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 (b) 0.1056.

Step-by-step solution

Calculate Moles of Solute

First, we need to find the number of moles of acetic acid, \(\mathrm{CH}_3\mathrm{COOH}\). The molar mass of \(\mathrm{CH}_3\mathrm{COOH}\) is approximately \(12.01 + 1.01 \times 3 + 12.01 + 16.00 \times 2 + 1.01 = 60.05\, \text{g/mol}\). So, the number of moles is calculated as follows:\[\text{moles of } \mathrm{CH}_3\mathrm{COOH} = \frac{5\, \mathrm{g}}{60.05\, \mathrm{g/mol}} \approx 0.0833\, \text{mol}.\]

Calculate Mass of Solvent

Using the density of ethanol, we calculate the mass. Given that the density is \(0.789\, \text{g/mL}\), and we have \(1000\, \text{mL}\) (since it's one litre), the mass is:\[\text{mass of ethanol} = 1000\, \mathrm{mL} \times 0.789\, \mathrm{g/mL} = 789\, \text{g}.\]

Calculate Molality

Molality is calculated using the following formula:\[\text{molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}.\]Convert the mass of solvent to kilograms: \(789\, \text{g} = 0.789\, \text{kg}\). Then calculate molality:\[\text{molality} = \frac{0.0833\, \text{mol}}{0.789\, \text{kg}} \approx 0.1056\, \text{mol/kg}.\]

Choose Correct Option

The calculated molality, \(0.1056\, \text{mol/kg}\), matches option (b). Therefore, the correct answer is (b) \(0.1056\).

Key concepts

Moles calculation

The concept of moles is fundamental in chemistry, as it allows us to quantify the amount of a substance. A mole is a standard unit of measurement used to express amounts of a chemical substance. It is defined as exactly 6.02214076×10²³ elementary entities (atoms, molecules, ions, or electrons), a value referred to as Avogadro's number.

When performing moles calculation, we use the formula:

• \( ext{Number of moles} = \frac{ ext{Given mass}}{ ext{Molar mass}} \)

To find the moles of a substance, you need two main things:

• The mass of the substance given in the problem.
• The molar mass of the substance, which can be found using the periodic table by adding up the atomic masses of its component atoms.

The formula requires dividing the given mass (in grams) by the molar mass (in g/mol).

In our exercise, we calculated the moles of acetic acid (\( \text{CH}_3\text{COOH} \)), using this formula, finding approximately \( 0.0833 \text{mol} \). This value is pivotal as it connects to other properties of the solution.

Density

Density is a physical property that defines the mass of a substance per unit volume. It is usually denoted by the symbol \( \rho \) and expressed in units such as grams per milliliter (g/mL) or kilograms per cubic meter (kg/m³). The formula to calculate density is:

• \( \text{Density} (\rho) = \frac{\text{mass}}{\text{volume}} \)

In this context, density helps us convert from volume to mass, which is necessary for further calculations like molality.

In our scenario, ethanol has a given density of \(0.789\, \text{g/mL}\). By knowing the volume of ethanol used (1000 mL or 1 L), we can determine its mass:

• \( \text{Mass} = \text{Volume} \times \text{Density} = 1000 \text{mL} \times 0.789\, \text{g/mL} = 789\, \text{g} \)

This calculated mass of the solvent (ethanol) is essential when we need to calculate the molality of the solution.

Molality formula

Molality is a measure of the concentration of a solute in a solution. It is different from molarity, as molality is based on the mass of the solvent, not the volume of the solution. Molality is expressed in moles of solute per kilogram of solvent as:

• \( \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)

The benefit of using molality is that it doesn't change with temperature, unlike molarity since it is based on mass rather than volume, which can expand or contract.

For the given exercise, once we determined the moles of the solute (acetic acid) and the mass of the solvent (ethanol), we converted the mass of the solvent from grams to kilograms (since \( 789 \text{g} = 0.789 \text{kg} \)). Then, inserting these values into the formula, we get:

• \( \text{Molality} = \frac{0.0833\, \text{mol}}{0.789\, \text{kg}} \approx 0.1056\, \text{mol/kg} \)

This final calculation gives us the molality of the solution, which is approximately \(0.1056\, \text{mol/kg}\), corresponding to option (b). Molality offers a direct insight into how concentrated our solution is, based on the weight of the solvent.