If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained? (i) 1.5

NCERT Exemplar · Q4

Q. If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?

(i) 1.5 M

(ii) 1.66 M

(iii) 0.017 M

(iv) 1.59 M

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Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

Imagine you're making a cup of sweet tea. You add sugar to hot water, stir, and taste. If you let the tea cool to room temperature, the amount of sugar hasn't changed — but the volume of the liquid has shrunk slightly. If you measured concentration as "grams of sugar per litre of solution," that number would change just because the temperature changed. That's annoying if you're a chemist who needs a reliable, temperature-independent way to describe how much solute is present.

Molality was invented to solve exactly this problem.

The Intuition

Instead of measuring the volume of the solution (which expands and contracts with temperature), molality measures the mass of the solvent. Mass doesn't change with temperature. So molality gives you a concentration that stays the same whether your solution is hot or cold.

Think of it this way:

Molarity = moles of solute per litre of solution (temperature-sensitive)
Molality = moles of solute per kilogram of solvent (temperature-independent)

The solvent is the substance doing the dissolving — usually water. The solute is what gets dissolved — sugar, salt, etc.

The Precise Definition

\text{Molality } (m) = \frac{\text{moles of solute}}{\text{kilograms of solvent}}

The symbol for molality is a lowercase m (not to be confused with M for molarity).

Key points to remember:

The denominator is solvent mass, not solution mass
The unit is mol/kg (often written as simply "m")
It is independent of temperature because mass doesn't change with temperature

Worked Example

Problem: 36 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180 g/mol) is dissolved in 500 g of water. Calculate the molality of the solution.

Step 1: Find moles of solute

\text{Moles of glucose} = \frac{36 \text{ g}}{180 \text{ g/mol}} = 0.2 \text{ mol}

Step 2: Convert solvent mass to kilograms

500 \text{ g} = 0.5 \text{ kg}

Step 3: Apply the formula

m = \frac{0.2 \text{ mol}}{0.5 \text{ kg}} = 0.4 \text{ m}

Note

The answer is 0.4 m (or 0.4 mol/kg). Notice we used the mass of water (500 g), not the mass of the solution (which would be 536 g).

Common Mistake to Avoid

Watch out

Do not use the mass of the solution in the denominator. The formula specifically asks for the mass of the solvent alone. If the problem gives you the total mass of the solution, subtract the mass of the solute to find the solvent mass.

When Do You Use Molality?

Molality is the star in two important situations: …

Why this formula?

Molality Calculation: Why the Formula Works

Molality is a measure of concentration that is temperature-independent — this is its key advantage over molarity. Let's understand why the formula takes the form it does.

The Definition First

Molality ( $m$ ) is defined as:

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

The unit is mol/kg, often written as m (e.g., 0.5 m glucose solution).

Why Mass of Solvent, Not Solution?

This is the critical conceptual point.

The Reasoning

Molarity uses volume of solution → volume changes with temperature (expansion/contraction). So molarity changes with temperature.
Molality uses mass of solvent → mass is invariant with temperature. So molality remains constant regardless of temperature changes.

Key insight: By using the solvent's mass (not the solution's volume), we eliminate temperature dependence. This is why molality is preferred for colligative properties (boiling point elevation, freezing point depression) — these properties depend on the number of solute particles, not on temperature.

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

If you have $w_{\text{solute}}$ grams of solute with molar mass $M_{\text{solute}}$ (g/mol):

\text{moles of solute} = \frac{w_{\text{solute}}}{M_{\text{solute}}}

Step 2: Mass of Solvent in kg

If the solvent mass is $W_{\text{solvent}}$ grams:

\text{mass of solvent in kg} = \frac{W_{\text{solvent}}}{1000}

Step 3: Putting It Together

m = \frac{\frac{w_{\text{solute}}}{M_{\text{solute}}}}{\frac{W_{\text{solvent}}}{1000}}

Simplifying:

m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}

The Final Formula (Exam-Ready)

\boxed{m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}}

Where:

$w_{\text{solute}}$ = mass of solute in grams
$M_{\text{solute}}$ = molar mass of solute in g/mol
$W_{\text{solvent}}$ = mass of solvent in grams

Why the ×1000 Factor? …

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When a solution is diluted, the total moles of solute remain constant. Using the dilution formula $M_1V_1 = M_2V_2$ , the final molarity of the solution is calculated to be $\boxed{1.67 \text{ M}}$ .

When we dilute a solution, we are essentially adding more solvent (usually water) to it. The key idea here is that the amount of the substance dissolved (the solute) does not change; only the total volume of the solution increases, which in turn decreases the concentration of the solute.

Molarity ( $M$ ) is defined as the number of moles of solute per liter of solution.

M = \frac{\text{moles of solute}}{\text{volume of solution (in Liters)}}

From this definition, we can express the moles of solute as:

\text{moles of solute} = M \times \text{Volume (in Liters)}

During dilution, the moles of solute before dilution must be equal to the moles of solute after dilution. This is the fundamental principle behind dilution calculations.

Let $M_1$ and $V_1$ be the initial molarity and volume, and $M_2$ and $V_2$ be the final molarity and volume.

Since the moles of solute remain constant:

\text{moles before dilution} = \text{moles after dilution}

M_1V_1 = M_2V_2

This is the dilution formula we will use.

M_1V_1 = M_2V_2

Let's apply this to the given problem:

Identify Initial Conditions:

We are given an initial solution with:
- Initial volume ( $V_1$ ) = $500 \text{ mL}$
- Initial molarity ( $M_1$ ) = $5 \text{ M}$
Identify Final Conditions:

The solution is diluted to a new total volume:
- Final volume ( $V_2$ ) = $1500 \text{ mL}$
- We need to find the final molarity ( $M_2$ ).
Apply the Dilution Formula:

Using the formula $M_1V_1 = M_2V_2$ , we can rearrange it to solve for $M_2$ :

M_2 = \frac{M_1V_1}{V_2}

Substitute Values and Calculate: …

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Common Mistakes in Molality & Dilution Problems

Students often confuse molality with molarity — this question is actually about molarity (M), not molality (m). Let's clarify first:

Molarity (M) = moles of solute per litre of solution
Molality (m) = moles of solute per kg of solvent

This is a dilution problem using the formula:

M_1 V_1 = M_2 V_2

Where:

$M_1 = 5 \, \text{M}$ , $V_1 = 500 \, \text{mL}$
$V_2 = 1500 \, \text{mL}$
$M_2 = ?$

✗ Mistake 1: Using the wrong formula (confusing with molality)

What students do:

They try to use molality formulas or incorrectly apply $M_1V_1 = M_2V_2$ with mass/volume units mixed up.

Why it happens:

The terms "molarity" and "molality" sound similar, and students memorise formulas without understanding the concept.

✓ How to avoid:

Read the question carefully — look for the word "molarity" (M) vs "molality" (m).
Remember: Molarity uses volume (L or mL), molality uses mass of solvent (kg).
If the question gives volumes (mL, L), it's almost always molarity.

✗ Mistake 2: Forgetting to convert mL to L

What students do:

They plug in $V_1 = 500$ and $V_2 = 1500$ directly without converting to litres.

Why it happens:

They think the formula works with any volume unit — but it only works if both volumes are in the same unit.

✓ How to avoid:

Always check units before substituting.
Since both volumes are in mL, you can keep them in mL — the ratio is what matters.
But if one volume is in L and the other in mL, convert first.

✗ Mistake 3: Dividing the wrong way

What students do:

They compute $M_2 = \frac{V_2}{V_1} \times M_1$ instead of $M_2 = \frac{M_1 V_1}{V_2}$ .

Why it happens:

They misremember the formula or think "dilution means multiply by the ratio".

✓ How to avoid:

Understand the logic: When you add water, moles of solute stay the same, but volume increases → concentration decreases.
So $M_2$ must be smaller than $M_1$ .
If your answer is larger than 5 M, you've divided wrong.

✗ Mistake 4: Incorrect arithmetic

What students do:

They compute $M_2 = \frac{5 \times 500}{1500} = \frac{2500}{1500} = 1.666...$ and then round incorrectly or misplace the decimal. …

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Q.If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?

Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

The Intuition

The Precise Definition

Worked Example

Common Mistake to Avoid

When Do You Use Molality?

Why this formula?

Molality Calculation: Why the Formula Works

The Definition First

Why Mass of Solvent, Not Solution?

The Reasoning

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

Step 2: Mass of Solvent in kg

Step 3: Putting It Together

The Final Formula (Exam-Ready)

Why the ×1000 Factor? …

Method: Dilution Formula (M₁V₁ = M₂V₂)

Steps:

Common Mistakes in Molality & Dilution Problems

✗ Mistake 1: Using the wrong formula (confusing with molality)

✗ Mistake 2: Forgetting to convert mL to L

✗ Mistake 3: Dividing the wrong way

✗ Mistake 4: Incorrect arithmetic

Q.If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?

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