Q.4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is _____________.

Concept: Molality After Dilution — molality depends on moles of solute and mass of solvent (in kg); dilution changes volume, not moles.

Step 1 – Moles of NaCl

Initial moles = $M \times V = 0.02 \times 4 = 0.08$ mol.

Step 2 – Mass of solvent after dilution

Total volume after adding 1 L water = $4 + 1 = 5$ L.

Assuming density of dilute aqueous solution ≈ 1 g/mL, mass of solution ≈ $5$ kg.

Mass of solute (NaCl) = $0.08 \times 58.5 \approx 4.68$ g = 0.00468 kg. …

Why Molality, Not Molarity?

This question tests a subtle but critical distinction. Molarity ($M$) is moles per litre of solution. Molality ($m$) is moles per kilogram of solvent. When you add water, the volume changes, but so does the mass of solvent. You cannot simply use the dilution formula $M_1V_1 = M_2V_2$ and then call that result "molality" — that would be a serious error.

Here, we start with an aqueous solution. The solvent is water. Adding 1 litre of water adds 1 kg of water (since density of water is 1 kg/L). The moles of NaCl stay fixed. So the problem reduces to: find the initial mass of water, add 1 kg to it, then compute moles per kg.

Step-by-Step Solution

1. Find the moles of NaCl initially.

We have 4 L of 0.02 M NaCl solution.

Molarity = moles / volume (in L). So:

These 0.08 moles will remain constant throughout — dilution does not change the amount of solute.

2. Find the mass of water in the original solution.

The solution is very dilute (0.02 M). For such dilute aqueous solutions, the volume contributed by the solute is negligible. So we can safely assume that the 4 L of solution is essentially 4 L of water.

Mass of 4 L of water = $4 \times 1000$ g = 4000 g = 4 kg. …

Method: Dilution + Molality Calculation

We'll use two steps — first find the new molarity after dilution, then convert to molality using density of water.

Step 1: Find new molarity after dilution

Formula:

$M_1 V_1 = M_2 V_2$

• Initial: $M_1 = 0.02 \, \text{M}$, $V_1 = 4 \, \text{L}$
• Final volume: $V_2 = 4 + 1 = 5 \, \text{L}$

So the molarity of the diluted solution is 0.016 M.

Step 2: Convert molarity to molality

Key assumption (standard for dilute aqueous solutions):

Density of water $\approx 1 \, \text{g/mL}$, so 1 L solution $\approx 1$ kg solvent.

Molality formula:

• Moles of NaCl in 1 L of 0.016 M solution = $0.016$ moles …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Molarity with Molality

The error: Students treat the problem as a simple dilution of molarity and directly calculate:

$M_1V_1 = M_2V_2$

$0.02 \times 4 = M_2 \times 5 \implies M_2 = 0.016$

Then they incorrectly mark option (iv) 0.016 as the molality.

Why it's wrong: Molarity ($M$) is moles per litre of solution. Molality ($m$) is moles per kg of solvent. They are not interchangeable, especially when water is added.

How to avoid: Always check: "Does the question ask for molarity or molality?" If it's molality, you need the mass of solvent in kg, not the volume of solution.

Mistake 2: Assuming Density of Water = 1 g/mL for the Final Solution

The error: Students take the final volume (5 L) and assume the mass of solvent = 5000 g = 5 kg, then compute:

$m = \frac{\text{moles}}{\text{kg solvent}} = \frac{0.08}{5} = 0.016$

Again landing on option (iv).

Why it's wrong: The density of the final solution is not 1 g/mL — it contains dissolved NaCl. The solvent mass is not the same as the solution mass.

How to avoid: Remember:

• Solvent mass = mass of water present (not the solution)
• The original 4 L of 0.02 M solution already contains water as solvent
• Adding 1 L water adds exactly 1 kg of solvent (since density of pure water ≈ 1 g/mL)

Mistake 3: Forgetting to Account for Solvent Already Present

The error: Students calculate moles of NaCl correctly:

$\text{Moles of NaCl} = 0.02 \times 4 = 0.08 \text{ mol}$

But then they only consider the added 1 L water as solvent (1 kg), giving:

$m = \frac{0.08}{1} = 0.08$ (not even in the options)

Why it's wrong: The original 4 L solution already contains water as solvent. You must find the total solvent mass after dilution.

How to avoid: Break it into steps:

1. Find moles of solute (NaCl) — done ✓
2. Find mass of solvent in the original solution — this requires knowing the mass of the original solution and subtracting the mass of NaCl
3. Add the mass of added water (1 kg)
4. Divide moles by total solvent mass in kg

Mistake 4: Not Using Density Information (When Needed)

The error: Students try to compute solvent mass without density of the original 0.02 M NaCl solution.

Why it's tricky: The problem doesn't give density of 0.02 M NaCl. But for such a dilute solution, we can approximate:

• Mass of 4 L solution ≈ 4000 g (since density ≈ 1 g/mL for very dilute aq. solutions)
• Mass of NaCl = $0.08 \times 58.5 = 4.68$ g
• Mass of solvent (water) in original = $4000 - 4.68 \approx 3995.32$ g ≈ 3.995 kg …