by-step solution

Calculate the moles of solute

The solute here is acetic acid (\(\text{CH}_{3}\text{COOH}\)), and we need to find out how many moles are present in 5 grams. To find the moles, we use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]The molar mass of acetic acid \(\text{CH}_{3}\text{COOH}\) is calculated as:\[ 2 \times \text{C (12 g/mol)} + 4 \times \text{H (1 g/mol)} + 2 \times \text{O (16 g/mol)} = 60 \text{ g/mol} \]Now, calculate the moles:\[ \text{moles of } \text{CH}_{3}\text{COOH} = \frac{5 \text{ g}}{60 \text{ g/mol}} = 0.0833 \text{ mol} \]

Convert solvent mass to kilograms

Molality is defined based on the mass of the solvent in kilograms. The solvent here is ethanol.First, convert the volume of ethanol to mass using its density. The density of ethanol is given as \(0.789 \text{ g/mL}\).Since the solution contains 1 litre (or 1000 mL) of ethanol:\[ \text{Mass of ethanol} = 1000 \text{ mL} \times 0.789 \text{ g/mL} = 789 \text{ g} \]Convert the mass of ethanol to kilograms:\[ 789 \text{ g} = 0.789 \text{ kg} \]

Calculate the molality of the solution

Molality \(m\) is defined as the number of moles of solute per kilogram of solvent. Using the moles of acetic acid and the mass of ethanol:\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]\[ m = \frac{0.0833 \text{ mol}}{0.789 \text{ kg}} \approx 0.1056 \text{ mol/kg} \]

Key concepts

Acetic Acid

Acetic acid, with the chemical formula \(\text{CH}_3\text{COOH}\), is a common organic acid known for giving vinegar its distinct sour taste and strong odor. It is a simple carboxylic acid.

• Acetic acid is a weak acid, which means it partially dissociates in solution.
• It is soluble in water and many organic solvents such as ethanol.
• When dissolved in ethanol, as in our problem, it does not react but forms a uniform solution.

Often used in chemical synthesis and industrial applications, understanding its properties, such as molality, is crucial for practical and lab applications.

Ethanol

Ethanol, often simply called alcohol, is a clear, colorless, and volatile liquid. It's also used extensively in industries and laboratories both as a solvent and in chemical reactions.

• Ethanol's chemical formula is \(\text{C}_2\text{H}_5\text{OH}\).
• It is known for its ability to dissolve various substances and is miscible with water.
• Its density is typically around 0.789 g/mL, as used in our exercise, indicating that one milliliter of ethanol weighs 0.789 grams.

These properties make ethanol a preferred solvent in chemical processes and its density and volumetric properties are often required to calculate other characteristics like molality.

Density

Density is a key concept in chemistry, defined as mass per unit volume of a substance. For ethanol, this density is crucial when converting volumes of solutions to mass for calculations.

• Density is represented in units such as g/mL or kg/L, depending on the scale.
• Understanding density allows chemists to accurately determine the mass of a liquid from its volume, essential for calculating concentrations like molality.
• In our problem, ethanol has a density of 0.789 g/mL, used to convert the given volume of ethanol (1 L = 1000 mL) into mass (789 g), and then into kilograms (0.789 kg).

Using density makes the process of handling chemical solutions and deriving characteristics from them, such as molality, more manageable.

Molar Mass

Molar mass is the mass of a given substance (chemical element or compound) divided by the amount of substance, measured in moles. Knowing the molar mass allows us to convert between mass and moles.

• It is expressed in units of grams per mole (g/mol).
• For acetic acid, calculated as \(2 \times 12 \text{ g/mol} + 4 \times 1 \text{ g/mol} + 2 \times 16 \text{ g/mol} = 60 \text{ g/mol}\).
• Using the molar mass, we can determine the number of moles present in a sample from its mass. For example, 5 grams of acetic acid equals 0.0833 moles using its molar mass of 60 g/mol.

Understanding and accurately utilizing molar mass is a fundamental skill in chemistry, critical for calculating concentrations in solutions like molality.