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In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how the physical properties of a solution depend on the number of dissolved solute particles rather than their chemical identity. The topic explains lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. Students also apply colligative-property equations to calculate molar mass, understand dilute solutions, and use the van’t Hoff factor to interpret association or dissociation of solute particles.
TOPIC PRACTICE
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Up to 25 questions from this page. Select your focus, then start.
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Easy · Level 2View options
\(i=1\)
\(i=2\)
\(i=0\)
\(i=4\)
Easy · Level 2View options
60 units
75 units
80 units
106.7 units
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Surface tension
Viscosity
Depression in freezing point
Refractive index
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\(0.2\,\mathrm{mol\,kg^{-1}}\)
\(0.5\,\mathrm{mol\,kg^{-1}}\)
\(1.8\,\mathrm{mol\,kg^{-1}}\)
\(5.0\,\mathrm{mol\,kg^{-1}}\)
Easy · Level 2View options
\(0.52\,\mathrm{K}\)
\(1.04\,\mathrm{K}\)
\(1.52\,\mathrm{K}\)
\(2.00\,\mathrm{K}\)
Easy · Level 2View options
\(0.372\,\mathrm{K}\)
\(1.116\,\mathrm{K}\)
\(1.86\,\mathrm{K}\)
\(5.58\,\mathrm{K}\)
Easy · Level 2View options
\(88.2^{\circ}\mathrm{C}\)
\(90.0^{\circ}\mathrm{C}\)
\(91.8^{\circ}\mathrm{C}\)
\(92.8^{\circ}\mathrm{C}\)
Easy · Level 2View options
Osmotic pressure
Boiling point
Molar mass
Colour
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\(4:1\)
\(1:2\)
\(1:4\)
\(2:1\)
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More dilute
More concentrated
Completely pure
Particle-free
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Approximately 2.5 times as large.
Approximately half as large.
Zero.
The same.
Easy · Level 2View options
Relative lowering of vapour pressure
Elevation in boiling point
Osmotic pressure
Viscosity
Easy · Level 2View options
0.2
0.4
0.8
1.0
Easy · Level 2View options
\(0.12\)
\(0.88\)
\(1.12\)
\(12\)
Easy · Level 2View options
Osmotic pressure
Viscosity
Surface tension
Refractive index
Easy · Level 2View options
0.5 mol kg⁻¹
1.0 mol kg⁻¹
2.0 mol kg⁻¹
4.0 mol kg⁻¹
Easy · Level 2View options
0.2 mol kg⁻¹
0.4 mol kg⁻¹
1.6 mol kg⁻¹
2.5 mol kg⁻¹
Easy · Level 2View options
−0.72 °C
0 °C
0.72 °C
1.44 °C
Easy · Level 2View options
यह घटेगा
यह बढ़ेगा
यह शून्य हो जाएगा
इसमें कोई परिवर्तन नहीं होगा
Easy · Level 2View options
It depends on the number of solute particles, not on their chemical nature.
It depends only on the chemical nature of the solute, not on the number of particles.
It depends only on the boiling point of the pure solvent.
It is determined by the colour and smell of the solute.
Easy · Level 2View options
Movement of solute from a concentrated solution to a dilute solution
Movement of solvent from a dilute solution to a concentrated solution
Movement of both solute and solvent in the same direction
Conversion of the solvent into vapour
Easy · Level 2View options
Hypertonic
Hypotonic
Isotonic
Unsaturated
Easy · Level 2View options
They mainly depend on the chemical nature of the solute.
They depend on the number of solute particles present in the solution.
They depend only on the colour of the solvent.
They are found only for solid solutes.
Easy · Level 2View options
It is completely dissociated
It is associated by forming dimers
It gives three ions in solution
It has become a gas
Easy · Level 2View options
0.5
1.0
1.5
2.0
Question 1EasyLevel 2
In an exam question, the solute is described as non-dissociating and non-volatile. What is the safest initial value to use while solving colligative-property questions?
Correct answer: A
The van’t Hoff factor \(i\) compares the actual number of particles in solution with the number expected from the formula units dissolved. A non-dissociating solute does not split into ions, and no association is stated either. Thus each dissolved formula unit remains one effective particle, so the initial value is \(i=1\). Volatility does not change this particle-count factor.
If the vapour pressure of the pure solvent is 80 units and the mole fraction of solvent is 0.75, what will be the vapour pressure of the solution?
Correct answer: A
For a solution with a non-volatile solute, Raoult’s law states that the partial vapour pressure of the solvent is \(p=x_1p_1^0\). Substituting the given values gives \(p=0.75\times80=60\) units. The pure-solvent pressure is not used unchanged because the solute lowers the solvent’s mole fraction and therefore lowers its vapour pressure.
Which of the following properties depends only on the number of solute particles present in a solution and not on their nature?
Correct answer: C
Depression in freezing point is a colligative property. For dilute solutions it is expressed as \(\Delta T_f=iK_fm\), so it depends on the effective number of dissolved particles, represented by \(i\), rather than on the chemical identity of those particles. Surface tension, viscosity, and refractive index generally depend on the nature and interactions of the substances as well.
If 0.3 mole of solute is dissolved in 600 grams of solvent, what will be the molality?
Correct answer: B
Molality is defined as moles of solute divided by the mass of solvent in kilograms: \(m=n_{solute}/mass_{solvent}\). Convert 600 grams to 0.600 kilograms, then calculate \(m=0.3/0.600=0.5\,\mathrm{mol\,kg^{-1}}\). The mass of the solution is not used; only the mass of the solvent belongs in the denominator.
If \(i=2\), \(K_b=0.52\), and \(m=1\), what will be \(\Delta T_b\)?
Correct answer: B
The elevation in boiling point for a solution is calculated using \(\Delta T_b=iK_bm\). Substituting the values gives \(\Delta T_b=2\times0.52\times1=1.04\,\mathrm{K}\). A temperature interval in kelvin has the same numerical size as an interval in degrees Celsius, so 1.04 K is also a rise of 1.04 °C. Therefore, option B is correct.
If \(i=3\), \(K_f=1.86\), and \(m=0.2\), what will be \(\Delta T_f\)?
Correct answer: B
Depression in freezing point is calculated from \(\Delta T_f=iK_fm\). Substitution gives \(\Delta T_f=3\times1.86\times0.2=1.116\,\mathrm{K}\). The factor \(i=3\) indicates that the solute produces three effective particles per formula unit under the stated conditions, so omitting it would give the smaller and incorrect value 0.372 K.
If the boiling point of the pure solvent is \(90^{\circ}\mathrm{C}\) and \(\Delta T_b=1.8^{\circ}\mathrm{C}\), what will be the boiling point of the solution?
Correct answer: C
The symbol \(\Delta T_b\) represents the elevation, or increase, in the boiling point caused by the solute. Therefore, the boiling point of the solution is found by adding the elevation to the pure-solvent boiling point: \(T_b(solution)=90+1.8=91.8^{\circ}\mathrm{C}\). Subtraction would describe a decrease and is inappropriate here.
Isotonic solutions are solutions that have equal osmotic pressure at the same temperature. Osmotic pressure depends on the number of dissolved solute particles and, for dilute solutions, is expressed as \(\pi=CRT\). Isotonicity does not require equal boiling points, molar masses, colours, or chemical identities. Therefore, the property common to isotonic solutions is osmotic pressure, so A is correct.
At the same temperature, osmotic pressures of two solutions are in the ratio \(1:4\). What will be the ratio of their molar concentrations?
Correct answer: C
For dilute solutions, osmotic pressure follows \(\pi=CRT\). Comparing two solutions at the same temperature means that both \(R\) and \(T\) are common constants. Consequently, \(\pi_1/\pi_2=C_1/C_2\). Since the osmotic-pressure ratio is \(1:4\), the molar-concentration ratio must also be \(1:4\). Therefore, option C is correct.
If a solution has higher osmotic pressure than another solution at the same temperature, how will the first solution be?
Correct answer: B
For dilute solutions at the same temperature, osmotic pressure is given by \(\pi=CRT\). Because \(R\) and \(T\) are fixed, a higher value of \(\pi\) means a higher molar concentration \(C\), or more dissolved solute particles per unit volume. Therefore, the first solution is more concentrated. It is not pure or particle-free, so option B is correct.
If i = 2.5, how will the observed colligative property compare with that of a normal non-electrolyte at the same concentration?
Correct answer: A
For a colligative property, the van’t Hoff factor multiplies the ideal particle-based effect. For example, ΔTf = iKf m, ΔTb = iKb m, and π = iCRT. If i = 2.5 while concentration and the solvent remain unchanged, the effective number of particles and therefore the observed effect are about 2.5 times those for a normal non-electrolyte. Thus A is correct.
Which of the following is not a colligative property of a solution?
Correct answer: D
Colligative properties depend primarily on the number of dissolved solute particles, not on their chemical identity. The four standard examples are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Viscosity depends on particle size, shape, interactions, and solvent structure, so it is not colligative. Hence D is correct.
If a solution contains 0.2 mol of solute and 0.8 mol of solvent, what is the mole fraction of the solvent?
Correct answer: C
The mole fraction of a component equals its number of moles divided by the total number of moles of all components. Here, total moles = 0.2 + 0.8 = 1.0 mol. Therefore, the solvent mole fraction is Xsolvent = 0.8/1.0 = 0.8. The mole fractions of solute and solvent together must add to one, confirming option C.
For a dilute solution containing a non-volatile solute, if the mole fraction of the solute is \(0.12\), what is the relative lowering of vapour pressure?
Correct answer: A
For a dilute solution of a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as \(\frac{p^0-p}{p^0}=x_{solute}\), where \(p^0\) is the vapour pressure of the pure solvent and \(p\) is the vapour pressure of the solution. Since \(x_{solute}=0.12\), the relative lowering is 0.12. Thus option A is correct; it is not the solvent mole fraction, 0.88.
Which of the following is a colligative property of a solution that depends mainly on the number of solute particles?
Correct answer: A
Osmotic pressure is a colligative property because, for a dilute solution, it is described by π = iCRT. Thus, at fixed temperature and concentration, the effective number of solute particles determines the osmotic pressure. Viscosity, surface tension, and refractive index depend substantially on the nature and interactions of the substances, so A is the only correct answer.
If 0.5 mol of solute dissolves in 250 g of solvent, what is the molality of the solution?
Correct answer: C
Molality is defined as the number of moles of solute divided by the mass of solvent in kilograms: m = moles of solute / kilograms of solvent. Convert 250 g of solvent to 0.250 kg. Therefore, m = 0.5/0.250 = 2.0 mol kg⁻¹. The correct answer is option C.
If Kf = 2.0 K kg mol⁻¹ and ΔTf = 0.8 K for a non-electrolyte solution, what is its molality?
Correct answer: B
For a non-electrolyte, the freezing-point depression is ΔTf = Kf m because i = 1. Solving for molality gives m = ΔTf/Kf. Substituting the values, m = 0.8/2.0 = 0.4 mol kg⁻¹. The unit K cancels appropriately with the unit in Kf, leaving mol kg⁻¹. Therefore, option B is correct.
If the freezing point of a pure solvent is 0 °C and that of its solution is −0.72 °C, what is the depression in freezing point, ΔT_f?
Correct answer: C
The depression in freezing point is defined as the difference between the freezing point of the pure solvent and that of the solution: ΔT_f = T_f° − T_f. Substituting the given values, ΔT_f = 0 − (−0.72) = 0.72 °C. The negative sign shows that the solution freezes at a lower temperature, while the depression itself is reported as a positive magnitude.
If the concentration of a dilute solution remains constant and its temperature is increased, what happens to its osmotic pressure?
Correct answer: B
For a dilute solution, the osmotic-pressure equation is π = iCRT. If concentration C and the van’t Hoff factor i remain constant, osmotic pressure is directly proportional to the absolute temperature T. Thus increasing temperature increases π. Temperature must be expressed in kelvin in the equation, although the direction of change is the same when temperature rises in Celsius.
Which of the following statements correctly describes a colligative property?
Correct answer: A
A colligative property depends primarily on the number of dissolved particles present in a solution, rather than on the chemical identity of those particles. Important examples are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Electrolytes may produce more particles because they dissociate into ions, so their effects can be larger at the same concentration.
What type of flow through a semipermeable membrane is called osmosis?
Correct answer: B
Osmosis is the spontaneous movement of solvent molecules through a semipermeable membrane from the side containing a more dilute solution, or nearly pure solvent, to the side containing a more concentrated solution. The membrane permits solvent molecules to pass but prevents solute particles from crossing. This movement continues until the chemical potential difference is balanced or external pressure stops it.
If two solutions separated by a semipermeable membrane show no net flow of solvent, what are they called?
Correct answer: C
The net movement of solvent across a semipermeable membrane is driven by a difference in osmotic pressure or effective solute-particle concentration. When two solutions have no net solvent flow under the stated conditions, their osmotic pressures are equal. Such solutions are called isotonic. Hypertonic and hypotonic describe unequal effective concentrations relative to one another, while unsaturated refers to solubility, not osmotic balance.
Which statement correctly identifies colligative properties?
Correct answer: B
The word colligative means related to collecting or counting particles. Thus, these properties depend on the total number of dissolved solute particles relative to the amount of solvent, not mainly on the particles’ chemical identity. Vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure are standard examples. Dissociation or association changes the effective particle count and therefore changes the magnitude of the effect.
A solute is found to have a van’t Hoff factor i = 0.5. Which conclusion is most appropriate?
Correct answer: B
The van’t Hoff factor compares the actual number of particles in solution with the number expected if no association or dissociation occurred. A value below 1 means that the number of independent particles has decreased. This commonly occurs when two solute molecules associate to form one dimer. Thus, association by dimer formation is the most appropriate conclusion.
A 0.1 m solution has ΔTb = 0.052 K. If Kb = 0.52 K kg mol⁻¹ for the solvent, what is the van’t Hoff factor i?
Correct answer: B
Use the relation ΔTb = iKb m. Substituting the given values gives 0.052 = i × 0.52 × 0.1. The product 0.52 × 0.1 equals 0.052, so i = 0.052/0.052 = 1. A van’t Hoff factor of 1 indicates that the solute behaves as though it neither dissociates into additional particles nor associates significantly in the solution.
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