Molality Definition, Formula, MCQs, Solved Examples, Practice Problems, Temperature Independent

ANAND CLASSES Study Material and Notes to explore the concept of Molality with clear definitions, formula derivation, detailed solved examples, MCQs, and practice problems designed for JEE, NEET, and CBSE Class 11 Chemistry.

Q1: What is Molality (m) and why is it important?

A: Molality (m) is a measure of concentration used in chemistry, specifically in solutions. It is defined as the number of moles of solute per kilogram of solvent.

The formula for molality is: $$m = \frac{\text{No. of moles of solute}}{\text{Mass of solvent in kg}}$$

Or, mathematically: $$m = \frac{n_{\text{solute}}}{W_{\text{solvent}}} \quad \text{mol/kg}$$

Units of molality : mol/kg

Molality is crucial because, unlike molarity, it is independent of temperature and is used in colligative property calculations, such as boiling point elevation and freezing point depression.

Q2: How can we express molality in terms of solute mass and molar mass?

A: The number of moles of the solute $(n_{\text{solute}})$ can be found using its mass (w) and molar mass (M) as follows: $$n_{\text{solute}} = \frac{\text{Mass of solute} (w)}{\text{Molar mass of solute} (M)}$$

Substituting this into the molality formula:

$$m = \frac{\left( \frac{w}{M} \right)}{W} = \frac{w}{M \times W} \quad \text{mol/kg}$$

Units of molality : mol/kg

Q3: How does unit conversion affect molality calculation?

A: The calculation of molality depends on how the units of mass of solute and solvent are expressed. Let’s explore two common cases:

Case I: When the solvent mass is given in kilograms

• w = Mass of the solute in grams
• M = Molar mass of the solute in g/mol
• W = Mass of the solvent in kg

Then, the molality formula simplifies to: $$m = \frac{w}{M \times W} \quad \text{mol/kg}$$

Case II: When the solvent mass is given in grams

• w = Mass of the solute in grams
• M = Molar mass of the solute in g/mol
• W = Mass of the solvent in grams

Since 1 kg = 1000 g, we convert W into kilograms: $$m = \frac{w}{M \times (W / 1000)} mol/kg$$

$$m = \frac{1000 w}{M \times W} \quad \text{mol/kg}$$

Q4: Why does molality remain unchanged with temperature?

A: Molality is based on mass, and mass does not change with temperature. Unlike molarity, which depends on volume and can fluctuate with temperature changes (due to thermal expansion or contraction of liquids), molality remains constant.

This makes molality more reliable in experiments involving temperature changes, such as boiling point elevation and freezing point depression.

Numerical Examples of Molality

Example 1: Calculating Molality for a Sugar Solution

Q: A solution is prepared by dissolving 10 g of glucose ($C_6H_{12}O_6$, molar mass = 180 g/mol) in 200 g of water. Find the molality of the solution.

Solution:

We use the formula for molality: $$m = \frac{w}{M \times W}$$

Step 1: Calculate the moles of glucose: $$n_{\text{solute}} = \frac{w}{M} = \frac{10}{180} = 0.0556 \text{ moles}$$

Step 2: Convert the solvent mass into kg: $$W = 200 \text{ g} = 0.200 \text{ kg}$$

Step 3: Calculate molality: $$m = \frac{0.0556}{0.200} = 0.278 \text{ mol/kg}$$

📌 Answer: The molality of the solution is 0.278 mol/kg.

Example 2: Finding Mass of Solute from Given Molality

Q: What mass of NaCl (sodium chloride, molar mass = 58.5 g/mol) should be dissolved in 500 g of water to make a 2 molal solution?

Solution:

We use the formula: $$m = \frac{w}{M \times W}$$

Step 1: Rearranging the formula to solve for w: $$w = m \times M \times W$$

Step 2: Substituting the values: $$w = (2) \times (58.5) \times \left(\frac{500}{1000}\right)$$ $$w = 2 \times 58.5 \times 0.5 = 58.5 \text{ g}$$

📌 Answer: 58.5 g of NaCl should be dissolved to make a 2 molal solution.

Practical Applications of Molality

Pharmaceutical Industry (Drug Formulations & Concentration Control)

• Many drugs and medicines are formulated using molality-based concentrations to ensure consistent and precise dosages.
• For example, IV fluids and electrolyte solutions are designed to have a specific molality to maintain osmotic balance in the human body.

Chemical & Industrial Applications (Electrolyte Solutions & Chemical Reactions)

• In industries, molality is preferred over molarity because it is independent of temperature changes.
• Electrolytes in batteries are often described in terms of molality because temperature variations can change the volume of the solution but not its mass.
• Example: The concentration of sulfuric acid in lead-acid batteries is measured in molality for better performance and stability.

Oceanography & Environmental Science (Salinity Measurement)

• The salinity of seawater is expressed in terms of molality to determine the concentration of dissolved salts.
• Changes in ocean water salinity affect marine life, weather patterns, and ocean currents.

📌 Example: If ocean salinity increases, the freezing point of seawater drops, affecting the formation of polar ice caps.

Food Science (Preservation & Flavor Enhancements)

• The molality of sugar or salt solutions is used in food preservation techniques like making jams, pickles, and brine solutions.
• In soft drinks and fruit juices, the sugar concentration is controlled using molality-based calculations to maintain sweetness and consistency.

Problem 1: Finding Molality from Given Mass of Solute and Solvent

Q: A solution is prepared by dissolving 15 g of urea ($CO(NH_2)_2$, molar mass = 60 g/mol) in 250 g of water. Calculate the molality of the solution.

Solution:

We use the molality formula: $$m = \frac{w}{M \times W}$$

Step 1: Calculate the number of moles of urea: $$n_{text{solute}} = frac{w}{M} = frac{15}{60} = 0.25 text{ moles}$$

Step 2: Convert the solvent mass into kg: $$W = 250 \text{ g} = 0.250 \text{ kg}$$

Step 3: Calculate molality: $$m = \frac{0.25}{0.250} = 1.0 \text{ mol/kg}$$

📌 Answer: The molality of the solution is 1.0 mol/kg.

Problem 2: Finding Mass of Solute from Given Molality

Q: What mass of glucose ($C_6H_{12}O_6$, molar mass = 180 g/mol) should be dissolved in 500 g of water to make a 0.5 molal solution?

Solution:

We use the molality formula and solve for w: $$m = \frac{w}{M \times W}$$

Rearranging: $$w = m \times M \times W$$

Step 1: Substituting the values: $$w = (0.5) \times (180) \times \left(\frac{500}{1000}\right)$$ $$w = 0.5 \times 180 \times 0.5 = 45 \text{ g}$$

📌 Answer: 45 g of glucose should be dissolved to prepare the required solution.

Problem.3: Finding the Mass of Solvent Required

Q: How much water (in grams) is required to dissolve 20 g of KCl (potassium chloride, molar mass = 74.5 g/mol) to make a 0.2 molal solution?

Solution:

We use the molality formula: $$m = \frac{w}{M \times W}$$

Rearranging for W: $$W = \frac{w}{M \times m}$$

Step 1: Substituting values: $$W = \frac{20}{(74.5) \times (0.2)} $$

$$W = \frac{20}{14.9} = 1.34 \text{ kg} = 1340 \text{ g}$$

📌 Answer: 1340 g (1.34 kg) of water is required.

Multiple Choice Questions (MCQs) on Molality

Basic Concept of Molality

1. Molality (m) is defined as:
A) Moles of solute per liter of solution
B) Moles of solute per kg of solvent
C) Moles of solute per mole of solvent
D) Grams of solute per liter of solvent

✅ Answer: B) Moles of solute per kg of solvent
📌 Explanation: Molality is the number of moles of solute present in 1 kg (1000 g) of solvent. It is expressed in mol/kg.

Conversion and Calculation Based Questions

2. The molality of a solution prepared by dissolving 0.5 moles of NaCl in 1 kg of water is:
A) 0.5 mol/kg
B) 1.0 mol/kg
C) 2.0 mol/kg
D) 5.0 mol/kg

✅ Answer: A) 0.5 mol/kg
📌 Explanation: Using the formula, $$m = \frac{n}{W} = \frac{0.5}{1} = 0.5 \text{ mol/kg}$$

3. How much urea ($CO(NH_2)_2$, molar mass = 60 g/mol) should be dissolved in 500 g of water to prepare a 2 molal solution?
A) 30 g
B) 60 g
C) 120 g
D) 240 g

✅ Answer: C) 120 g
📌 Explanation: Using the formula, $$m = \frac{w}{M \times W}$$

Rearranging for w, $$w = m \times M \times W = (2) \times (60) \times \left(\frac{500}{1000}\right) = 120 \text{ g}$$

Conceptual Questions on Molality and its Applications

4. Which of the following is true regarding molality?
A) It depends on the volume of the solution
B) It depends on temperature
C) It depends only on the mass of the solvent
D) It depends on pressure

✅ Answer: C) It depends only on the mass of the solvent
📌 Explanation: Unlike molarity, which depends on volume and can change with temperature, molality depends only on the mass of the solvent, which remains constant.

5. If the solvent evaporates from a solution, the molality will:
A) Increase
B) Decrease
C) Remain constant
D) First increase, then decrease

✅ Answer: A) Increase
📌 Explanation: Since molality is defined as moles of solute per kg of solvent, as the solvent evaporates, its mass decreases, leading to an increase in molality.

Molality in Colligative Properties

6. The freezing point of a solution decreases when:
A) Solvent is added
B) Solute is added
C) The solution is heated
D) Pressure is increased

✅ Answer: B) Solute is added
📌 Explanation: Adding a solute lowers the freezing point of a solution due to the freezing point depression ($\Delta T_f = i K_f \:m$).

7. The boiling point of a solution is higher than that of the pure solvent because:
A) The molality of the solution is less than 1
B) The molality of the solution is greater than 1
C) Solute particles increase the vapor pressure
D) Solute particles lower the vapor pressure

✅ Answer: D) Solute particles lower the vapor pressure
📌 Explanation: Solute particles interfere with solvent molecules, reducing vapor pressure. Since boiling occurs when vapor pressure = atmospheric pressure, a lower vapor pressure requires a higher temperature to boil.

Numerical-Based Questions

8. What is the molality of a solution containing 90 g of glucose ($C_6H_{12}O_6$, molar mass = 180 g/mol) in 500 g of water?
A) 0.5 mol/kg
B) 1.0 mol/kg
C) 2.0 mol/kg
D) 5.0 mol/kg

✅ Answer: A) 0.5 mol/kg
📌 Explanation:
Moles of glucose: $$n = \frac{90}{180} = 0.5 \text{ moles}$$

Mass of solvent in kg: $$W = 500 \text{ g} = 0.5 \text{ kg}$$

Molality: $$m = \frac{0.5}{0.5} = 0.5 \text{ mol/kg}$$

Comparison with Molarity

9. Which of the following is an advantage of using molality instead of molarity?
A) Molality changes with temperature, while molarity does not
B) Molality is independent of temperature and pressure
C) Molality depends on the volume of the solution
D) Molality is more difficult to measure than molarity

✅ Answer: B) Molality is independent of temperature and pressure
📌 Explanation: Molality is based on mass of solvent, which does not change with temperature or pressure. Molarity, however, is based on volume, which can change with temperature.

Real-Life Applications of Molality

10. Why is molality preferred over molarity in industrial and scientific applications?
A) It is easier to measure
B) It is independent of temperature changes
C) It is used only for gases
D) It is more accurate for dilute solutions

✅ Answer: B) It is independent of temperature changes
📌 Explanation: Since molality depends on mass rather than volume, it remains constant even when temperature changes, making it more reliable in industrial and scientific applications.

Challenge Question

11. A solution of 10 g of KCl (molar mass = 74.5 g/mol) is dissolved in 200 g of water. What is its molality?
A) 0.25 mol/kg
B) 0.5 mol/kg
C) 1.0 mol/kg
D) 2.0 mol/kg

✅ Answer: B) 0.5 mol/kg
📌 Explanation:
Moles of KCl: $$n = \frac{10}{74.5} = 0.134 \text{ moles}$$

Mass of solvent in kg: $$W = 200 \text{ g} = 0.200 \text{ kg}$$

Molality: $$m = \frac{0.134}{0.200} = 0.5 \text{ mol/kg}$$

Final Takeaways

✅ Molality is expressed as m (mol/kg)
✅ Molality is independent of temperature and pressure
✅ It is useful for colligative properties like freezing and boiling point changes
✅ Used in scientific, industrial, and pharmaceutical applications

🔹 Molality (m) measures the number of moles of solute per kilogram of solvent.
🔹 It is calculated using: $$m = \frac{w}{M \times W} \quad \text{mol/kg}$$

🔹 Molality is independent of temperature, making it ideal for studying colligative properties.
🔹 It differs from molarity (M), which is based on solution volume and varies with temperature.

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