Question

A solution is prepared by dissolving ethanol in water. The mole fraction of ethanol in this solution is 0.04. What is the molarity (in $\text{mol L}^{-1}$) of the solution? (density of water is $1\text{ g mL}^{-1}$. Neglect the volume of ethanol)

• A. 0.96
• B. 2.31
• C. 1.96
• D. 3.31

Hint:

When mole fraction is small, you can relate molarity directly to molality because the volume of solute is negligible.
Formula: $M \approx m = \frac{x_2 \times 1000}{x_1 \times M_1}$.
Here, $M \approx \frac{0.04 \times 1000}{0.96 \times 18} \approx 2.31\text{ M}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question is about liquid solution concentration terms.
We need to convert the mole fraction of ethanol in an aqueous solution into its molarity ($M$), given the density of water and neglecting the volume of ethanol.

Step 2: Key Formula or Approach:
The mole fraction of a component in a binary solution is:
\[ x_2 = \frac{n_2}{n_1 + n_2} \] Molarity ($M$) is given by:
\[ M = \frac{\text{moles of solute } (n_2)}{\text{Volume of solution in L } (V)} \] Since the volume of ethanol is neglected, the volume of the solution is equal to the volume of the water solvent:
\[ V_{\text{soln}} \approx V_{\text{water}} = \frac{\text{Mass of water}}{\text{Density of water}} \]

Step 3: Detailed Explanation:

• Let us assume a total of 1 mole of the solution.
The mole fraction of ethanol ($x_{\text{ethanol}}$) = 0.04.
The mole fraction of water ($x_{\text{water}}$) = $1 - 0.04 = 0.96$.
Therefore, in 1 mole of solution:
Moles of ethanol ($n_2$) = 0.04 mol.
Moles of water ($n_1$) = 0.96 mol.

• Let us calculate the mass and volume of the water:
Molar mass of water ($\text{H}_2\text{O}$) = $18\text{ g/mol}$.
Mass of water = $\text{moles} \times \text{molar mass} = 0.96\text{ mol} \times 18\text{ g/mol} = 17.28\text{ g}$.
Given density of water = $1\text{ g/mL}$.
Volume of water = $\frac{\text{Mass}}{\text{Density}} = \frac{17.28\text{ g}}{1\text{ g/mL}} = 17.28\text{ mL}$.

• Convert the volume of the solution to liters:
\[ V_{\text{soln}} = 17.28\text{ mL} = 0.01728\text{ L} \]

• Now, we calculate the molarity of the solution:
\[ M = \frac{n_2}{V_{\text{soln}}} = \frac{0.04\text{ mol}}{0.01728\text{ L}} \approx 2.3148\text{ mol L}^{-1} \]

• This matches the value 2.31.


Step 4: Final Answer:
The molarity of the ethanol solution is 2.31 mol L$^{-1}$.