Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how the molar mass of a solute can be determined from the measurable properties of a solution. The topic connects mass, moles, concentration, and colligative properties such as relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Students also practise selecting suitable formulas, interpreting experimental data, and recognising how observed results can indicate association or dissociation of solute particles.
TOPIC PRACTICE
Quiz this set
Up to 12 questions from this page. Select your focus, then start.
12 questions
Choose questions
Medium · Level 9View options
60 g mol⁻¹
40 g mol⁻¹
30 g mol⁻¹
20 g mol⁻¹
Medium · Level 9View options
(p⁰ − p)/p⁰ = w₂M₁/(w₁M₂)
(p⁰ − p)/p⁰ = w₁M₂/(w₂M₁)
(p⁰ − p)/p⁰ = M₁M₂/(w₁w₂)
(p⁰ − p)/p⁰ = w₁w₂M₁M₂
Medium · Level 9View options
Its molar mass may be lower
Its molar mass must be higher
It contains no solute
Its temperature is zero
Medium · Level 9View options
Whether the solution is sufficiently dilute, completely dissolved, and homogeneous
Whether the solvent has a dark colour
Whether the container is made of glass
Whether the solute is shiny
Medium · Level 9View options
Dissociation
Association
Complete ionisation
Evaporation of the solvent
Medium · Level 9View options
Osmotic pressure
Elevation in boiling point
Relative lowering of vapour pressure
Depression in freezing point
Medium · Level 9View options
It will be higher
It will be lower
It will remain unchanged
It will become zero
Medium · Level 9View options
The molar mass will be higher
The molar mass will be lower
There will be no effect
The molar mass will become negative
Medium · Level 9View options
The calculated value may be wrong because the actual pure-solute amount differs
There will be no effect
The molar mass will always be exact
The osmotic pressure will become zero
Medium · Level 9View options
Because interparticle interactions are no longer negligible
Because temperature disappears
Because solvent mass is always zero
Because solute becomes colourless
Medium · Level 9View options
It can indicate abnormality such as association or dissociation
It necessarily changes the solvent
It converts molar mass into colour
It makes temperature irrelevant
Medium · Level 9View options
First check method, given quantities, unit conversion, and need for i
Directly write any formula
Keep all temperatures in Celsius
Interchange solute and solvent masses
Question 1MediumLevel 9
If 0.5 g of a solute in 250 mL solution gives an osmotic pressure of 0.82 atm at 27 °C, find its molar mass. Use R = 0.082 L atm K⁻¹ mol⁻¹.
Correct answer: A
For a dilute non-electrolyte solution, osmotic pressure is π = wRT/(MV), so M = wRT/(πV). Convert 27 °C to 300 K and 250 mL to 0.250 L. Therefore M = (0.5 × 0.082 × 300)/(0.82 × 0.250) = 12.3/0.205 = 60 g mol⁻¹. Hence option A is correct. Correct unit conversion is essential in this calculation.
Which is the correct relation for molar-mass determination using relative lowering of vapour pressure?
Correct answer: A
For a dilute solution containing a non-volatile solute, Raoult’s law gives (p⁰ − p)/p⁰ ≈ x₂, the mole fraction of the solute. If w₁ and M₁ represent the mass and molar mass of the solvent, and w₂ and M₂ those of the solute, then x₂ ≈ (w₂/M₂)/(w₁/M₁) = w₂M₁/(w₁M₂). Thus option A is the correct mass-form relation.
A solution has higher osmotic pressure than another solution at the same temperature. If both contain the same mass of solute, what is the most suitable conclusion about the first solute?
Correct answer: A
For a dilute solution, osmotic pressure is π = cRT, or π = nRT/V. At the same temperature and comparable volume, higher osmotic pressure means a greater concentration of effective solute particles. With equal solute masses, the substance having a lower molar mass provides more moles because n = w/M, so it can produce the higher osmotic pressure. Dissociation could also contribute, so “may be lower” is the suitably cautious conclusion.
When a solute is very sparingly soluble in a solvent, what should be checked most carefully before determining its molar mass by a colligative-property method?
Correct answer: A
Colligative-property equations assume that the solute is uniformly distributed in a sufficiently dilute solution. If a sparingly soluble substance remains undissolved, the measured mass is greater than the mass actually present as dissolved particles. The calculated molality and colligative effect will then be wrong, leading to an incorrect apparent molar mass. Therefore, complete dissolution, homogeneity, and suitable dilution must be checked carefully.
A solution has a van’t Hoff factor i = 0.5. What type of particle behaviour does this most clearly indicate?
Correct answer: B
The van’t Hoff factor i compares the actual number of particles in solution with the number expected if no association or dissociation occurred. When i is less than 1, separate solute molecules combine to form larger associated units, so the effective particle number decreases. An ideal dimerisation example gives i close to 0.5 when association is essentially complete. Therefore, i = 0.5 indicates association, not dissociation.
Which colligative property is most suitable for determining the molar mass of a high-molar-mass polymer?
Correct answer: A
A polymer has a very large molar mass, so a given mass contains only a very small number of moles. Consequently, changes in boiling point or freezing point are often extremely small and difficult to measure accurately. Osmotic pressure can be measured for dilute polymer solutions at room temperature and is sufficiently sensitive for determining number-average molar mass. It also avoids heating that might degrade the polymer, making it the preferred colligative method.
In an experiment, the mass of the solvent is accidentally recorded as higher than its actual value. How will the molar mass calculated from freezing-point depression be affected?
Correct answer: B
For a dilute solution, the molar mass of the solute is calculated using M = Kf × w₂ × 1000 /(ΔTf × w₁), where w₁ is the mass of solvent. Since w₁ is in the denominator, recording it as too large makes the calculated value of M smaller than the correct value. The actual molar mass does not change; only the experimental calculation is affected.
If the freezing-point depression is mistakenly measured as lower than its true value, what is the mathematical effect on the calculated molar mass?
Correct answer: A
The molar mass relation is M = Kf × w₂ × 1000 /(ΔTf × w₁). The freezing-point depression, ΔTf, occurs in the denominator. Therefore, if ΔTf is recorded smaller than its true value while the other quantities remain unchanged, the denominator becomes smaller and the calculated molar mass becomes larger. This is a mathematical measurement error, not a change in the substance.
If the solute sample contains an impurity during molar-mass determination, what is the usual effect?
Correct answer: A
Molar-mass determination assumes that the weighed sample contains a known amount of pure solute. An impurity changes the relationship between the measured sample mass and the number of moles of the actual solute particles. Consequently, the colligative effect may not correspond to the assumed composition, and the calculated molar mass can be inaccurate. The direction of error depends on the nature and amount of impurity.
Why does the reliability of simple colligative formulas decrease if the solution is too concentrated during molar mass determination?
Correct answer: A
Simple colligative equations are derived for dilute, nearly ideal solutions, where solute particles interact only weakly and the solvent behaves regularly. In a concentrated solution, solute–solute and solute–solvent interactions become important, so the measured property may not follow the simple proportional relation. Therefore, option A is correct; temperature, colour, and solvent mass do not explain this deviation.
Why is comparing results from different colligative properties useful for accurately determining the molar mass of an unknown solute?
Correct answer: A
Each colligative property responds to the effective number of solute particles. If vapour-pressure, boiling-point, freezing-point, or osmotic-pressure methods give consistent molar masses, the result is more trustworthy. A systematic difference can reveal association, dissociation, or an experimental error. Thus A is correct; the other choices are unrelated to method comparison.
What is the best exam strategy while determining the molar mass of an unknown solute?
Correct answer: A
A reliable solution begins by identifying the method: vapour pressure, boiling-point elevation, freezing-point depression, or osmotic pressure. Then label solute and solvent quantities, convert units consistently, use kelvin temperature where required, and decide whether the van’t Hoff factor is needed. This prevents formula and substitution errors; the other strategies deliberately create them. Thus A is correct.
Google Analytics helps us understand site usage. Google may send limited cookie-free signals before your choice. The Live Visitors widget operates independently of this analytics choice; see the privacy policy for its provider and fallback details. Essential site features work without analytics cookies. You can change your choice later in Privacy choices. Privacy policy