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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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Easy · Level 6View options
Relative lowering of vapour pressure
Elevation in boiling point
Depression in freezing point
Osmotic pressure
Easy · Level 6View options
0.123 atm
0.246 atm
0.492 atm
0.984 atm
Easy · Level 6View options
\(45\,\mathrm{g\,mol^{-1}}\)
\(60\,\mathrm{g\,mol^{-1}}\)
\(90\,\mathrm{g\,mol^{-1}}\)
\(180\,\mathrm{g\,mol^{-1}}\)
Easy · Level 6View options
80 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
160 g mol⁻¹
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0.093 K
0.186 K
0.279 K
0.372 K
Easy · Level 6View options
0.052 K
0.104 K
0.156 K
0.208 K
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The volume of solvent only
The volume of the entire solution
The volume of solute
The empty volume of the container
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The van’t Hoff factor is usually 1
The van’t Hoff factor is always 2
The molar mass is always halved
All solvents form ions
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The difference between the boiling points of the solution and the pure solvent
The colour of the solute
The odour of the solvent
The shape of the solute particles
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A larger freezing-point depression is produced for the same molality
The mass of the solute becomes zero
The solvent automatically becomes ionic
The molar mass always becomes 1 g mol⁻¹
Easy · Level 6View options
It remains unchanged
It becomes double
It becomes half
It becomes four times
Easy · Level 6View options
The moles of solute must be doubled
The moles of solute must be halved
The solute must be removed
The solvent must be converted into a solid
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It will become double
It will become half
It will remain the same
It will become one-fourth
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It will be much lower
It will be much higher
It will remain the same
It will be negative
Easy · Level 6View options
Osmotic pressure
Colour
Smell
Taste
Question 1EasyLevel 6
Which of the following colligative properties is most suitable for determining the molar mass of a high-molar-mass solute such as a polymer?
Correct answer: D
For polymers and other high-molar-mass solutes, osmotic pressure is the most useful colligative property because it can be measured in dilute solutions at ordinary temperature. The corresponding changes in boiling point, freezing point, or vapour pressure are often too small for precise measurement. Osmotic-pressure data can therefore provide a more dependable estimate of polymer molar mass.
If a 0.01 M non-dissociating solute has an osmotic pressure of 0.246 atm at 300 K, what will be the osmotic pressure of a 0.02 M solution of the same solute at the same temperature?
Correct answer: C
For a non-dissociating solute, osmotic pressure is given by π = CRT. At constant temperature, with the same solute and solvent conditions, π is directly proportional to concentration. The concentration changes from 0.01 M to 0.02 M, so it doubles. Therefore, the osmotic pressure also doubles: π2 = (0.02/0.01) × 0.246 = 0.492 atm. The value 0.246 atm would apply only to the original concentration.
If \(0.10\,\mathrm{mol}\) of solute has a mass of \(9.0\,\mathrm{g}\), and the same amount of solute is obtained from a colligative-property measurement, what is its molar mass?
Correct answer: C
Molar mass is the mass of a substance divided by the amount of substance in moles: \(M=m/n\). Here, \(m=9.0\,\mathrm{g}\) and \(n=0.10\,\mathrm{mol}\). Therefore, \(M=9.0/0.10=90\,\mathrm{g\,mol^{-1}}\). The reference to the colligative property simply supplies the amount of solute; it does not change the definition. Hence, option C is correct.
If a 0.2 m solution contains 0.04 mol of solute and the mass of the solute is 4.8 g, what is its molar mass?
Correct answer: C
The amount of solute is already given directly as 0.04 mol, so the stated molality is not needed for finding molar mass. Apply M = mass/moles: M = 4.8 g ÷ 0.04 mol = 120 g mol⁻¹. Therefore, the correct choice is option C; no additional solvent-mass calculation is required.
For a non-dissociated solute, K_f = 1.86 K kg mol⁻¹, the solute mass is 2.5 g, the solvent mass is 250 g, and the molar mass is 100 g mol⁻¹. What is the depression in freezing point, ΔT_f?
Correct answer: B
For a non-dissociated solute, the van’t Hoff factor is i = 1, so the freezing-point depression is ΔT_f = iK_fm = K_fm. Moles of solute = 2.5/100 = 0.025 mol. The solvent mass is 250 g = 0.250 kg, giving molality m = 0.025/0.250 = 0.100 mol kg⁻¹. Therefore, ΔT_f = 1.86 × 0.100 = 0.186 K, so option B is correct.
For a non-dissociated solute, K_b = 0.52 K kg mol⁻¹, the solute mass is 1.2 g, the solvent mass is 100 g, and the molar mass is 60 g mol⁻¹. What is the elevation in boiling point, ΔT_b?
Correct answer: B
For a non-dissociated solute, i = 1 and the boiling-point elevation is ΔT_b = iK_bm = K_bm. The amount of solute is 1.2/60 = 0.020 mol. The solvent mass is 100 g = 0.100 kg, so molality is m = 0.020/0.100 = 0.200 mol kg⁻¹. Thus, ΔT_b = 0.52 × 0.200 = 0.104 K. Therefore, option B is correct.
In the osmotic-pressure method, using M = wRT/(πV), what does V represent?
Correct answer: B
The osmotic-pressure equation is obtained from πV = nRT, where V is the volume occupied by the solution containing the dissolved solute. Since osmotic pressure depends on the concentration of solute particles in the entire solution, V must be the total solution volume, not the solvent volume alone or the container’s empty space. Thus option B is correct.
For molar-mass determination of a non-electrolyte solute, which statement is correct?
Correct answer: A
A non-electrolyte normally neither dissociates into ions nor associates significantly in the solution. Consequently, the number of solute particles remains close to the number predicted from its formula, so the van’t Hoff factor i is approximately 1. Colligative-property equations can then be used with the ordinary molar mass. The word “usually” is appropriate because experimental deviations may occur, but i is not automatically 2.
In the boiling-point-elevation method for determining the molar mass of an unknown solute, which quantity is measured directly?
Correct answer: A
In this method, the boiling point of the pure solvent and the boiling point of the solution are measured experimentally. Their difference is the elevation in boiling point, written as ΔTb = Tb(solution) − Tb(pure solvent). For a dilute solution, ΔTb is related to molality by ΔTb = Kb m, so this measured temperature difference is then used to calculate the amount of solute and its molar mass. The other properties are irrelevant to the method.
What is the advantage of using a solvent with a larger Kf value in freezing-point-depression measurements?
Correct answer: A
For a dilute nonelectrolyte solution, the depression in freezing point is given by ΔTf = Kf m, or by ΔTf = iKf m when abnormal particle behaviour is included. If molality and the van’t Hoff factor remain unchanged, a larger cryoscopic constant Kf produces a larger temperature change. The depression is therefore easier to observe accurately, improving experimental sensitivity. It does not alter the solute mass or molar mass.
If both the mass of the solute and the mass of the solvent are doubled, what happens to the freezing-point depression, assuming the solute remains non-electrolytic and the solution remains dilute?
Correct answer: A
Freezing-point depression is proportional to molality: ΔTf = Kf m. Molality is the number of moles of solute divided by the kilograms of solvent. Doubling the solute mass doubles its moles, while doubling the solvent mass doubles the denominator. Their ratio therefore remains unchanged, so the molality and hence the freezing-point depression remain unchanged, provided the solution stays dilute and no association or dissociation changes occur.
To double the osmotic pressure of a solution while keeping temperature and volume constant, what change is required in the amount of solute?
Correct answer: A
For a dilute solution, osmotic pressure is described by πV = nRT, or π = nRT/V. If temperature T and volume V are constant, osmotic pressure is directly proportional to the number of moles n of solute. Consequently, doubling n doubles π. For an electrolyte or an associating solute, the effective relation includes the van’t Hoff factor, but with unchanged particle behaviour the required change is still to double the amount of solute.
If the value of K_b is mistakenly taken as double, how will the molar mass calculated from boiling-point elevation change?
Correct answer: A
For a dilute solution, the molar mass is calculated using M = K_b w_2 × 1000/(ΔT_b w_1), where the measured elevation, solute mass, and solvent mass are kept fixed. Since K_b appears in the numerator, replacing it by 2K_b multiplies the calculated M by 2. Thus the reported molar mass becomes double. This is a calculation error, not a real change in the solute’s molar mass; options B and D reverse or overstate the proportional effect.
A student uses 27 K instead of 27 °C in the osmotic-pressure method. How will the calculated molar mass compare with the correct value?
Correct answer: A
The osmotic-pressure relation for molar mass is M = wRT/(πV), so temperature must be expressed on the absolute Kelvin scale and it is directly proportional to the calculated M. The correct temperature is 27 °C + 273 ≈ 300 K, whereas the student uses only 27 K. Therefore the calculated value is multiplied by 27/300, about 0.09, and is much lower than the correct value. It is not negative because all quantities in the formula remain positive.
Which colligative property can be used to find abnormal molecular mass?
Correct answer: A
Osmotic pressure is a colligative property because, for a dilute solution at fixed temperature, it depends on the concentration of dissolved particles: π = iCRT. Association or dissociation changes the effective particle concentration, so measuring π can reveal an abnormal molecular mass or allow its correction using i. Colour, smell, and taste are not colligative properties and cannot provide this basis.
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