Q.34.2 g of sugar is present in 234.2 g of its aqueous solution. Then its molal concentration is
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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
Molality (m)=kilograms of solventmoles of solute
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 (C6H12O6, molar mass = 180 g/mol) is dissolved in 500 g of water. Calculate the molality of the solution.
Step 1: Find moles of solute
Moles of glucose=180 g/mol36 g=0.2 mol
Step 2: Convert solvent mass to kilograms
500 g=0.5 kg
Step 3: Apply the formula
m=0.5 kg0.2 mol=0.4 m
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
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=mass of solvent in kgmoles of solute
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 wsolute grams of solute with molar mass Msolute (g/mol):
moles of solute=Msolutewsolute
Step 2: Mass of Solvent in kg
If the solvent mass is Wsolvent grams:
mass of solvent in kg=1000Wsolvent
Step 3: Putting It Together
m=1000WsolventMsolutewsolute
Simplifying:
m=Msolute×Wsolventwsolute×1000
The Final Formula (Exam-Ready)
m=Msolute×Wsolventwsolute×1000
Where:
- wsolute = mass of solute in grams
- Msolute = molar mass of solute in g/mol
- Wsolvent = mass of solvent in grams
Why the ×1000 Factor? …
Finding the molality here means converting the sugar's mass to moles, finding the solvent's mass by subtracting the solute from the total solution mass, a …
Moles of sugar = 34.2/342 = 0.1; mass of solvent (water) = 234.2 - 34.2 = 200 g = 0.2 kg; molality = 0.1/0.2 = 0.5 m.
Molar mass of sugar (sucrose) = 342 g/mol.
Moles of sugar = 34.2/342 = 0.1 mol. …
Showing the 12 most recent of 31 on this concept.
- CBSE 2026Set 56/3/11 markMCQQ.The mole fraction of a solute in 2.0 molal aqueous solution is : (A) 1.87 (B) 0.347 (C) 0.0347 (D) 0.00347
›Reveal solutionSolution
A 2.0 molal solution means 2 moles of solute in 1 kg of water. Convert the solvent mass to moles, then apply the mole fraction formula: χsolute=nsolute+nsolventnsolute. The answer is 0.0347.
Molality is defined as moles of solute per kilogram of solvent, not per kilogram of solution. This distinction matters because we need to count the moles of both solute and solvent separately to find the mole fraction.
When we say a solution is 2.0 molal, we're saying there are 2.0 moles of solute dissolved in exactly 1000 g (1 kg) of water. The mole fraction then asks: what fraction of the total number of particles (molecules) in the solution comes from the solute?
Let me work through the calculation systematically.
1. Identify what we know from "2.0 molal aqueous solution"
The molality m=2.0 tells us:
- Moles of solute: nsolute=2.0 mol
- Mass of water (solvent): 1000 g
2. Convert the mass of water to moles
Water has a molar mass of 18 g/mol, so:
nwater=18 g/mol1000 g=55.56 mol
3. Calculate the total moles in the solution
ntotal=nsolute+nwater=2.0+55.56=57.56 mol
4. Apply the mole fraction formula
The mole fraction of solute is:
χsolute=ntotalnsolute=57.562.0=0.03474 …
- CBSE 2026Set A1 markMCQQ.34.2 g of sugar is present in 234.2 g of its aqueous solution. Then its molal concentration is(a) 0.1(b) 0.5(c) 5.5(d) 55.0
›Reveal solutionSolution
Moles of sugar = 34.2/342 = 0.1; mass of solvent (water) = 234.2 - 34.2 = 200 g = 0.2 kg; molality = 0.1/0.2 = 0.5 m.
Molar mass of sugar (sucrose) = 342 g/mol.
Moles of sugar = 34.2/342 = 0.1 mol. …
- CBSE 2026Set ANNUAL1 markMCQQ.What will be the molarity of 30 ml of 0.5 M H2SO4 solution diluted to 50 ml?(a) 0.3 M(b) 0.03 M(c) 3 M(d) 0.13 M
›Reveal solutionSolution
Dilution does not change the number of moles of solute, so M1V1 (before) = M2V2 (after).
Given: M1 = 0.5 M, V1 = 30 mL, final volume V2 = 50 mL.
…
- CBSE 2026Set ANNUAL1 markQ.Define mole fraction.
›Reveal solutionSolution
Mole fraction expresses a component's amount relative to the total moles in a mixture, independent of temperature.
For a solution containing components with n1, n2, n3, ... moles, the mole fraction of component 1 is defined as:
x1 = n1 / (n1 + n2 + n3 + ...)
…
- CBSE 2026Set ANNUAL1 markMCQQ.In an acid-base titrimetric analysis, the concentration of a sulphuric acid analyte is found to be 0.044 M. The strength of the acid in g/L is –(a) 0.44(b) 4.31(c) 2.15(d) 44.00
›Reveal solutionSolution
Strength in g/L = molarity × molar mass = 0.044 × 98 ≈ 4.31 g/L, so option (B).
The strength of a solution in grams per litre is related to its molarity by
Strength (g/L)=Molarity (mol/L)×Molar mass (g/mol).
…
- CBSE 2025Set ANNUAL1 markQ.Write the definition of molality.
›Reveal solutionSolution
Molality expresses concentration as moles of solute per kilogram of SOLVENT (not solution), and unlike molarity it does not change with temperature.
Definition:
Molality (m) = (moles of solute) / (mass of solvent in kg)
Unit: mol kg^-1 (also written 'molal', symbol m)
…
- CBSE 2025Set ANNUAL1 markMCQQ.What is the molarity of a solution with a mass of solute 10 kg mass and 100 litre volume?(a) 0.1 molar(b) 1 molar(c) 10 molar(d) 100 molar
›Reveal solutionSolution
Molarity is defined as moles of solute per litre of solution; dividing the given amount of solute by the solution volume gives 0.1 M.
Molarity is defined as:
M=volume of solution in litresmoles of solute
Note: as printed, the question states the solute quantity as '10 kg mass'; for the arithmetic to match any of the given options (0.1, 1, 10, 100 M) the intended quantity is 10 moles of solute (a common wording/printing slip in this recurring question, where 'kg' should read 'mol') - with n …
- CBSE 2025Set ANNUAL1 markMCQQ.A solution contains 8 moles of solute and the mass of solvent is 4 kg. What is the molality of this solution?(a) 5 mol/kg(b) 8 mol/kg(c) 4 mol/kg(d) 2 mol/kg
›Reveal solutionSolution
Molality is moles of solute per kilogram of solvent (not solution); here it works out to 8/4 = 2 mol/kg.
Molality (m) is defined as:
m = (moles of solute) / (mass of solvent in kg)
Given:
- moles of solute = 8 mol
- mass of solvent = 4 kg
m = 8 mol / 4 kg = 2 mol/kg
…
- CBSE 2025Set ANNUAL1 markQ.Define molality.
›Reveal solutionSolution
Molality is defined as moles of solute per kilogram of solvent; unlike molarity, it does not depend on temperature since it is based on mass, not volume.
Molality (denoted m) of a solution is defined as the number of moles of solute dissolved in one kilogram (1000 g) of solvent:
Molality (m) = (number of moles of solute) / (mass of solvent in kg)
Unit: mol kg^-1 (also written as 'm', e.g., a '1 molal' or '1 m' solution).
…
- CBSE 2024Set A11 markQ.The number of moles of solute present in one kilogram of the solvent is called \rule{2cm}{0.4pt}.
›Reveal solutionSolution
Moles of solute per kilogram of solvent defines molality.
Molality (m) is a concentration term that depends only on the mass of solvent (and is therefore temperature-independent):
m=mass of solvent in kgmoles of solute(mol kg−1) …
- CBSE 2024Set ANNUAL1 markMCQQ.Which of the following has no unit?(a) Molarity(b) Molality(c) Normality(d) Molar Fraction
›Reveal solutionSolution
Mole fraction is a dimensionless ratio, so it has no unit.
…
- CBSE 2024Set ANNUAL1 markQ.Write down the formula of Molarity.
›Reveal solutionSolution
Molarity is the number of moles of solute dissolved per litre of solution.
Molarity is one of the most widely used units of concentration. It is defined as the number of moles of solute dissolved in one litre (one cubic decimetre) of solution:
M=volume of solution in litres (V)moles of solute (n) …
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