Problem 81 - A Complete Resource Book in Chemistry for JEE Main

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) \(0.2076\)

Step-by-Step Solution

Answer

The molality of the solution is approximately 0.1056 mol/kg (Option b).

Step 1: Calculate Moles of Solute

First, calculate the moles of acetic acid (\(\mathrm{CH}_3\mathrm{COOH}\)) using its molar mass.The molar mass of \(\mathrm{CH}_3\mathrm{COOH}\) is \(2(12.01) + 4(1.01) + 16.00 + 1.01 = 60.05 \, \text{g/mol}\).Thus, the moles of \(\mathrm{CH}_3\mathrm{COOH}\) is given by:\[moles = \frac{5 \, \text{g}}{60.05 \, \text{g/mol}} = 0.0833 \, \text{mol}.\]

Step 2: Determine Mass of Solvent

Convert the volume of ethanol to mass using its density.Ethanol's density is \(0.789 \, \text{g/mL}\) and the volume is given as \(1 \, \text{L} = 1000 \, \text{mL}\).Thus, the mass of ethanol is:\[mass = 0.789 \, \text{g/mL} \times 1000 \, \text{mL} = 789 \, \text{g}.\]

Step 3: Calculate Molality

Molality is the number of moles of solute per kilogram of solvent.The formula is:\[\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}.\]From previous steps:- Moles of \(\mathrm{CH}_3\mathrm{COOH}\) are \(0.0833 \, \text{mol}\).- Mass of ethanol is \(789 \, \text{g} = 0.789 \, \text{kg}\).Thus, the molality is:\[\text{Molality} = \frac{0.0833 \, \text{mol}}{0.789 \, \text{kg}} \approx 0.1056 \, \text{mol/kg}.\]

Key Concepts

Moles of Solute: A Key Ingredient

Grasping the concept of "moles of solute" is essential in chemistry, especially when discussing solutions. Moles measure the amount of a substance. Think of it as a set number of molecules. For example, when we say there are "moles of acetic acid" in a solution, we are referring to the number of molecules of acetic acid present.
The number of moles can be calculated using the formula: \[\text{moles} = \frac{\text{mass of solute in grams}}{\text{molar mass of solute in g/mol}}.\]
In our exercise, for \(\mathrm{CH}_3\mathrm{COOH}\), we know the mass of the solute is 5 grams and the molar mass is 60.05 \(\mathrm{g/mol}\). Using the formula above: \[\text{moles} = \frac{5}{60.05} \approx 0.0833 \, \text{mol}.\]
This tells us how much of the acetic acid we have, relative to its molecular composition.

Molar Mass: What’s in a Molecule?

Molar mass is vital for converting between grams and moles. It represents the mass of one mole of a given substance. It is calculated by summing the atomic masses of each element within a molecule, derived from the periodic table.
For acetic acid, \(\mathrm{CH}_3\mathrm{COOH}\), we perform the following calculation:- Carbon (C): 12.01 \(\mathrm{g/mol}\) and there are 2 carbon atoms.- Hydrogen (H): 1.01 \(\mathrm{g/mol}\) and there are 4 hydrogen atoms.- Oxygen (O): 16.00 \(\mathrm{g/mol}\) with one atom included in the calculation.- Another hydrogen contributes an additional 1.01 \(\mathrm{g/mol}\).
The summation leads to: \(2 \times 12.01 + 4 \times 1.01 + 16.00 + 1.01 = 60.05 \, \mathrm{g/mol}\).
Understanding molar mass allows us to easily convert from mass (in grams) to moles, which is crucial in solution chemistry.

Density of Solvent: Understanding Volume and Mass

Density connects a material's mass with its volume. In simple terms, density tells us how "packed" the atoms or molecules in a material are. It is often expressed in \(\mathrm{g/mL}\) or \(\mathrm{g/cm^3}\).
In our problem, ethanol is used as the solvent and its density is given as 0.789 \(\mathrm{g/mL}\). This implies each milliliter of ethanol weighs 0.789 grams.
To find the mass of our ethanol solvent, we use the volume (1 liter) converted to milliliters (1000 ml), and multiply by the density: \(0.789 \, \mathrm{g/mL} \times 1000 \, \mathrm{mL} = 789 \, \mathrm{g}\).
Understanding density is essential for accurately determining how much of a solvent is used in creating a solution, which in turn affects calculations like the one required for determining molality.