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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
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Medium · Level 7View options
\(100\,g\,mol^{-1}\)
\(150\,g\,mol^{-1}\)
\(200\,g\,mol^{-1}\)
\(250\,g\,mol^{-1}\)
Medium · Level 7View options
100 g mol⁻¹
150 g mol⁻¹
200 g mol⁻¹
250 g mol⁻¹
Medium · Level 7View options
A measurable osmotic pressure can be obtained even in very dilute solutions.
The solute must be burnt before measurement.
The solvent must always be coloured.
No calculation is needed because molality is never used.
Medium · Level 7View options
0.1488 K
0.186 K
0.2976 K
0.372 K
Medium · Level 7View options
18 g mol⁻¹
36 g mol⁻¹
72 g mol⁻¹
90 g mol⁻¹
Medium · Level 7View options
50 g mol⁻¹
75 g mol⁻¹
100 g mol⁻¹
150 g mol⁻¹
Medium · Level 7View options
600 g mol⁻¹
1200 g mol⁻¹
2400 g mol⁻¹
4800 g mol⁻¹
Medium · Level 7View options
50 g mol⁻¹
100 g mol⁻¹
150 g mol⁻¹
200 g mol⁻¹
Medium · Level 7View options
80 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
150 g mol⁻¹
Medium · Level 7View options
20.8 g mol⁻¹
50 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
Medium · Level 7View options
60 g mol⁻¹
90 g mol⁻¹
120 g mol⁻¹
150 g mol⁻¹
Medium · Level 7View options
80 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
150 g mol⁻¹
Medium · Level 7View options
40 g mol⁻¹
60 g mol⁻¹
80 g mol⁻¹
100 g mol⁻¹
Medium · Level 7View options
0.04 m
0.08 m
0.16 m
0.32 m
Medium · Level 7View options
Molality will be extremely low and the calculated molar mass may be much too high
Molality will remain correct
The molar mass will always be zero
There will be no effect
Medium · Level 7View options
1000 times larger
1000 times smaller
Equal
250 times larger
Medium · Level 7View options
60 g mol⁻¹
80 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
Medium · Level 7View options
50 g mol⁻¹
60 g mol⁻¹
80 g mol⁻¹
100 g mol⁻¹
Medium · Level 7View options
19.6 g mol⁻¹
39.2 g mol⁻¹
49.0 g mol⁻¹
98.0 g mol⁻¹
Medium · Level 7View options
100 g mol⁻¹
160 g mol⁻¹
200 g mol⁻¹
240 g mol⁻¹
Medium · Level 7View options
60 g mol⁻¹
75 g mol⁻¹
90 g mol⁻¹
120 g mol⁻¹
Medium · Level 7View options
i = 0.67, association
i = 1.5, dissociation
i = 1, normal behaviour
i = 2.5, association
Medium · Level 7View options
80 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
150 g mol⁻¹
Medium · Level 7View options
i = 1.5, dissociation
i = 0.5, association
i = 1, normal
i = 3, complete dissociation
Medium · Level 7View options
200 g mol⁻¹
300 g mol⁻¹
400 g mol⁻¹
500 g mol⁻¹
Question 1MediumLevel 7
When \(10\,g\) of an unknown solute is dissolved in \(1\,kg\) of solvent, the freezing-point depression is \(\Delta T_f=0.093\,K\). If \(K_f=1.86\,K\,kg\,mol^{-1}\), what is the molar mass of the solute? Assume no association or dissociation.
Correct answer: C
For a nonelectrolyte, \(\Delta T_f=K_fm\), so the molality is \(m=0.093/1.86=0.05\,mol\,kg^{-1}\). Since the solvent mass is exactly \(1\,kg\), the amount of solute is \(0.05\,mol\). The molar mass is therefore mass divided by moles: \(M=10/0.05=200\,g\,mol^{-1}\). The assumption of no association or dissociation means \(i=1\). Hence, option C is correct.
If the osmotic-pressure method gives a molarity of 0.015 mol L⁻¹ and 400 mL of the solution contains 1.2 g of solute, what is the molar mass of the solute?
Correct answer: C
Convert the volume first: 400 mL = 0.400 L. The amount of solute is n = molarity × volume = 0.015 mol L⁻¹ × 0.400 L = 0.006 mol. Molar mass is mass divided by amount: M = 1.2 g ÷ 0.006 mol = 200 g mol⁻¹. Therefore, option C is correct. The osmotic-pressure method is useful because osmotic pressure can provide the solution concentration, from which molar mass is calculated when the solute mass and solution volume are known.
Why is the osmotic-pressure method considered more reliable for determining the molar mass of large molecules?
Correct answer: A
Large molecules have high molar masses, so a given mass contains very few moles. Consequently, their boiling-point elevation or freezing-point depression may be too small for accurate measurement. Osmotic pressure is measurable even at very low concentrations and is proportional to molar concentration, making it especially useful for biomolecules and polymers. Thus, option A is correct.
For the same solute, 1.0 g dissolved in 100 g of solvent gives ΔTf = 0.186 K. If 2.0 g of the same solute is dissolved in 250 g of the same solvent, what will be the new value of ΔTf?
Correct answer: A
For the same solute, solvent, and temperature conditions, ΔTf is proportional to the ratio of solute mass to solvent mass. The ratio of the new concentration effect to the old one is (2.0/250)/(1.0/100) = 0.8. Therefore, the new depression is 0.186 × 0.8 = 0.1488 K. Both solute and solvent masses must be included in the comparison; comparing solute masses alone would be incorrect.
In the vapour-pressure-lowering method, 1.8 g of a solute is dissolved in 9.0 g water. The relative lowering of vapour pressure is 0.10. If the molar mass of water is 18 g mol⁻¹, what is the approximate molar mass of the solute?
Correct answer: B
For a dilute solution containing a non-volatile solute, the relative lowering of vapour pressure is approximately equal to the mole fraction of the solute, and therefore Δp/p⁰ ≈ n₂/n₁. The moles of water are n₁ = 9.0/18 = 0.50 mol. Hence n₂ = 0.10 × 0.50 = 0.050 mol. The solute molar mass is M₂ = mass/moles = 1.8/0.050 = 36 g mol⁻¹. Thus option B is correct.
A solution contains 1.5 g solute dissolved in 0.25 kg solvent. If its molality is 0.06 m, what is the molar mass of the solute?
Correct answer: C
Molality is defined as the number of moles of solute present in one kilogram of solvent. Therefore, moles of solute = molality × mass of solvent in kilograms = 0.06 × 0.25 = 0.015 mol. The molar mass is then the mass of solute divided by its amount: M = 1.5 g/0.015 mol = 100 g mol⁻¹. The solvent mass, not solution mass, must be used in molality. Hence option C is correct.
A polymer solution contains 0.2 g polymer in 200 mL solution. At 300 K, its osmotic pressure is 0.0205 atm. If R = 0.082 L atm K⁻¹ mol⁻¹, what is the molar mass of the polymer?
Correct answer: B
For a dilute polymer solution, osmotic pressure is given by π = (w/MV)RT, where w is the polymer mass, M its molar mass, and V the solution volume in litres. Rearranging gives M = wRT/(πV). Substituting w = 0.2 g, T = 300 K, V = 0.200 L, π = 0.0205 atm, and R = 0.082 gives M = (0.2 × 0.082 × 300)/(0.0205 × 0.200) ≈ 1200 g mol⁻¹. Therefore, option B is correct.
When 1.0 g of a nonelectrolyte is dissolved in 50 g camphor, the freezing-point depression is 8 K. If K_f = 40 K kg mol⁻¹ for camphor, what is the molar mass of the substance?
Correct answer: B
For a nonelectrolyte, the freezing-point relation is ΔT_f = K_f m, where m is the molality of the solute. Using the molar-mass form, M = K_f × w₂ × 1000/(ΔT_f × w₁), with w₂ = 1.0 g solute and w₁ = 50 g camphor. Therefore M = 40 × 1.0 × 1000/(8 × 50) = 100 g mol⁻¹. This is the principle used in the Rast method. Hence option B is correct.
When 1.5 g of a non-electrolyte substance is dissolved in 250 g of a solvent, the depression in freezing point is 0.1116 K. If the cryoscopic constant of the solvent is 1.86 K kg mol⁻¹, what is the molar mass of the substance?
Correct answer: B
For a non-electrolyte, the depression in freezing point is given by ΔTf = Kf m, where m is the molality of the solution. Using the mass form, M = (Kf × w₂ × 1000)/(ΔTf × w₁), where w₂ is the mass of solute in grams and w₁ is the mass of solvent in grams. Substituting the data: M = (1.86 × 1.5 × 1000)/(0.1116 × 250) = 2790/27.9 = 100 g mol⁻¹. Therefore, option B is correct. The calculation assumes no association or dissociation of solute particles.
The observed molar mass of a substance is 50 g mol⁻¹ and its van’t Hoff factor is 2.4. What is its normal molar mass?
Correct answer: D
The van’t Hoff factor connects normal and observed molar masses through i = Mnormal/Mobserved. Hence, Mnormal = i × Mobserved. Using i = 2.4 and Mobserved = 50 g mol⁻¹, Mnormal = 2.4 × 50 = 120 g mol⁻¹. The factor greater than one indicates dissociation or an increase in the number of solute particles, so option D is correct.
A 0.05 mol kg⁻¹ solution is prepared by dissolving 3.0 g of a nonelectrolyte in 500 g of solvent. What is the molar mass of the solute?
Correct answer: C
Molality is the number of moles of solute per kilogram of solvent. The solvent mass is 500 g = 0.500 kg. Therefore, moles of solute = molality × mass of solvent in kilograms = 0.05 × 0.500 = 0.025 mol. The molar mass is mass/moles = 3.0/0.025 = 120 g mol⁻¹, so option C is correct.
A 1.0 L solution is prepared by dissolving 2.2 g of a substance. Its osmotic pressure at 300 K is 0.451 atm. If R = 0.082 L atm K⁻¹ mol⁻¹, what is the approximate molar mass of the substance?
Correct answer: C
For a dilute solution of a non-electrolyte, osmotic pressure is given by π = (w/MV)RT, where w is the mass of solute, M is its molar mass, V is the solution volume in litres, and T is the absolute temperature. Rearranging gives M = wRT/(πV). Substituting the data, M = (2.2 × 0.082 × 300)/(0.451 × 1.0) = 54.12/0.451 ≈ 120 g mol⁻¹. Therefore, option C is correct.
A solution contains 4.0 g of solute in 250 g of solvent. If Kf = 2.0 K kg mol⁻¹ and the freezing-point depression is 0.4 K, what is the molar mass of the solute?
Correct answer: C
For a dilute nonelectrolyte solution, ΔTf = Kf m, where m is molality. Using m = (w₂/M)/(w₁ in kg), the molar mass formula is M = Kf w₂ × 1000/(ΔTf w₁), with w₁ in grams. Substitution gives M = (2.0 × 4.0 × 1000)/(0.4 × 250) = 80 g mol⁻¹. Hence option C is correct.
A 0.04 m solution is formed by dissolving 2.0 g of a nonelectrolyte in 0.5 kg solvent. If 4.0 g of the same solute is dissolved in 0.25 kg of the same solvent, what will be the new molality?
Correct answer: C
For the same solute, molality is proportional to the mass of solute and inversely proportional to the mass of solvent: m = (mass of solute/M) ÷ mass of solvent. The solute mass changes from 2.0 g to 4.0 g, so it doubles, while the solvent changes from 0.5 kg to 0.25 kg, so it is halved. Therefore the molality becomes 2 × 2 = 4 times the original value. New molality = 0.04 × 4 = 0.16 m.
In molar mass determination, a student treats 100 g solvent as 100 kg while calculating molality. What will happen to the result?
Correct answer: A
The conversion 100 g = 0.100 kg is essential in the molality formula, m = moles of solute per kilogram of solvent. If 100 kg is used instead of 0.100 kg, the solvent mass is taken as 1000 times too large, so the calculated molality becomes 1000 times too small. Any molar mass obtained from this erroneous molality will consequently be unreliable; in the usual rearrangement, the molar mass may appear excessively high.
A student mistakenly takes 250 g solvent as 250 kg. In the freezing-point method, how will the calculated molar mass compare with the correct value?
Correct answer: B
The correct solvent mass is 250 g = 0.250 kg. Treating it as 250 kg makes the solvent mass 250/0.250 = 1000 times too large. In the freezing-point method, ΔTf = Kf m and the number of solute moles is calculated from m × kilograms of solvent. Thus the erroneous calculation gives 1000 times too many moles. Since molar mass equals mass of solute divided by its moles, the calculated molar mass becomes 1000 times smaller than the correct value.
When 4 g of a solute is dissolved in 500 g of water, the depression in freezing point is 0.279 K. If the van’t Hoff factor is 1.5, what is the true molar mass?
Correct answer: B
For freezing-point depression, ΔT_f = iK_fm, where m is the actual molality of solute particles. Therefore, m = 0.279/(1.5 × 1.86) = 0.100 mol kg⁻¹. The solvent mass is 500 g = 0.5 kg, so moles of solute = 0.100 × 0.5 = 0.050 mol. Hence, true molar mass = 4/0.050 = 80 g mol⁻¹. Thus option B is correct.
When 6 g of a solute is dissolved in 300 g of solvent, the elevation in boiling point is 0.156 K. If K_b = 0.52 K kg mol⁻¹ and i = 0.75, what is the true molar mass?
Correct answer: A
The boiling-point relation is ΔT_b = iK_bm. Thus the actual molality is m = 0.156/(0.75 × 0.52) = 0.400 mol kg⁻¹. The solvent mass is 300 g = 0.300 kg, so the amount of solute is 0.400 × 0.300 = 0.120 mol. Therefore, the true molar mass is 6/0.120 = 50 g mol⁻¹, which is option A.
When 2 g of a non-volatile solute is dissolved in 90 g of water, the relative lowering of vapour pressure is 0.02. What is the approximate molar mass of the solute?
Correct answer: A
For a non-volatile solute, relative lowering of vapour pressure equals the solute mole fraction: x₂ = n₂/(n₁+n₂) = 0.02. Water moles are n₁ = 90/18 = 5 mol. Thus n₂ = 0.02(5+n₂), giving n₂ = 0.1/0.98 = 0.102 mol approximately. The molar mass is 2/0.102 ≈ 19.6 g mol⁻¹, so option A is correct.
A 250 mL solution is prepared from 3 g of solute. At 300 K, its osmotic pressure is 1.845 atm. If i = 1.5, what is the true molar mass?
Correct answer: D
Osmotic pressure follows π = iCRT, where C is the true molarity. Therefore, C = 1.845/(1.5 × 0.0821 × 300) ≈ 0.050 M. In 0.250 L, moles of solute are n = 0.050 × 0.250 = 0.0125 mol. Hence the true molar mass is 3/0.0125 = 240 g mol⁻¹. Therefore, option D is correct.
When 1.5 g of a solute is dissolved in 0.3 kg of solvent, the observed (effective) molality is 0.05 mol kg⁻¹. If the solute has a van’t Hoff factor of 0.75, what is its true molar mass?
Correct answer: B
The effective molality is related to the true molality by m_effective = i × m_true. Therefore, m_true = 0.05/0.75 = 0.0667 mol kg⁻¹. The solvent mass is 0.3 kg, so moles of solute = 0.0667 × 0.3 = 0.020 mol. Hence, molar mass = 1.5 g/0.020 mol = 75 g mol⁻¹. Thus, option B is correct. Since i is less than 1, the solute undergoes association.
The true molar mass of a solute is 150 g mol⁻¹, but its molar mass determined by a colligative-property method is 100 g mol⁻¹. What is the van’t Hoff factor, and what is the probable behaviour of the solute?
Correct answer: B
For abnormal molar masses, the relation is observed molar mass = true molar mass/i. Therefore, i = true molar mass/observed molar mass = 150/100 = 1.5. A van’t Hoff factor greater than 1 means that the number of solute particles increases in solution. This increase is normally caused by dissociation, such as an electrolyte separating into ions. Hence, option B is correct.
A solute has van’t Hoff factor i = 2.4 and an observed molar mass of 50 g mol⁻¹. What is its true molar mass?
Correct answer: C
Colligative properties give an observed molar mass related to the true molar mass by M observed = M true/i. Rearranging gives M true = i × M observed. Substitution gives M true = 2.4 × 50 = 120 g mol⁻¹. Because i is greater than one, the solute produces additional particles, consistent with dissociation. Therefore option C is correct.
A 0.01 M solute has an osmotic pressure of 0.369 atm at 300 K. If R = 0.082 L atm mol⁻¹ K⁻¹, what are i and the likely behaviour of the solute?
Correct answer: A
The osmotic-pressure equation is π = iCRT, so i = π/(CRT). Here CRT = 0.01 × 0.082 × 300 = 0.246 atm, and i = 0.369/0.246 = 1.5. Since i is greater than one, the number of solute particles has increased, indicating dissociation rather than association or normal non-electrolyte behaviour. Thus option A is correct.
If 0.2 g solute in 100 mL solution gives an osmotic pressure of 0.123 atm at 300 K and i = 1, what is the molar mass?
Correct answer: C
For osmotic pressure, π = iCRT. Since i = 1, C = π/(RT) = 0.123/(0.082 × 300) = 0.005 mol L⁻¹. The solution volume is 0.100 L, so solute moles equal 0.005 × 0.100 = 0.0005 mol. Hence molar mass = mass/moles = 0.2/0.0005 = 400 g mol⁻¹, making C correct.
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