Q.Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

The key idea is Molarity from Mole Fraction.

To calculate the molarity, we use the mole fraction to fix the ratio of ethanol to water, then use the density of water (given as 1) to convert the water into a volume — for this dilute solution the volume of the solution is approximately the volume of the water in it.

1. Take a convenient basis of $1$ total mole of solution. With $x_{C_2H_5OH} = 0.040$, this contains $0.040$ mol ethanol and $0.960$ mol water.
2. Mass of the water: $0.960 \text{ mol} \times 18.015 \text{ g/mol} = 17.29 \text{ g}$. …

Molarity and mole fraction are two ways to express the concentration of a solution. To convert between them, we need to relate the moles of solute and solvent to the total volume of the solution.

• Molarity ($M$) is defined as the number of moles of solute per liter of solution.

• Mole fraction ($X$) of a component is the ratio of the moles of that component to the total moles of all components in the solution.

The problem provides the mole fraction of ethanol and the density of water. We need to find the molarity. The key challenge is to determine the volume of the solution, as the density of the solution itself is not given. For dilute aqueous solutions, a common and reasonable approximation is to assume that the volume of the solution is approximately equal to the volume of the solvent (water). We will proceed with this assumption, as the mole fraction of ethanol (0.040) indicates a relatively dilute solution.

Here's how we can calculate the molarity:

1. Assume a basis for calculation.

   To work with mole fractions, it's convenient to assume a total amount of solution. Let's assume we have a total of $1 \text{ mole}$ of the solution. This allows us to directly use the mole fraction to find the moles of each component.

2. Calculate moles of ethanol and water.

   Given the mole fraction of ethanol ($X_{\text{ethanol}}$) is $0.040$:

   Moles of ethanol ($n_{\text{ethanol}}$) = $X_{\text{ethanol}} \times \text{total moles of solution}$

   $n_{\text{ethanol}} = 0.040 \times 1 \text{ mol} = 0.040 \text{ mol}$

   The mole fraction of water ($X_{\text{water}}$) is $1 - X_{\text{ethanol}}$:

   $X_{\text{water}} = 1 - 0.040 = 0.960$

   Moles of water ($n_{\text{water}}$) = $X_{\text{water}} \times \text{total moles of solution}$

   $n_{\text{water}} = 0.960 \times 1 \text{ mol} = 0.960 \text{ mol}$

3. Determine the molar masses of ethanol and water.

   Using standard atomic masses (C=12.011 g/mol, H=1.008 g/mol, O=15.999 g/mol):

   • For ethanol ($C_2H_5OH$): $M_{\text{ethanol}} = (2 \times 12.011) + (6 \times 1.008) + (1 \times 15.999) = 24.022 + 6.048 + 15.999 = 46.069 \text{ g/mol}$
   • For water ($H_2O$): $M_{\text{water}} = (2 \times 1.008) + (1 \times 15.999) = 2.016 + 15.999 = 18.015 \text{ g/mol}$

4. Calculate the mass of water.

   Mass of water ($m_{\text{water}}$) = $n_{\text{water}} \times M_{\text{water}}$

   $m_{\text{water}} = 0.960 \text{ mol} \times 18.015 \text{ g/mol} = 17.2944 \text{ g}$

5. Calculate the volume of water.

   The problem states the density of water is one. We will interpret this as $1.00 \text{ g/mL}$. …

Method: Mole Fraction to Molarity Conversion

This method uses the definition of mole fraction and the definition of molarity, linking them through the mass and volume of the solvent.

Step-by-step solution

Step 1: Understand the given data

• Mole fraction of ethanol ($C_2H_5OH$), $x_{\text{ethanol}} = 0.040$
• Density of water = $1 \text{ g/mL}$ (so $1 \text{ mL}$ water = $1 \text{ g}$ water)
• We need molarity = moles of solute per litre of solution

Step 2: Choose a convenient basis

Take 1 mole of solution as the basis.

In 1 mole of solution:

• Moles of ethanol = $0.040$ mol
• Moles of water = $1 - 0.040 = 0.960$ mol

Step 3: Find mass of water

Mass of water = moles of water × molar mass of water

$= 0.960 \times 18 \text{ g/mol} = 17.28 \text{ g}$

Since density of water is $1 \text{ g/mL}$,

Volume of water = $17.28 \text{ mL} = 0.01728 \text{ L}$

Step 4: Assume volume of solution ≈ volume of water

For dilute solutions (ethanol mole fraction is small), the volume of solute is negligible compared to solvent. …

Common Mistakes in Molarity Calculation from Mole Fraction (Ethanol in Water)

Students often struggle with this problem because it requires converting between concentration units. Here are the most frequent errors and how to avoid them.

Mistake 1: Confusing Mole Fraction with Molarity

The error: Students treat the mole fraction (0.040) as if it were already the molarity, writing $M = 0.040$.

Why it's wrong: Mole fraction is a ratio of moles of one component to total moles. Molarity is moles of solute per litre of solution — a completely different quantity.

How to avoid: Always write the definitions side-by-side:

• Mole fraction of ethanol: $x_{\text{ethanol}} = \frac{n_{\text{ethanol}}}{n_{\text{ethanol}} + n_{\text{water}}}$
• Molarity: $M = \frac{n_{\text{ethanol}}}{V_{\text{solution (in L)}}}$

Key insight: You must find the volume of the solution, not just the moles.

Mistake 2: Forgetting to Account for the Solvent's Contribution to Volume

The error: Students calculate moles of ethanol, then divide by the volume of water only (assuming volume of solution = volume of water).

Why it's wrong: When ethanol dissolves in water, the total volume changes. The problem gives density of water as 1 g/mL, but the solution volume is not the same as water volume.

How to avoid:

• Remember: $V_{\text{solution}} \neq V_{\text{water}}$ (unless stated otherwise)
• You must find the mass of the solution first, then use density to find volume
• Since density of water is given as 1 g/mL, you can find mass of water from its volume, but the solution's mass = mass of water + mass of ethanol

Mistake 3: Incorrectly Setting Up the Mole Fraction Equation

The error: Writing $x_{\text{ethanol}} = \frac{n_{\text{ethanol}}}{n_{\text{water}}}$ instead of $\frac{n_{\text{ethanol}}}{n_{\text{ethanol}} + n_{\text{water}}}$.

Why it's wrong: Mole fraction uses total moles in the denominator, not just the solvent.

How to avoid: Memorise the formula with a mnemonic: "Part over Whole" — the mole fraction of any component is its moles divided by total moles of all components.

Mistake 4: Assuming 1 Mole of Solution

The error: Students set $n_{\text{ethanol}} + n_{\text{water}} = 1$ and then take $n_{\text{ethanol}} = 0.040$ directly.

Why it's wrong: While this is mathematically valid for finding the ratio, students then forget to scale to a real volume. The mole fraction 0.040 means: for every 1 mole of total mixture, 0.040 moles are ethanol. But you need to convert this to a real mass and volume.

How to avoid:

• Yes, you can assume total moles = 1 (or any convenient number)
• But then calculate the actual masses and actual volume of that mixture
• Only then compute molarity

Mistake 5: Using Wrong Molar Masses

The error: Using incorrect molar masses for ethanol ($C_2H_5OH$) or water.

Why it's wrong: Even a small error propagates through the calculation.

How to avoid: Write the calculation clearly:

• Ethanol ($C_2H_5OH$): $2(12) + 6(1) + 16 = 24 + 6 + 16 = 46 \text{ g/mol}$
• Water ($H_2O$): $2(1) + 16 = 18 \text{ g/mol}$ …