Problem 238 - A Complete Resource Book in Chemistry for JEE Main

Question

\(10 \mathrm{~mL}\) of pure ethanol of density \(0.785 \mathrm{~g} / \mathrm{mL}\) was diluted with water to a final volume of \(100 \mathrm{~mL}\). The density of the resulting solution was \(0.9866 \mathrm{~g} / \mathrm{mL}\). The percentage by weight of ethanol is about

Step-by-Step Solution

Answer

The percentage by weight of ethanol is approximately 7.96%.

Step 1: Calculate the Mass of Pure Ethanol

Given the density of ethanol is \(0.785 \mathrm{~g} / \mathrm{mL}\) and the volume is \(10 \mathrm{~mL}\), you can calculate the mass of ethanol using the formula: \( \text{Mass} = \text{Volume} \times \text{Density} \). So, the mass of ethanol is \(10 \mathrm{~mL} \times 0.785 \mathrm{~g} / \mathrm{mL} = 7.85 \mathrm{~g}\).

Step 2: Calculate the Total Mass of the Solution

The total volume of the solution after dilution is \(100 \mathrm{~mL}\) and its density is \(0.9866 \mathrm{~g} / \mathrm{mL}\). Use the formula \( \text{Mass} = \text{Volume} \times \text{Density} \) to find the total mass of the solution: \(100 \mathrm{~mL} \times 0.9866 \mathrm{~g} / \mathrm{mL} = 98.66 \mathrm{~g}\).

Step 3: Calculate the Mass of Water in the Solution

Subtract the mass of ethanol from the total mass of the solution to find the mass of water: \(98.66 \mathrm{~g} - 7.85 \mathrm{~g} = 90.81 \mathrm{~g}\). This represents the mass of water added.

Step 4: Compute the Percentage by Weight of Ethanol

To find the percentage by weight of ethanol in the solution, use the formula: \( \text{Percentage by weight} = \left( \frac{\text{Mass of ethanol}}{\text{Total mass of solution}} \right) \times 100 \% \). Plugging in the values gives: \( \left( \frac{7.85 \mathrm{~g}}{98.66 \mathrm{~g}} \right) \times 100 \% \approx 7.96 \%\).

Key Concepts

Mass Calculation

Whenever we want to determine the amount of a substance, we need to calculate its mass. Mass helps us understand how much material is present. Let's break down how to find the mass of ethanol in our example:

First, identify the given data:

• Density of ethanol: \(0.785 \, \text{g/mL}\)
• Volume of ethanol: \(10 \, \text{mL}\)

To find the mass, we employ the formula:\[ \text{Mass} = \text{Volume} \times \text{Density} \]Applying it to our case:\[ 10 \, \text{mL} \times 0.785 \, \text{g/mL} = 7.85 \, \text{g} \]Thus, the mass of ethanol is \(7.85 \, \text{g}\). By understanding this process, you can easily calculate the mass whenever density and volume are provided.

Density Concepts

Density is a crucial concept that tells us how much mass is in a specific volume. It's like knowing how tightly packed the molecules are in the space provided. In our example, density helps us find the total mass of the resulting solution after dilution:

Several key pieces of information are needed:

• Total volume after dilution: \(100 \, \text{mL}\)
• The density of the solution: \(0.9866 \, \text{g/mL}\)

Use the same mass formula to find the total mass:\[ \text{Mass} = \text{Volume} \times \text{Density} \]Here, we compute:\[ 100 \, \text{mL} \times 0.9866 \, \text{g/mL} = 98.66 \, \text{g} \]This tells us that the total mass of the solution is \(98.66 \, \text{g}\). Understanding density allows you to estimate how much material will end up in a given volume.

Percentage by Weight

Percentage by weight is an intuitive way to express concentration, showing how much of the total solution is due to a particular substance. Let's find the percent weight of ethanol in the solution:

First, recall the essential equation:\[ \text{Percentage by weight} = \left( \frac{\text{Mass of ethanol}}{\text{Total mass of solution}} \right) \times 100\% \]In our case, calculate it as follows:

• Mass of ethanol: \(7.85 \, \text{g}\)
• Total mass of solution: \(98.66 \, \text{g}\)

Substituting these values:\[ \left( \frac{7.85}{98.66} \right) \times 100 \% \approx 7.96\% \]This means ethanol is about \(7.96\%\) of the solution's mass. Knowing this calculation helps you determine the concentration of any solute in a solution, useful for many applications in chemistry and industry.