Q.If the density of some lake water is 1.25 g mL$^{-1}$ and contains 92 g of $Na^+$ ions per kg of water, calculate the molarity of $Na^+$ ions in the lake.

Concept: Molarity from Mass, Given Density — find moles of Na⁺ per kg of water, then use the given density to convert to solution volume and molarity.

Step 1: Moles of Na⁺

Molar mass of Na = 23 g mol⁻¹

Moles of Na⁺ = $\frac{92}{23} = 4$ mol per kg of water.

Step 2: Mass of solution

Water = 1000 g, Na⁺ = 92 g → total mass = 1092 g. …

1. Moles of Na$^+$ (atomic mass $23$):

2. Mass of solution (water $+$ Na$^+$):

3. Volume from density ($1.25\ \text{g mL}^{-1}$): …

Method: Density-Assisted Molality-to-Molarity Conversion

This method converts molality (moles per kg of solvent) into molarity (moles per litre of solution) using the density of the solution.

Step 1: Find moles of Na⁺ ions

Given:

• Mass of Na⁺ = 92 g per kg of water
• Molar mass of Na = 23 g mol⁻¹

So, molality of Na⁺ = 4 mol kg⁻¹ (since it's per kg of water).

Step 2: Find total mass of solution

We have:

• Mass of water (solvent) = 1000 g = 1 kg
• Mass of Na⁺ ions = 92 g

Step 3: Use density to find volume of solution

Density = 1.25 g mL⁻¹

Convert to litres: …

Great — let’s break this down. The question is a classic molality-to-molarity conversion problem, and students often slip on the density and mass relationships.

🧠 The Core Concept

We are given:

• Density of lake water = $1.25 \, \text{g mL}^{-1}$
• Mass of $\text{Na}^+$ ions = $92 \, \text{g}$ per kg of water
• We need molarity = moles of $\text{Na}^+$ per litre of solution

The trap: “per kg of water” ≠ “per kg of solution”.

You must convert mass of solvent → mass of solution → volume of solution.

✗ Common Mistake #1

Treating “per kg of water” as “per kg of solution”

🧩 What students do wrong:

They take $92 \, \text{g Na}^+$ in $1 \, \text{kg}$ of solution and directly compute moles per litre using density.

✓ How to avoid:

• Read carefully: “per kg of water” means solvent mass = 1 kg, not solution mass.
• Solution mass = mass of water + mass of solute Here: $1000 \, \text{g water} + 92 \, \text{g Na}^+ = 1092 \, \text{g solution}$

✗ Common Mistake #2

Forgetting to convert density units properly

🧩 What students do wrong:

They use density $1.25 \, \text{g mL}^{-1}$ directly without converting to $\text{g L}^{-1}$ or to volume in litres.

✓ How to avoid:

• Always convert density to g per litre for molarity: $1.25 \, \text{g mL}^{-1} = 1250 \, \text{g L}^{-1}$
• Or, compute volume in mL first, then convert to L.

✗ Common Mistake #3

Using wrong molar mass or forgetting to convert grams to moles

🧩 What students do wrong:

They might use atomic mass of Na as $23 \, \text{g mol}^{-1}$ correctly, but then forget to divide $92 \, \text{g}$ by $23$ — or they use $22.99$ and round poorly.

✓ How to avoid:

• Molar mass of Na$^+$ = $23 \, \text{g mol}^{-1}$ (same as Na atom)
• Moles of Na$^+$ = $\frac{92}{23} = 4 \, \text{mol}$

✗ Common Mistake #4

Miscalculating volume from mass and density

🧩 What students do wrong: …