Q.If 4 g of NaOH dissolves in 36 g of $H_2O$, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is $1\ \text{g mL}^{-1}$).

The key idea here is to understand and apply the definitions of mole fraction and molarity, which are fundamental ways to express solution concentration.

First, we determine the moles of each component.

The molar mass of NaOH is $23 + 16 + 1 = 40\ \text{g mol}^{-1}$.

The molar mass of $H_2O$ is $2 \times 1 + 16 = 18\ \text{g mol}^{-1}$.

Moles of NaOH ($n_{\text{NaOH}}$) $= \frac{4\ \text{g}}{40\ \text{g mol}^{-1}} = 0.1\ \text{mol}$.

Moles of $H_2O$ ($n_{H_2O}$) $= \frac{36\ \text{g}}{18\ \text{g mol}^{-1}} = 2.0\ \text{mol}$.

Total moles ($n_{\text{total}}$) $= 0.1\ \text{mol} + 2.0\ \text{mol} = 2.1\ \text{mol}$.

Next, we calculate the mole fraction of each component.

Mole fraction of NaOH ($X_{\text{NaOH}}$) $= \frac{n_{\text{NaOH}}}{n_{\text{total}}} = \frac{0.1\ \text{mol}}{2.1\ \text{mol}} = \frac{1}{21} \approx 0.0476$.

Mole fraction of $H_2O$ ($X_{H_2O}$) $= \frac{n_{H_2O}}{n_{\text{total}}} = \frac{2.0\ \text{mol}}{2.1\ \text{mol}} = \frac{20}{21} \approx 0.9524$.

Finally, we determine the molarity of the solution.

The total mass of the solution is $4\ \text{g (NaOH)} + 36\ \text{g (H}_2\text{O)} = 40\ \text{g}$. …

When we talk about the composition of a solution, we often need ways to quantify how much of each substance is present. Two common measures are mole fraction and molarity, each serving a different purpose and requiring a slightly different approach to calculation.

Mole fraction ($X$) tells us the proportion of moles of a particular component relative to the total moles of all components in the solution. It's a dimensionless quantity and is particularly useful when studying colligative properties, as it directly reflects the number of particles. Since it's a ratio, the sum of mole fractions of all components in a solution must always be equal to 1.

Molarity ($M$) expresses the concentration of a solution in terms of moles of solute per liter of solution. It's a very common unit for laboratory work because it relates directly to the volume of solution used in reactions. The key distinction here is that molarity uses the volume of the entire solution, not just the solvent. This is where density (or specific gravity) becomes crucial if the volume isn't directly given.

Let's break down the calculation step-by-step.

1. Identify the given information and what needs to be calculated.

   We are given:

   • Mass of solute (NaOH) = $4\ \text{g}$
   • Mass of solvent ($H_2O$) = $36\ \text{g}$
   • Specific gravity of solution = $1\ \text{g mL}^{-1}$ (This means the density of the solution is $1\ \text{g mL}^{-1}$)

   We need to calculate:

   • Mole fraction of NaOH ($X_{NaOH}$)
   • Mole fraction of $H_2O$ ($X_{H_2O}$)
   • Molarity of the solution ($M$)

2. Calculate the molar mass of each component.

   To find the number of moles, we first need the molar mass of each substance.

   • For Sodium Hydroxide (NaOH):

     Molar mass of Na = $23\ \text{g mol}^{-1}$

     Molar mass of O = $16\ \text{g mol}^{-1}$

     Molar mass of H = $1\ \text{g mol}^{-1}$

     Molar mass of NaOH $= (23 + 16 + 1)\ \text{g mol}^{-1} = 40\ \text{g mol}^{-1}$

   • For Water ($H_2O$):

     Molar mass of H = $1\ \text{g mol}^{-1}$

     Molar mass of O = $16\ \text{g mol}^{-1}$

     Molar mass of $H_2O = (2 \times 1 + 16)\ \text{g mol}^{-1} = 18\ \text{g mol}^{-1}$

3. Calculate the number of moles for each component.

   Using the formula $n = \frac{\text{mass}}{\text{molar mass}}$:

   • Moles of NaOH ($n_{NaOH}$):

     $n_{NaOH} = \frac{4\ \text{g}}{40\ \text{g mol}^{-1}} = 0.1\ \text{mol}$

   • Moles of $H_2O$ ($n_{H_2O}$):

     $n_{H_2O} = \frac{36\ \text{g}}{18\ \text{g mol}^{-1}} = 2.0\ \text{mol}$

4. Calculate the total number of moles in the solution.

   The total moles ($n_{total}$) is the sum of the moles of all components:

   $n_{total} = n_{NaOH} + n_{H_2O} = 0.1\ \text{mol} + 2.0\ \text{mol} = 2.1\ \text{mol}$

5. Calculate the mole fraction of each component.

   The mole fraction of a component A is given by $X_A = \frac{n_A}{n_{total}}$.

   • Mole fraction of NaOH ($X_{NaOH}$):

     $X_{NaOH} = \frac{n_{NaOH}}{n_{total}} = \frac{0.1\ \text{mol}}{2.1\ \text{mol}} \approx 0.0476$

   • Mole fraction of $H_2O$ ($X_{H_2O}$):

     $X_{H_2O} = \frac{n_{H_2O}}{n_{total}} = \frac{2.0\ \text{mol}}{2.1\ \text{mol}} \approx 0.9524$

   Tip

   Always check that the sum of mole fractions is approximately 1. …

Method: Stepwise Composition Analysis (Mole Fraction + Molarity)

We’ll solve this in two clear parts — first mole fraction, then molarity.

Part 1 — Mole Fraction

Step 1: Find moles of each component

• Molar mass of NaOH = $23 + 16 + 1 = 40\ \text{g mol}^{-1}$

  Moles of NaOH = $\frac{4}{40} = 0.1\ \text{mol}$

• Molar mass of $\text{H}_2\text{O}$ = $18\ \text{g mol}^{-1}$

  Moles of $\text{H}_2\text{O}$ = $\frac{36}{18} = 2\ \text{mol}$

Step 2: Apply mole fraction formula

• Mole fraction of NaOH ($x_{\text{NaOH}}$)

• Mole fraction of $\text{H}_2\text{O}$ ($x_{\text{H}_2\text{O}}$)

Result:

$x_{\text{NaOH}} = \frac{1}{21}$, $x_{\text{H}_2\text{O}} = \frac{20}{21}$

Part 2 — Molarity

Step 3: Find total mass and volume of solution

• Total mass = mass of NaOH + mass of water = $4 + 36 = 40\ \text{g}$
• Specific gravity = $1\ \text{g mL}^{-1}$ → density = $1\ \text{g/mL}$ …

Here are the common mistakes students make on this exact problem, along with how to avoid each — broken down concept-first.

1. Confusing Molality with Molarity

Mistake:

Students often use the mass of solvent (36 g) to calculate molarity, or use the volume of solution to calculate molality.

Why it happens:

The terms sound similar and both involve moles of solute, but the denominator is different:

• Molality ($m$) = moles of solute / kg of solvent
• Molarity ($M$) = moles of solute / L of solution

How to avoid:

Always ask: “Am I dividing by solvent mass or solution volume?”

• If the problem gives mass of solvent → you are likely finding molality or mole fraction.
• If the problem gives volume of solution or density → you are likely finding molarity.

2. Using the Wrong Mass for Mole Fraction Denominator

Mistake:

Students write mole fraction of NaOH as:

but then incorrectly compute $n_{\text{H}_2\text{O}}$ using 36 g as if it were the total mass of solution.

Why it happens:

They forget that 36 g is only the solvent, not the solution.

How to avoid:

• Step 1: Identify each component separately.
  • Solute: NaOH, mass = 4 g
  • Solvent: H₂O, mass = 36 g
• Step 2: Find moles of each:
  • $n_{\text{NaOH}} = \frac{4}{40} = 0.1$ mol
  • $n_{\text{H}_2\text{O}} = \frac{36}{18} = 2$ mol
• Step 3: Total moles = $0.1 + 2 = 2.1$ mol
• Step 4:

Check: Mole fractions must add to 1.

3. Forgetting to Convert Mass to kg for Molality

Mistake:

Using 36 g directly in the molality formula:

Why it happens:

Molality is defined per kg of solvent, not per gram.

How to avoid:

Always convert solvent mass to kg:

Quick tip: Write the unit as mol/kg to remind yourself.

4. Misusing Specific Gravity to Find Volume

Mistake:

Assuming specific gravity = 1 g/mL means the volume of solution equals the mass of solvent (36 mL).

Why it happens:

They confuse “specific gravity of solution” with “density of water.”

How to avoid:

• Specific gravity = 1 g/mL means density of solution = 1 g/mL. …