• Amal Augustine
• January 21, 2026

Molality Explained in Simple Words to Master Chemistry with Confidences

Molality explained in simple words, molaltiy is an important concept in chemistry that helps aspirants understand how concentrated a solution is. In very simple terms, molality tells us how much solute is dissolved in a given amount of solvent. Molality explained in simple words, it is defined as the number of moles of solute present in one kilogram of solvent. The symbol used for molality is m.

In molality explained in simple words,in order to understand molality clearly, we must first know what a solution is. A solution is made up of two parts: the solute and the solvent. The solute is the substance that gets dissolved, such as salt or sugar, while the solvent is the substance that does the dissolving, such as water. Molality focuses only on the mass of the solvent, not on the total volume of the solution.

One of the main reasons molality is so important is because it does not change with temperature. When temperature increases or decreases, the volume of a solution may change, but the mass of the solvent remains the same. Since molality depends on mass and not volume, it remains constant even when temperature changes. This makes molality more reliable than molarity, which depends on the volume of the solution and therefore changes with temperature.

In molality explained in simple words, molality is especially useful when studying colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure. These properties depend on the number of solute particles present in the solvent, and molality gives a clear and accurate way to measure this. Because of this, molality is widely used in physical chemistry and competitive exams like NEET, JEE, and CUET.

In short, molality is a concentration unit that connects the amount of solute directly to the mass of solvent, making it simple, stable, and very useful for calculations involving temperature changes.

MCQs on Molality Explained in Simple Words-

1.

At 298 K, vapour pressures of pure liquids A and B are 200 mm Hg and 400 mm Hg respectively. If mole fractions of A and B in solution are 0.7 and 0.3, the mole fraction of B in vapour phase is:

a) 0.279
b) 0.721
c) 0.53
d) 0.46

Answer: d

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2.

3.1 g of compound X (molar mass = 62 g/mol) is dissolved in 19.5 g of compound Y (molar mass = 78 g/mol). The ratio of mole fractions of X and Y is:

a) 1 : 5
b) 5 : 1
c) 4 : 1
d) 1 : 4

Answer: a

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3.

Assertion (A): Molality of a solution increases with temperature.
Reason (R): Molality expression does not involve any volume term.

a) Both A and R are correct and R is the correct explanation of A
b) Both A and R are correct but R is not the explanation
c) A is correct, R is not correct
d) A is not correct, R is correct

Answer: d

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4.

The molality and molarity of a glucose solution labeled as 10% (w/w) (density = 1.2 g/mL) are:

a) 0.57 m, 0.517 M
b) 0.67 m, 0.617 M
c) 0.617 m, 0.67 M
d) 0.517 m, 0.57 M

Answer: c

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5.

Which of the following are correct for an ideal solution?
(i) ΔVₘᵢₓ = 0
(ii) Vₛₒₗᵥₑₙₜ + Vₛₒₗᵤₜₑ = Vₛₒₗᵤₜᵢₒₙ
(iii) ΔHₘᵢₓ = 0
(iv) H₂O + CO₂ → H₂CO₃ is an ideal solution example

a) (i), (ii)
b) (ii), (iii)
c) All are correct
d) (i), (ii), (iii)

Answer: d

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6.

If 2 g of NaOH is dissolved to make 200 mL solution at 25°C, the molarity at 90°C is:

a) M < 0.25
b) 0.5 > M > 0.25
c) M = 0.25
d) 0.5 < M < 1.0

Answer: a

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7.

A solvent freezes at 17°C and has latent heat of fusion 180 J g⁻¹. The molal depression constant (Kf) is:

a) 3.88
b) 3.55
c) 3.77
d) 4.77

Answer: a

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8.

A drinking water sample contains 15 ppm (by mass) of a carcinogen (molar mass 120 g mol⁻¹). Its molality is:

a) 2.5 × 10⁻⁴
b) 2.5 × 10⁻³
c) 1.25 × 10⁻⁴
d) 1.25 × 10⁻³

Answer: c

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9.

The molarity of 10% (w/w) aqueous NaOH (density = 1.11 g mL⁻¹) is:

a) 2.50 M
b) 3.25 M
c) 2.78 M
d) 3.78 M

Answer: c

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10.

50% reagent is used for dehydrohalogenation of 6.45 g CH₃CH₂Cl. The mass of major product formed is:

a) 0.7 g
b) 1.4 g
c) 2.8 g
d) 5.6 g

Answer: b

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11.

Boiling points of aqueous sucrose and urea solutions are equal. If 3 g urea is dissolved in 1 L, the mass of sucrose in 1 L is:

a) 34.2 g
b) 17.1 g
c) 3.0 g
d) 1.7 g

Answer: b

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12.

Which concentration unit is independent of temperature?

a) Normality
b) Molarity
c) Molality
d) ppm

Answer: c

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13.

250 mL Na₂CO₃ solution contains 2.65 g Na₂CO₃. If 10 mL is diluted to 1 L, the concentration is:

a) 0.1 M
b) 0.01 M
c) 0.001 M
d) 10⁻⁴ M

Answer: c

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14.

What amount of Cl₂ gas is liberated when 1 A current flows for 30 min?

a) 0.66 mol
b) 0.33 mol
c) 0.66 g
d) 0.33 g

Answer: c

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15.

Molality is defined as:

a) Moles of solute in 1 dm³ of solution
b) Moles of solute in 1 kg of solvent
c) Moles of solute in 100 dm³ of solution
d) Moles of solute in 1 kg of solution

Answer: a

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16.

Mole fraction of solute in a 1.00 m aqueous solution is:

a) 1.77
b) 0.0354
c) 0.0177
d) 0.177

Answer: c

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17.

Which is independent of temperature?

a) Normality
b) Molarity
c) Molality
d) Weight–volume %

Answer: c

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18.

Volume of oxygen at STP required to burn 39 g benzene is:

a) 112 L
b) 84 L
c) 22.4 L
d) 168 L

Answer: b

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19.

Mass of oxalic acid (MW = 126) needed for 100 mL of 0.1 N solution is:

a) 12.6 g
b) 0.126 g
c) 0.63 g
d) 0.063 g

Answer: c

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20.

If glucose concentration in blood is 0.72 g L⁻¹, its molarity is ___ ×10⁻³ M:

a) 1
b) 2
c) 3
d) 4

Answer: d

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21.

15 mL Fe²⁺ reacts with 20 mL of 0.03 M Cr₂O₇²⁻. Molarity of Fe²⁺ is ___ ×10⁻² M:

a) 24
b) 12
c) 10
d) 20

Answer: a

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22.

Mole fraction of solute in 100 m aqueous solution is ___ ×10⁻²:

a) 8
b) 64
c) 81
d) 72

Answer: b

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23.

CO₂ dissolved in 0.9 L water at 298 K under 0.835 bar pressure is:

a) 25 × 10⁻³ mol
b) 3 × 10⁻³ mol
c) 5 × 10⁻³ mol
d) PO₄²⁻

Answer: a

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24.

Vapour pressures of A and B are 90 and 15 mm Hg. If xA = 0.6, mole fraction of B in vapour phase is x × 10⁻¹:

a) 1
b) 2
c) 3
d) 4

Answer: a

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25.

At 20°C vapour pressures of benzene and methyl benzene are 70 and 20 torr. Mole fraction of benzene in vapour phase is ___ ×10⁻²:

a) 78
b) 87
c) 80
d) 8

Answer: a

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26.

Titre values: 4.8, 4.9, 5.0, 5.0, 5.0 mL. Concentration of Na₂CO₃ is:

a) 50 mM
b) 25 mM
c) 75 mM
d) 5 mM

Answer: a

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27.

20 g Na₂O dissolved in 500 mL water gives NaOH concentration ___ ×10⁻¹ M:

a) 13
b) 23
c) 33
d) 3

Answer: a

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28.

100 g propane reacts with 1000 g oxygen. Mole fraction of CO₂ is x ×10⁻². Value of x is:

a) 19
b) 25
c) 19
d) 15

Answer: a

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29.

80 g CuSO₄·5H₂O in 5 L gives concentration x ×10⁻³ M. Value of x is:

a) 64
b) 32
c) 70
d) 7

Answer: a

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30.

Molarity of solution prepared using 6.3 g oxalic acid dihydrate in 250 mL is x ×10⁻² M. Value of x is:

a) 10
b) 20
c) 30
d) 40

Answer: b

Conclusion on Molality Explained in Simple Words

Molality is one of the most straightforward and dependable ways to express the concentration of a solution. By defining concentration in terms of moles of solute per kilogram of solvent, molality avoids many complications caused by temperature changes. This is its biggest advantage over molarity, which depends on the volume of the solution and can vary when temperature changes.

Another important benefit of molality is its usefulness in numerical problems and real-life applications. Since many chemical properties depend on the number of solute particles rather than the type of solute, molality provides a direct and accurate measurement. This is why it is used extensively in calculating freezing point depression, boiling point elevation, and osmotic pressure.

For aspirants, in molality explained in simple words , understanding molality also helps in building strong basics of solution chemistry. Once the difference between molality and molarity is clear, many numerical problems become easier to solve. Remembering that molality depends only on the mass of the solvent makes it simpler to apply formulas correctly without confusion.

In conclusion, molality explained in simple words, molality is a simple yet powerful concept in chemistry. Its independence from temperature, clear definition, and wide application make it an essential topic for both academic learning and competitive exam preparation. By understanding molality well, aspirants gain better control over solution calculations and develop confidence in physical chemistry concepts.