Q.Calculate the molarity of each of the following solutions:

Concept: Molarity Calculation — Molarity is moles of solute per litre of solution.

(a) Molar mass of $Co(NO_3)_2 \cdot 6H_2O$:

$59 + 2(14+48) + 6(18) = 59 + 124 + 108 = 291\ \text{g/mol}$

Moles = $\frac{30}{291} \approx 0.1031\ \text{mol}$

Molarity = $\frac{0.1031}{4.3} \approx 0.0240\ \text{M}$

(b) Using dilution formula $M_1V_1 = M_2V_2$: …

The Core Idea

Molarity ($M$) is defined as the number of moles of solute dissolved in one litre of solution. The formula is:

The trick in part (a) is that the solute is a hydrated salt — $Co(NO_3)_2 \cdot 6H_2O$. The water of crystallisation is part of the compound’s formula mass, so we must include it when calculating the molar mass. Many students forget this and use the mass of the anhydrous salt, which gives a wrong answer.

Part (b) is a straightforward dilution: when you add solvent, the number of moles of solute stays the same, so $M_1V_1 = M_2V_2$.

Step-by-Step Solution

Part (a): 30 g of $Co(NO_3)_2 \cdot 6H_2O$ in 4.3 L of solution

1. Find the molar mass of the hydrated salt.

We need the atomic masses (rounded to one decimal place, as is standard for such problems):

• Co = 58.9 g/mol
• N = 14.0 g/mol
• O = 16.0 g/mol
• H = 1.0 g/mol

First, the anhydrous part $Co(NO_3)_2$:

• 1 Co: $1 \times 58.9 = 58.9$
• 2 N: $2 \times 14.0 = 28.0$
• 6 O: $6 \times 16.0 = 96.0$
• Total for $Co(NO_3)_2$ = $58.9 + 28.0 + 96.0 = 182.9$ g/mol

Now the water of crystallisation: $6H_2O$:

• 12 H: $12 \times 1.0 = 12.0$
• 6 O: $6 \times 16.0 = 96.0$
• Total for $6H_2O$ = $12.0 + 96.0 = 108.0$ g/mol

So the molar mass of $Co(NO_3)_2 \cdot 6H_2O$ is:

2. Calculate the number of moles of solute.

Let’s compute:

3. Apply the molarity formula.

Volume of solution = 4.3 L.

$M = \frac{0.1031 \text{ mol}}{4.3 \text{ L}} \approx 0.0240 \text{ M}$ …

Method: Molarity Calculation via Moles and Volume

Concept first: Molarity ($M$) is the number of moles of solute per litre of solution (not solvent).

Formula:

(a) 30 g of $\text{Co(NO}_3\text{)}_2 \cdot 6\text{H}_2\text{O}$ in 4.3 L of solution

Step 1 — Find molar mass of the hydrated compound

• Co: 58.93 g/mol
• N: 2 × 14.01 = 28.02 g/mol
• O (from nitrate): 6 × 16.00 = 96.00 g/mol
• Water molecules: 6 × (2 × 1.008 + 16.00) = 6 × 18.016 = 108.096 g/mol

Total molar mass = 58.93 + 28.02 + 96.00 + 108.096 = 291.046 g/mol

(Exam tip: often rounded to 291 g/mol)

Step 2 — Calculate moles of solute

Step 3 — Apply molarity formula

Volume = 4.3 L

Answer (a): $\boxed{0.024 \text{ M}}$ (rounded to 2 significant figures)

(b) 30 mL of 0.5 M $\text{H}_2\text{SO}_4$ diluted to 500 mL

Step 1 — Identify the method: Dilution Formula

When a solution is diluted, moles of solute remain constant.

Where:

• $M_1$ = initial molarity = 0.5 M
• $V_1$ = initial volume = 30 mL
• $V_2$ = final volume = 500 mL
• $M_2$ = final molarity (what we need)

Step 2 — Plug into formula

$0.5 \times 30 = M_2 \times 500$ …

Common Mistakes in Molality & Molarity Calculations (with Fixes)

Students often confuse molality (moles of solute per kg of solvent) with molarity (moles of solute per litre of solution). The question here asks for molarity, but the same pitfalls apply to molality problems. Below are the most frequent errors and how to avoid them.

Mistake 1: Using the wrong mass for molar mass (hydrated vs anhydrous)

The error:

For part (a), students use the molar mass of anhydrous $Co(NO_3)_2$ instead of the hydrated form $Co(NO_3)_2 \cdot 6H_2O$.

Why it happens:

They forget that the water of crystallisation ($6H_2O$) is part of the compound’s formula mass.

How to avoid:

• Always check the formula given — if it includes $\cdot nH_2O$, include that water mass in the molar mass.
• Calculate molar mass step by step:

For $Co(NO_3)_2 \cdot 6H_2O$:

• Co: $58.93 \, \text{g/mol}$
• N: $2 \times 14.01 = 28.02$
• O (from nitrate): $6 \times 16.00 = 96.00$
• $6H_2O$: $6 \times (2 \times 1.008 + 16.00) = 6 \times 18.016 = 108.096$

Total: $58.93 + 28.02 + 96.00 + 108.096 = 291.046 \, \text{g/mol}$

Mistake 2: Confusing mass of solution with mass of solvent

The error:

In molality problems, students use the total mass of solution instead of the mass of solvent only.

Why it happens:

They misread “kg of solvent” as “kg of solution”.

How to avoid:

• Molarity uses volume of solution (litres).
• Molality uses mass of solvent (kg).
• For part (a), the volume is given directly — no solvent mass needed. But if it were a molality problem, you would subtract the solute mass from the solution mass to get solvent mass.

Mistake 3: Forgetting to convert volume units

The error:

In part (b), students use 30 mL directly without converting to litres, or they forget that dilution changes the total volume.

Why it happens:

They treat mL as if it were L, or they use the initial volume instead of the final diluted volume.

How to avoid:

• Always convert mL to L: $1 \, \text{mL} = 10^{-3} \, \text{L}$
• For dilution, use the dilution formula:

Where $M_1 = 0.5 \, \text{M}$, $V_1 = 30 \, \text{mL} = 0.030 \, \text{L}$, $V_2 = 500 \, \text{mL} = 0.500 \, \text{L}$.

So:

Mistake 4: Using volume of solvent instead of volume of solution

The error:

Students assume the volume of solution equals the volume of solvent (e.g., adding 30 g of solid to water and taking the water volume as the solution volume).

Why it happens:

They forget that the solute occupies space, so the final volume is not the same as the initial solvent volume.

How to avoid:

• In molarity, the volume is always the total volume of the final solution. …