Calculate the molarity of each of the following solutions: (a) 30 g of $Co(NO_3)_2 \cdot 6H_2O$ in 4.3 L of solution (b) 30 mL of 0.5 M $H_2SO_4$ diluted to 500 mL.

Question

Accepted Answer

> [!TLDR]
> Molarity is moles of solute per litre of solution. For (a), we convert the mass of the hydrated salt to moles and divide by the volume in litres. For (b), we use the dilution formula $M_1V_1 = M_2V_2$. The answers are **(a) 0.024 M** and **(b) 0.03 M**.

## The Core Idea

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

$$M = \frac{	ext{moles of solute}}{	ext{volume of solution in litres}}$$

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 $M_1V_1 = M_2V_2$.

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## 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: $1 	imes 58.9 = 58.9$
- 2 N: $2 	imes 14.0 = 28.0$
- 6 O: $6 	imes 16.0 = 96.0$
- 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: $12 	imes 1.0 = 12.0$
- 6 O: $6 	imes 16.0 = 96.0$
- 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:
$$182.9 + 108.0 = 290.9 	ext{ g/mol}$$

> [!WARNING]
> A common mistake is to use the molar mass of anhydrous $Co(NO_3)_2$ (182.9 g/mol) instead of the hydrated form. This would give a larger number of moles and a higher molarity — which is incorrect because the water molecules are part of the solute’s formula mass.

**2. Calculate the number of moles of solute.**

$$

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

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? …

The Core Idea

Step-by-Step Solution

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

Method: Molarity Calculation via Moles and Volume

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

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

Common Mistakes in Molality & Molarity Calculations (with Fixes)

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

Mistake 2: Confusing mass of solution with mass of solvent

Mistake 3: Forgetting to convert volume units

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

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

The Core Idea

Step-by-Step Solution

Part (a): 30 g of Co(NO3)2⋅6H2OCo(NO_3)_2 \cdot 6H_2OCo(NO3​)2​⋅6H2​O in 4.3 L of solution

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

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