Q.The density of 3 molal solution of NaOH is $1.110\ \text{g mL}^{-1}$. Calculate the molarity of the solution.

Concept: Molality Calculation — Molality uses mass of solvent (kg), while molarity uses volume of solution (L). Density links mass and volume.

Step 1: Base the calculation on 1 kg of solvent.

3 molal NaOH means 3 moles of NaOH in 1 kg of water.

Mass of NaOH = $3 \times 40 = 120\ \text{g}$.

Total mass of solution = $1000\ \text{g} + 120\ \text{g} = 1120\ \text{g}$.

Step 2: Find volume of solution using density.

Volume = $\frac{\text{mass}}{\text{density}} = \frac{1120\ \text{g}}{1.110\ \text{g mL}^{-1}} = 1009.01\ \text{mL} = 1.00901\ \text{L}$. …

1. Interpret 3 molal

A $3$ molal solution has $3$ mol $NaOH$ dissolved in $1$ kg $=1000$ g of water.

2. Mass of the solution

Molar mass of $NaOH = 23+16+1 = 40\ \text{g mol}^{-1}$, so mass of $NaOH = 3\times40 = 120$ g.

3. Volume of the solution

Using $\text{density}=1.110\ \text{g mL}^{-1}$:

4. Molarity …

Method: Density–Molality to Molarity Conversion

This method uses the definition of molality and density to find the volume of the solution, then converts moles per kg solvent to moles per litre solution.

Steps

Step 1: Understand the given data

• Molality ($m$) = 3 molal = 3 moles of NaOH per 1 kg of water
• Density ($\rho$) = $1.110\ \text{g mL}^{-1}$
• Molar mass of NaOH = $23 + 16 + 1 = 40\ \text{g mol}^{-1}$

Step 2: Find the mass of the solution

Mass of solute (NaOH) = $3 \times 40 = 120\ \text{g}$

Mass of solvent (water) = $1000\ \text{g}$ (since 1 kg)

Mass of solution = $120 + 1000 = 1120\ \text{g}$

Step 3: Find the volume of the solution

Using density:

Step 4: Calculate molarity

Molarity ($M$) = moles of solute per litre of solution …

Common Mistakes in Molality → Molarity Calculation

Students often struggle with this conversion because it requires careful handling of mass, volume, and the distinction between molality and molarity. Here are the most frequent errors and how to avoid them.

Mistake 1: Confusing molality with molarity

The error:

Students treat 3 molal as 3 moles per litre of solution, which is actually molarity.

Why it happens:

Both terms sound similar and both use "moles of solute," but:

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

How to avoid:

Always write the definition before starting:

• $m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$
• $M = \frac{\text{moles of solute}}{\text{volume of solution (in L)}}$

Mistake 2: Using density to directly convert molality to molarity

The error:

Students multiply: $3 \times 1.110 = 3.33\ \text{M}$ — this is wrong.

Why it happens:

Density is mass/volume, but molality uses solvent mass, not solution mass.

How to avoid:

Remember the stepwise logic:

1. Assume 1 kg of solvent (this is the standard starting point for molality)
2. Find moles of solute from molality
3. Find mass of solute
4. Find total mass of solution = mass of solvent + mass of solute
5. Use density to find volume of solution
6. Finally, molarity = moles of solute / volume of solution (in L)

Mistake 3: Forgetting to convert units

The error:

Using density in g/mL but volume in litres without conversion.

Example of wrong work:

Volume = $\frac{\text{mass of solution (g)}}{\text{density (g/mL)}}$ → then using this volume directly in litres.

How to avoid:

Always convert mL to L:

• $1\ \text{mL} = 10^{-3}\ \text{L}$
• So if volume comes out as $V\ \text{mL}$, then $V\ \text{mL} = \frac{V}{1000}\ \text{L}$

Mistake 4: Using wrong molar mass or forgetting it entirely

The error:

Students skip finding the mass of NaOH, or use an incorrect molar mass.

Why it happens:

They forget that to find the mass of solute from moles, you must know the molar mass.

How to avoid:

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