Q.The density of 3 M solution of NaCl is $1.25\ g\ mL^{-1}$. Calculate the molality of the solution.

Concept: Interconversion between molarity and molality using density.

We have a 3 M NaCl solution with density $1.25\ \text{ g mL}^{-1}$. Take a basis of 1 L (1000 mL) of solution.

Step 1: Mass of solution = $1000 \times 1.25 = 1250\ \text{g}$.

Step 2: Moles of NaCl = $3\ \text{mol}$ (from molarity). Mass of NaCl = $3 \times 58.5 = 175.5\ \text{g}$ (molar mass of NaCl = 58.5 g/mol).

Step 3: Mass of solvent (water) = $1250 - 175.5 = 1074.5\ \text{g} = 1.0745\ \text{kg}$. …

Why this conversion works

Molarity and molality measure concentration differently. Molarity ($M$) counts moles of solute per liter of solution, while molality ($m$) counts moles per kilogram of solvent. The density bridges these two worlds: it tells us the total mass of solution, from which we can extract the solvent mass.

The strategy is to pick a convenient sample—1 L of solution works beautifully—find how many moles of NaCl it contains (that's what molarity gives us), calculate the mass of just the water in that liter, then compute molality.

Step-by-step calculation

1. Choose a basis volume

Take exactly 1 L (or 1000 mL) of the 3 M NaCl solution. This contains:

2. Find the total mass of this solution

The density is $1.25\ \text{g mL}^{-1}$, so:

3. Calculate the mass of NaCl present

The molar mass of NaCl is $M_{\text{NaCl}} = 23 + 35.5 = 58.5\ \text{g mol}^{-1}$.

Mass of solute:

4. Determine the mass of water (solvent)

The solution mass equals solute plus solvent:

5. Apply the definition of molality

Molality is moles of solute per kilogram of solvent: …

Method: Density–Molarity to Molality Conversion

This method uses the density of the solution to find the mass of solvent, then applies the definition of molality.

Step-by-step solution

Step 1: Understand the given data

• Molarity ($M$) = $3\ \text{mol L}^{-1}$ → Means 3 moles of NaCl in 1 litre of solution
• Density ($\rho$) = $1.25\ \text{g mL}^{-1}$ → $1\ \text{mL}$ of solution weighs $1.25\ \text{g}$

Step 2: Find the mass of 1 litre of solution

Since $1\ \text{L} = 1000\ \text{mL}$:

Step 3: Find the mass of solute (NaCl)

Molar mass of NaCl = $23 + 35.5 = 58.5\ \text{g mol}^{-1}$

Mass of 3 moles of NaCl:

Step 4: Find the mass of solvent (water)

Convert to kg: …

Here are the most common mistakes students make when solving this exact type of problem, along with the concept-first reasoning to avoid each.

1. Confusing Molarity with Molality

The Mistake:

Students often treat the given $3\ M$ as if it were molality ($m$), or they try to directly convert without using mass of solvent.

Why it happens:

Both start with "m" and both involve moles of solute. But molarity uses volume of solution, while molality uses mass of solvent.

How to avoid:

• Molarity ($M$) = moles of solute per litre of solution
• Molality ($m$) = moles of solute per kg of solvent
• Always ask: “Do I have the mass of the solvent or the volume of the solution?”
• Here, you have volume of solution (from $3\ M$), so you must find the mass of solvent using density.

2. Forgetting to Subtract Mass of Solute from Mass of Solution

The Mistake:

Students calculate the mass of 1 L of solution using density, then directly use that as the mass of solvent.

Example of error:

Mass of solution = $1250\ g$ → they take this as mass of solvent.

Why it happens:

They forget that the solution contains both solute and solvent.

How to avoid:

• Mass of solution = mass of solute + mass of solvent
• Always compute: $\text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute}$
• In this problem:
  • Mass of 1 L solution = $1.25 \times 1000 = 1250\ g$
  • Mass of NaCl (3 moles) = $3 \times 58.5 = 175.5\ g$
  • Mass of solvent = $1250 - 175.5 = 1074.5\ g = 1.0745\ kg$

3. Using Wrong Molar Mass of NaCl

The Mistake:

Using $Na = 23$, $Cl = 35.5$, but then writing $23 + 35.5 = 58$ (rounding incorrectly) or using $58.5$ but then making arithmetic errors.

Why it happens:

Rushing or not double-checking the periodic table values.

How to avoid:

• Memorise common molar masses:
  • $Na = 23\ g/mol$
  • $Cl = 35.5\ g/mol$
  • $\therefore NaCl = 58.5\ g/mol$
• Write it clearly: $M_{NaCl} = 23 + 35.5 = 58.5\ g\ mol^{-1}$
• Double-check multiplication: $3 \times 58.5 = 175.5$ (not $175$ or $176$)

4. Unit Conversion Errors (g to kg)

The Mistake:

Using mass of solvent in grams directly in the molality formula.

Why it happens:

Molality requires kg of solvent, but density gives mass in grams.

How to avoid:

• Molality formula:

• After finding solvent mass in grams, divide by 1000
  • $1074.5\ g = 1.0745\ kg$
• If you forget, your answer will be 1000 times too large.

5. Assuming 1 L of Solution Contains 1 kg of Solvent

The Mistake:

Thinking that because density is $1.25\ g/mL$, 1 L of solution weighs $1250\ g$, and assuming that is the solvent mass.

Why it happens:

Confusing "solution mass" with "solvent mass" — a very common conceptual leap.

How to avoid:

• Draw a mental picture:
  • 1 L of solution = solute particles + water molecules …