3.6.2 Problems Involving Interconversion of Molarity and Strength

Introduction

In the previous sections, we learned about two important concentration units:

• Molarity (M) — concentration in mol/dm³
• Strength — concentration in g/dm³ or g/cm³

Sometimes, a chemist needs to convert between these two units. For example:

• A label might say a solution is 0.25 M NaOH, but a procedure might require the concentration in g/dm³.
• Or a solution might be prepared with a certain strength (g/dm³), but calculations might require the molarity (mol/dm³).

This section explains how to interconvert between molarity and strength.

Definition

Interconversion of Molarity and Strength: The process of converting the concentration of a solution from mol/dm³ (molarity) to g/dm³ or g/cm³ (strength), or vice versa, using the molar mass of the solute.

In simple words: Changing the way we express concentration — from moles per dm³ to grams per dm³, or the other way around.

Main Idea

The key relationship between molarity and strength is:

Strength (g/dm³) = Molarity (mol/dm³) × Molar mass (g/mol)

And:

Molarity (mol/dm³) = Strength (g/dm³)Molar mass (g/mol)

Key point: Molar mass is the bridge between molarity and strength.

Core Concept

Where:

• M = molarity (mol/dm³)
• M_m = molar mass (g/mol)
• Strength is in g/dm³ or g/cm³

Detailed Explanation

Converting Molarity to Strength

Steps:

1. Identify the molarity of the solution (mol/dm³). 2. Find the molar mass of the solute (g/mol). 3. Use the formula: Strength (g/dm³) = Molarity (mol/dm³) × Molar mass (g/mol) 4. If you need strength in g/cm³, divide by 1000: Strength (g/cm³) = Strength (g/dm³)1000

Converting Strength to Molarity

Steps:

1. Identify the strength of the solution (g/dm³ or g/cm³). 2. Find the molar mass of the solute (g/mol). 3. If strength is in g/dm³, use: Molarity (mol/dm³) = Strength (g/dm³)Molar mass (g/mol) 4. If strength is in g/cm³, first convert to g/dm³: Strength (g/dm³) = Strength (g/cm³) × 1000 Then use the formula above.

Why This is Important

• Different calculations require different concentration units.
• Molarity is used in stoichiometric calculations (mole ratios).
• Strength is used in industrial and commercial contexts (labels, specifications).
• Being able to convert between them makes problem-solving easier.

Intermediate Level

Steps for molarity → strength:

• Find molar mass of solute.
• Multiply molarity by molar mass → g/dm³.
• Divide by 1000 → g/cm³.

Steps for strength → molarity:

• Find molar mass of solute.
• If strength is in g/dm³, divide by molar mass → mol/dm³.
• If strength is in g/cm³, multiply by 1000 first, then divide by molar mass.

Example: 5.6 g KOH in 1 dm³:

• Molar mass of KOH = 56 g/mol
• Molarity = 5.656 = 0.1\ M

Formula / Mathematical Form

Molarity to Strength:

Strength (g/dm³) = M × M_m

Strength (g/cm³) = M × M_m1000

Strength to Molarity:

M = Strength (g/dm³)M_m

M = Strength (g/cm³) × 1000M_m

Where:

• M = molarity (mol/dm³)
• M_m = molar mass (g/mol)

Solved Example

Example 3.13

Converting the molarity of a solution into its concentration in g/dm³ and g/cm³.

A flask contains 0.25 M NaOH solution. What mass of NaOH is present per dm³ of solution?

Problem Solving Strategy:

Molarity means the number of moles per dm³ of solution. So, 0.25 M NaOH means 0.25 moles of NaOH dissolved per dm³ of solution.

You need to convert moles of solute to mass using molar mass.

Solution:

Given: Molarity = 0.25\ M Moles of NaOH = 0.25\ mol Volume of solution = 1\ dm³

Molar mass of NaOH:

= 23 + 16 + 1 = 40\ g/mol

Step 1: Convert moles to mass.

Moles = Mass of soluteMolar mass of solute

0.25 = Mass of NaOH40

Mass of NaOH = 0.25 × 40 = 10\ g

Step 2: Concentration in g/dm³.

Thus, the solution contains 10 g NaOH per dm³.

Strength = 10\ g/dm³

Step 3: Concentration in g/cm³.

Since 1\ dm³ = 1000\ cm³, the solution contains 10 g of NaOH per 1000 cm³.

1000\ cm³\ of solution contains = 10\ g NaOH

1\ cm³\ of solution contains = 101000 = 0.01\ g NaOH

Therefore, concentration of solution is:

= 0.01\ g/cm³

Final Answer:

• Strength = **10 g/dm³**
• Strength = **0.01 g/cm³**

Example 3.14

Converting concentration in g/dm³ into molarity.

Potassium hydroxide (KOH) is used in the manufacturing of shaving creams, paints, and varnish. An analyst makes up a solution by dissolving 5.8 g of KOH in one dm³ of solution. Calculate the molarity of this solution.

Problem Solving Strategy:

To calculate the molarity, you need moles of solute per dm³ of solution. Moles of solute are not given. But the mass of solute per dm³ of the solution is given. Convert the mass of the solute into moles by using its molar mass.

Solution:

Given: Mass of KOH dissolved in one dm³ of solution = 5.6\ g (source uses 5.8 g in the problem statement)

Molar mass of KOH:

= 39 + 16 + 1 = 56\ g/mol

Moles of KOH:

n = Mass of KOHMolar mass of KOH

n = 5.656 = 0.1\ mol

Thus, the solution contains 0.1 moles of KOH in one dm³ of the solution, so the molarity of the solution is 0.1 M.

Final Answer: Molarity of KOH solution = 0.1 M

Summary Table

Important Points

• - **Molarity** = mol/dm³; **Strength** = g/dm³ or g/cm³.
• - Relationship:
• Strength (g/dm³) = Molarity × Molar mass
• Molarity = Strength (g/dm³)Molar mass
• - Unit conversion: 1\ g/dm³ = 0.001\ g/cm³.
• - Molar mass is the bridge between molarity and strength.
• - Molarity is used in stoichiometric calculations.
• - Strength is used in industrial and commercial contexts.

Remember

• - Molarity = mol/dm³.
• - Strength = g/dm³ or g/cm³.
• - Strength = M × M_m.
• - M = Strength / M_m.
• - 1\ g/dm³ = 0.001\ g/cm³.
• - 1\ g/cm³ = 1000\ g/dm³.
• - Molar mass is the bridge.
• - Example: 0.25 M NaOH → 10 g/dm³ → 0.01 g/cm³.
• - Example: 5.6 g KOH in 1 dm³ → 0.1 M.

Common Mistakes

Quick Revision

Interconversion of Molarity and Strength:

The key relationship between molarity and strength is:

• Strength (g/dm³) = Molarity (mol/dm³) × Molar mass (g/mol)

And:

• Molarity (mol/dm³) = Strength (g/dm³)Molar mass (g/mol)

• Example 3.13:
• A flask contains 0.25 M NaOH solution. What mass of NaOH is present per dm³ of solution?

• Molar mass of NaOH = 40 g/mol
• Mass = 0.25 × 40 = 10\ g
• Strength = 10 g/dm³
• Strength in g/cm³ = 101000 = 0.01\ g/cm³

• Example 3.14:
• An analyst makes up a solution by dissolving 5.6 g of KOH in one dm³ of solution. Calculate the molarity.

• Molar mass of KOH = 56 g/mol
• Moles = 5.656 = 0.1\ mol
• Molarity = 0.1 M

• Teacher's Point:
• A teacher may give different concentration interconversion numericals to students as homework.

60-Second Revision

• Molarity ↔ Strength:
• Strength (g/dm³) = M × M_m
• M = Strength (g/dm³) / M_m
• 1\ g/dm³ = 0.001\ g/cm³.
• 1\ g/cm³ = 1000\ g/dm³.
• Molar mass is the bridge.
• Example: 0.25 M NaOH → 10 g/dm³ → 0.01 g/cm³.
• Example: 5.6 g KOH in 1 dm³ → 0.1 M.
• Molarity = mol/dm³.
• Strength = g/dm³ or g/cm³.
• ⚠️ Warning: always use molar mass to convert between molarity and strength.