CBSE Class 12 Chemistry Chapter 1 Solutions Competency Based Questions Set 01

Competency Based Questions (MCQs)

Question 1. Mole fraction of glycerine \( \text{C}_3\text{H}_5(\text{OH})_3 \) in solution containing 36 g of water and 46 g of glycerine is
(a) 0.46
(b) 0.40
(c) 0.20
(d) 0.36
Answer: (c) 0.20
In simple words: To get the mole fraction, we divide the moles of glycerine by the total moles in the mixture. This gives us 0.20.

Exam Tip: Always calculate the molar masses of all components carefully before computing their moles to avoid simple math errors.

Question 2. Out of molality (m), molarity (M), formality (F) and mole fraction (x), those which are independent of temperature are
(a) M, m
(b) F, x
(c) m, x
(d) M, x
Answer: (c) m, x
In simple words: Temperature changes can affect volume, but not mass. Since molality and mole fraction only use mass, they stay the same when temperature changes.

Exam Tip: Remember that any concentration term containing volume is temperature-dependent, while those with mass only are temperature-independent.

Question 3. Which of the following condition is not satisfied by an ideal solution?
(a) \( \Delta H_{\text{mixing}} = 0 \)
(b) \( \Delta V_{\text{mixing}} = 0 \)
(c) Raoult’s Law is obeyed
(d) Formation of an azeotropic mixture
Answer: (d) Formation of an azeotropic mixture
In simple words: Ideal solutions mix without any temperature or volume change and follow Raoult's law. They do not form azeotropes.

Exam Tip: Azeotropic mixtures have constant boiling points and are characteristic of non-ideal solutions with large deviations.

Question 4. The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol. The mixture shows
(a) no deviation from Raoult’s Law.
(b) positive deviation from Raoult’s Law.
(c) negative deviation from Raoult’s Law.
(d) that the solution is unsaturated.
Answer: (b) positive deviation from Raoult’s Law.
In simple words: When a mixture boils at a lower temperature than its components, it means the particles escape more easily, showing a positive deviation.

Exam Tip: Minimum boiling azeotropes correspond to maximum vapour pressure and positive deviations from Raoult's law.

Question 5. Which has the lowest boiling point at 1 atm pressure?
(a) 0.1 M KCl
(b) 0.1 M Urea
(c) 0.1 M CaCl2
(d) 0.1 M AlCl3
Answer: (b) 0.1 M Urea
In simple words: Urea does not break down into ions in water, so it has the fewest particles and the lowest boiling point among these solutions.

Exam Tip: Boiling point elevation is a colligative property; more dissociated ions mean a higher boiling point, so non-electrolytes have the lowest boiling point.

Question 6. Osmotic pressure of a solution is 0.0821 atm at a temperature of 300 K. The concentration in moles/litre will be
(a) 0.33
(b) 0.666
(c) \( 0.3 \times 10^{-2} \)
(d) 3
Answer: (c) \( 0.3 \times 10^{-2} \)
In simple words: Using the formula for osmotic pressure, we find the concentration is about 0.0033 moles per litre.

Exam Tip: Always match your units carefully, especially ensuring the gas constant \( R \) matches the units of pressure (atm) and temperature (K).

Question 7. People add sodium chloride to water while boiling eggs. This is to
(a) decrease the boiling point.
(b) increase the boiling point.
(c) prevent the breaking of eggs.
(d) make eggs tasty.
Answer: (b) increase the boiling point.
In simple words: Salt raises the temperature at which water boils, helping the eggs cook more quickly.

Exam Tip: Adding any non-volatile solute elevates the boiling point of the solvent.

Question 8. The van’t Hoff factor (i) accounts for
(a) degree of solubilisation of solute.
(b) the extent of dissociation of solute.
(c) the extent of dissolution of solute.
(d) the degree of decomposition of solution.
Answer: (b) the extent of dissociation of solute.
In simple words: This factor tells us how many pieces a molecule breaks into when dissolved in water.

Exam Tip: Remember that \( i > 1 \) indicates dissociation, while \( i < 1 \) indicates association.

Question 9. Which relationship is not correct?
(a) \( \Delta T_b = \frac{K_b \cdot 1000 \cdot W_2}{M_2 \cdot W_1} \)
(b) \( M_2 = \frac{K_f \cdot 1000 \cdot W_1}{W_2 \cdot \Delta T_b} \)
(c) \( \pi = \left( \frac{n_2}{V} \right) RT \)
(d) \( \frac{p^\circ - p_s}{p^\circ} = \frac{W_2}{M_2} \times \frac{M_1}{W_1} \)
Answer: (b) \( M_2 = \frac{K_f \cdot 1000 \cdot W_1}{W_2 \cdot \Delta T_b} \)
In simple words: The second formula is wrong because it mixes up the terms for boiling point elevation and freezing point depression.

Exam Tip: Pay close attention to subscripts representing solute (2) and solvent (1) in colligative property formulas.

Question 10. The molal elevation constant depends upon
(a) nature of solute.
(b) nature of the solvent.
(c) vapour pressure of the solution.
(d) enthalpy change.
Answer: (b) nature of the solvent.
In simple words: This constant is determined entirely by the type of liquid you use as the solvent, not the solute.

Exam Tip: \( K_b \) is calculated solely using solvent parameters: \( K_b = \frac{R T_b^2 M}{1000 \Delta H_{\text{vap}}} \).

Assertion-Reason Type Questions

Direction: Mark the option which is most suitable:
(a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

Question 1. Assertion: The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol.
Reason: Azeotropic mixture of water and ethanol show positive deviation from Raoult’s law.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
In simple words: Both statements are correct. When they mix, they push each other away, which raises the pressure and lowers the boiling point.

Exam Tip: Positive deviations lead to minimum boiling azeotropes where the boiling point is lower than both pure liquids.

Question 2. Assertion: On adding a non-volatile solute to a solvent, the vapour pressure of the solution get lowered.
Reason: A solution is said to be ideal if it strictly obeys Raoult’s law at all concentrations and temperatures.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.
In simple words: Both facts are true, but the second one does not explain why the first one happens.

Exam Tip: Vapour pressure lowering is explained by the reduced surface fraction of solvent molecules.

Question 3. Assertion: An isotonic solution exerts same osmotic pressure under similar conditions.
Reason: Solute-solvent dipolar interactions exist in the pair of isotonic solution.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.
In simple words: Both statements are correct, but the second statement is not the reason why they have the same osmotic pressure.

Exam Tip: Isotonic solutions must have equal molar concentrations at the same temperature.

Question 4. Assertion: Barium Chloride is more effective in causing coagulation than Potassium Chloride.
Reason: \( \text{Ba}^{2+} \) has greater valency of 2 than one valency of \( \text{K}^+ \).
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
In simple words: Both statements are correct. Barium has a higher charge, so it is much better at clotting or grouping particles.

Exam Tip: Coagulating power increases rapidly with the valency of the active ion (Hardy-Schulze rule).

Question 5. Assertion: Molality is independent of temperature whereas molarity in a function of temperature.
Reason: Volume depends on temperature and mass does not depend on temperature.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
In simple words: Both statements are correct. Volume changes when heated, which alters molarity, but mass stays constant.

Exam Tip: Use molality instead of molarity when experiments involve temperature variations.

Question 6. Assertion: The aquatic species feel more comfortable in winter due to single space low temperature.
Reason: Solubility of gases increases with increase of temperature.
Answer: (c) Assertion is correct statement but Reason is wrong statement.
In simple words: The first statement is true, but the second is false. Gases dissolve better in cold water, not warm water.

Exam Tip: Gas dissolution is typically exothermic, so solubility decreases with increasing temperature.

Question 7. Assertion: Soft drink and soda water bottles are sealed under high pressure.
Reason: The dissolution of gas in liquid is an endothermic process.
Answer: (c) Assertion is correct statement but Reason is wrong statement.
In simple words: The assertion is true, but the reason is incorrect. High pressure helps dissolve the gas, which is an exothermic process.

Exam Tip: Henry's law states that solubility increases with partial pressure of the gas above the liquid.

Question 8. Assertion: An aqueous solution of NaCl freezes below 273 K.
Reason: Vapour pressure of the solution is less than that of the pure solvent.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
In simple words: Both statements are correct. Lower vapour pressure means the liquid needs to be colder to freeze.

Exam Tip: Freezing point depression is directly caused by the lowering of vapour pressure when a solute is added.

Question 9. Assertion: 1 M glucose will have a higher boiling point than 2 M glucose.
Reason: Elevation in boiling point is a colligative property which depends upon the number of particles of solute in the solution.
Answer: (d) Assertion is wrong statement but reason is correct statement.
In simple words: The assertion is false because more glucose means a higher boiling point, but the reason is correct.

Exam Tip: Colligative properties like boiling point elevation increase with higher solute concentrations.

Question 10. Assertion: The Semipermeable membrane made of copper (II) ferrocyanide \( \text{Cu}_2[\text{Fe}(\text{CN})_6] \) is not used for studying osmosis in a non-aqueous solution.
Reason: Copper (II) ferrocyanide is soluble in non-aqueous medium and insoluble in water.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
In simple words: Both statements are true. This membrane would melt or dissolve in non-water liquids, so we cannot use it there.

Exam Tip: Semipermeable membranes must be completely insoluble in the specific solvent used for osmosis.

Case Based Questions

Case 1. Fuel cells: Fuel cells are galvanic cells in which the chemical energy of fuel is directly converted into electrical energy. A type of fuel cell is a hydrogen-oxygen fuel cell. It consists of two electrodes made up of two porous graphite impregnated with a catalyst (platinum, silver, or metal oxide). The electrodes are placed in aqueous solution of NaOH. Oxygen and hydrogen are continuously fed into the cell. Hydrogen gets oxidized to \( \text{H}^+ \) which is neutralized by \( \text{OH}^- \), i.e., anodic reaction.
\( \text{H}_2 \xrightarrow{\quad} 2\text{H}^+ + 2\text{e}^- \)
\( 2\text{H}^+ + 2\text{OH}^- \xrightarrow{\quad} 2\text{H}_2\text{O} \)
\( \text{H}_2 + 2\text{OH}^- \xrightarrow{\quad} 2\text{H}_2\text{O} + 2\text{e}^- \)
At cathode, \( \text{O}_2 \) gets reduced to \( \text{OH}^- \)
i.e., \( \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \xrightarrow{\quad} 4\text{OH}^- \)
Hence, the net reaction is \( 2\text{H}_2 + \text{O}_2 \xrightarrow{\quad} 2\text{H}_2\text{O} \)
The overall reaction has \( \Delta H = -285.6\text{ kJ mol}^{-1} \) and \( \Delta G = -237.4\text{ kJ mol}^{-1} \) at \( 25^\circ\text{C} \).

Question (i) If the cell voltage is 1.23 V for the \( \text{H}_2 - \text{O}_2 \) fuel cell and for the half cell: \( \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \xrightarrow{\quad} 4\text{OH}^- \) has \( E^\circ = 0.40\text{V} \), then \( E^\circ \) for \( 2\text{H}_2\text{O} + 2\text{e}^- \xrightarrow{\quad} \text{H}_2 + 2\text{OH}^- \) will be
(a) 0.41V
(b) 0.83V
(c) –0.41V
(d) –0.83V
Answer: (d) –0.83V
In simple words: Using the cell voltage formula, the oxidation potential at the anode is calculated to be -0.83 V.

Exam Tip: Always remember that the cell voltage is the difference between the reduction potentials of cathode and anode.

Question (ii) What is the value of \( \Delta S^\circ \) for the fuel cell at \( 25^\circ\text{C} \)?
(a) –1600 JK–1
(b) –160 JK–1
(c) 160 JK–1
(d) 1600 JK–1
Answer: (b) –160 JK–1
In simple words: By applying the thermodynamic formula, we find that the entropy change is -160 Joules per Kelvin.

Exam Tip: Convert units of \( \Delta G \) and \( \Delta H \) from kJ to J before calculating entropy change.

Question (iii) Suppose the concentration of hydroxide ion is doubled, then the cell voltage will be
(a) Reduced by half
(b) Increased by a factor of 2
(c) Increased by a factor of 4
(d) Unchanged
Answer: (d) Unchanged
In simple words: Because hydroxide ions cancel out in the net chemical reaction, changing their concentration has no effect on the voltage.

Exam Tip: Concentration changes only affect cell potential if the ions remain in the net chemical equation.

Question (iv) A fuel cell is
I. A voltaic cell in which continuous supply of fuels are sent at anode to perform oxidation.
II. A voltaic cell in which fuels such as—\( \text{CH}_4, \text{H}_2 \), and CO are used up at anode.
III. One which involves the reaction of \( \text{H}_2 - \text{O}_2 \) fuel cell such as:
    Anode: \( 2\text{H}_2 + 4\text{OH}^- \xrightarrow{\quad} 4\text{H}_2\text{O}(l) + 4\text{e}^- \)
    Cathode: \( \text{O}_2 + 2\text{H}_2\text{O}(l) + 4\text{e}^- \xrightarrow{\quad} 4\text{OH}^- \)
IV. The efficiency of \( \text{H}_2 - \text{O}_2 \) fuel cell is 70%–75%.
(a) I, III
(b) I, III, IV
(c) I, II, III, IV
(d) I, II, III
Answer: (c) I, II, III, IV
In simple words: All four statements are correct descriptions of how fuel cells operate and their high performance.

Exam Tip: Fuel cells are highly valued because their efficiency is much higher than standard thermal power plants.

Question (v) The reaction occurring at the cathode of hydrogen-oxygen fuel cell is
(a) \( 2\text{H}_2 + 4\text{OH}^- \xrightarrow{\quad} 4\text{H}_2\text{O} + 4\text{e}^- \)
(b) \( 2\text{H}_2 + \text{O}_2 \xrightarrow{\quad} 2\text{H}_2\text{O}(l) \)
(c) \( \text{H}^+ + \text{OH}^- \xrightarrow{\quad} \text{H}_2\text{O} \)
(d) \( \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \xrightarrow{\quad} 4\text{OH}^- \)
Answer: (d) \( \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \xrightarrow{\quad} 4\text{OH}^- \)
In simple words: At the cathode, oxygen gas takes in electrons and combines with water to form hydroxide ions.

Exam Tip: Reduction always happens at the cathode, which means electron gain is observed there.

Case 2. The Properties of the solutions which depend only on the number of solute Particles but not on the nature of the solute are called colligative properties.
Relative lowering in vapour pressure is also an example of colligative properties. For an experiment, sugar solution is prepared for which lowering in vapour pressure was found to be 0.061 mm of Hg. (Vapour pressure of water at \( 20^\circ\text{C} \) is 17.5 mm of Hg).

Question (i) Relative lowering of vapour pressure for the given solution is
(a) 0.00348
(b) 0.061
(c) 0.122
(d) 1.75
Answer: (a) 0.00348
In simple words: We divide the pressure drop by the original pressure to get the relative lowering.

Exam Tip: Relative lowering of vapour pressure is a unitless ratio.

Question (ii) The vapour pressure (mm of Hg) of Solution will be
(a) 17.5
(b) 0.61
(c) 17.439
(d) 0.00348
Answer: (c) 17.439
In simple words: Subtracting the drop from the initial pressure gives the pressure of the solution.

Exam Tip: Vapour pressure of a solution containing a non-volatile solute is always lower than that of the pure solvent.

Question (iii) Mole fraction of sugar in the solution is
(a) 0.00348
(b) 0.9965
(c) 0.061
(d) 1.75
Answer: (a) 0.00348
In simple words: The mole fraction of sugar is exactly equal to the relative lowering we just calculated.

Exam Tip: Remember that \( \frac{p^\circ - p}{p^\circ} = x_{\text{solute}} \).

Question (iv) If weight of sugar taken is 5 g in 108 g of water then molar mass of sugar will be
(a) 358
(b) 120
(c) 240
(d) 400
Answer: (c) 240
In simple words: We can use the mole fraction and the masses of sugar and water to calculate the molar mass of sugar.

Exam Tip: Use the dilute solution approximation \( x_{\text{solute}} \approx \frac{w_2 \cdot M_1}{M_2 \cdot w_1} \) for quick calculations when solute moles are very small.

Question (v) The vapour pressure (mm of Hg) of water at 293 K when 25 g of glucose is dissolved in 450 g of water is
(a) 17.2
(b) 17.4
(c) 17.120
(d) 17.02
Answer: (b) 17.4
In simple words: Substituting the given weights into Raoult's law yields a solution vapour pressure of 17.4 mm Hg.

Exam Tip: Glucose has a molar mass of 180 g/mol, which must be substituted for the solute's molar mass.

Case 3. The concentration of a solute is very important in studying chemical reactions because it determines how often molecules collide in solution and thus indirectly determine the rate of reactions and the conditions at equilibrium.
There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. Concentration can be expressed in terms of molarity, molality, parts per million, mass percentage, volume percentage etc.

Question (i) A solution is prepared using aqueous KI which is turned out to be 20% w/w. Density of KI is 1.202 g/mL. The molality of the given solution and mole fraction of solute are respectively.
(a) 1.95 m, 0.120
(b) 1.5 m, 0.0263
(c) 2.5 m, 0.0569
(d) 3.0 m, 0.0352
Answer: (b) 1.5 m, 0.0263
In simple words: By setting the total solution mass to 100 grams, we find we have 20 grams of solute and 80 grams of solvent, which gives these values.

Exam Tip: Always start with a basis of 100 g of solution when dealing with weight percentage calculations.

Question (ii) The molarity (in mol L–1) of the given solution will be
(a) 1.56
(b) 1.89
(c) 0.263
(d) 1.44
Answer: (d) 1.44
In simple words: Using the solution's density, we convert the mass of the solution to volume and find the molarity is 1.44.

Exam Tip: Use the conversion formula \( M = \frac{\% \times d \times 10}{M_{\text{solute}}} \) for a quicker shortcut.

Question (iii) Which of the following is correct relationship between mole fraction and molality?
(a) \( x_2 = \frac{m M_1}{1 + m M_1} \)
(b) \( x_2 = \frac{m M_1}{1 - m M_1} \)
(c) \( x_2 = \frac{1 + m M_1}{m M_1} \)
(d) \( x_2 = \frac{1 - m M_1}{m M_1} \)
Answer: (a) \( x_2 = \frac{m M_1}{1 + m M_1} \)
In simple words: This algebraic formula directly links the mole fraction of the solute to its molality.

Exam Tip: Remember that \( M_1 \) in this formula must be expressed in kg/mol to keep units consistent.

Question (iv) Which of the following is temperature dependent?
(a) Molarity
(b) Molality
(c) Mole fraction
(d) Mass percentage
Answer: (a) Molarity
In simple words: Only molarity changes when the temperature goes up or down because liquids expand when heated.

Exam Tip: Temperature dependency is always linked to the presence of a volume term in the definition.

Question (v) Which of the following is true for an aqueous solution of the solute in term of concentration?
(a) 1 M = 1 m
(b) 1 M > 1 m
(c) 1 M < 1 m
(d) Cannot be predicted
Answer: (b) 1 M > 1 m
In simple words: Since 1 M contains less solvent for the same amount of solute, it is more concentrated than 1 m.

Exam Tip: For aqueous solutions, 1 M is always more concentrated than 1 m because the solute takes up some of the volume.

Case 4. At 298 K, the vapour pressure of pure benzene, \( \text{C}_6\text{H}_6 \) is 0.256 bar and the vapour pressure of pure toluene \( \text{C}_6\text{H}_5\text{CH}_3 \) is 0.0925 bar.
Two mixtures were prepared as follows:
(I) 7.8 g of \( \text{C}_6\text{H}_6 \) + 9.2 g of toluene
(II) 3.9 g of \( \text{C}_6\text{H}_6 \) + 13.8 g of toluene

Question (i) The total vapour pressure (bar) of Solution I is
(a) 0.128
(b) 0.174
(c) 0.198
(d) 0.258
Answer: (b) 0.174
In simple words: We calculate the mole fractions of both components and multiply them by their pure pressures to get 0.174 bar.

Exam Tip: Ensure you calculate the correct number of moles for benzene (M = 78) and toluene (M = 92).

Question (ii) Which of the given solutions have higher vapour pressure?
(a) I
(b) II
(c) Both have equal vapour pressure
(d) Cannot be predicted
Answer: (a) I
In simple words: Solution I contains more benzene, which evaporates more easily, so it has a higher overall vapour pressure.

Exam Tip: Mixtures with a higher proportion of the component with the greater vapour pressure will always have a higher total vapour pressure.

Question (iii) Mole fraction of benzene in vapour phase in Solution I is
(a) 0.128
(b) 0.174
(c) 0.734
(d) 0.266
Answer: (c) 0.734
In simple words: We divide the partial pressure of benzene by the total pressure to find its fraction in the vapour.

Exam Tip: The vapour phase is always richer in the more volatile component compared to the liquid phase.

Question (iv) Which of the following statements is/are correct?
I. Mole fraction of toluene in vapour phase is more in solution I.
II. Mole fraction of toluene in vapour phase is less in solution I.
III. Mole fraction of benzene in vapour phase is less in Solution I.
(a) Only II
(b) Only I
(c) I and III
(d) II and III
Answer: (a) Only II
In simple words: Only statement II is true because Solution I has a lower portion of toluene in its vapour.

Exam Tip: Remember that the sum of vapour mole fractions must always equal 1.

Question (v) Solution I is an example of a/an
(a) ideal solution
(b) non-ideal solution with positive deviation
(c) non-ideal solution with negative deviation
(d) cannot be predicted
Answer: (a) ideal solution
In simple words: Because benzene and toluene are very similar molecules, their mixture behaves ideally.

Exam Tip: Hydrocarbons of similar sizes typically form nearly ideal solutions.

Case 5. The solubility of gases increases with increase of pressure. William Henry made a systematic investigation of the solubility of a gas in a liquid.
According to Henry’s law “the mass of a gas dissolved per unit volume of the solvent at constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution”.
Dalton during the same period also concluded independently that the solubility of a gas in a liquid solution depends upon the partial pressure of the gas. If we use the mole fraction of gas in the solution as a measure of its solubility, then Henry’s law can be modified as “the partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution”.

Question (i) Henry’s law constant for the solubility of methane in benzene at 298 K is \( 4.27 \times 10^5\text{ mm Hg} \). The solubility of methane in benzene at 298 K under 760 mm Hg is
(a) \( 4.27 \times 10^{-5} \)
(b) \( 1.78 \times 10^{-3} \)
(c) \( 4.27 \times 10^{-3} \)
(d) \( 1.78 \times 10^{-5} \)
Answer: (b) \( 1.78 \times 10^{-3} \)
In simple words: Dividing the pressure by the constant gives the solubility as a mole fraction of 0.00178.

Exam Tip: Ensure both pressure and the constant are in the same units (mm Hg) before dividing.

Question (ii) The partial pressure of ethane over a saturated solution containing \( 6.56 \times 10^{-2}\text{ g} \) of ethane is 1 bar. If the solution contains \( 5.00 \times 10^{-2}\text{ g} \) of ethane then what will be the partial pressure (in bar) of the gas?
(a) 0.762
(b) 1.312
(c) 3.81
(d) 5.0
Answer: (a) 0.762
In simple words: Using the direct proportion of mass and pressure, we get a pressure of 0.762 bar.

Exam Tip: Henry's law can be written simply as \( \frac{m_1}{p_1} = \frac{m_2}{p_2} \) when temperature remains constant.

Question (iii) \( K_H \) (K bar) values for \( \text{Ar}_{(g)} \), \( \text{CO}_{2(g)} \), \( \text{HCHO}_{(g)} \) and \( \text{CH}_{4(g)} \) are 40.39, 1.67, \( 1.83 \times 10^{-5} \) and 0.413 respectively. Arrange these gases in the order of their increasing solubility.
(a) HCHO < CH4 < CO2 < Ar
(b) HCHO < CO2 < CH4 < Ar
(c) Ar < CO2 < CH4 < HCHO
(d) Ar < CH4 < CO2 < HCHO
Answer: (c) Ar < CO2 < CH4 < HCHO
In simple words: The lower the Henry's constant, the more soluble the gas is. Ar has the highest constant, so it is the least soluble.

Exam Tip: Always remember that a larger \( K_H \) value indicates lower gas solubility at a given pressure.

Question (iv) When a gas is bubbled through water at 298 K, a very dilute solution of the gas is obtained. Henry’s law constant for the gas at 298 K is 150 K bar. If the gas exerts a partial pressure of 2 bar, the number of millimoles of the gas dissolved in 1 L of water is
(a) 0.55
(b) 0.87
(c) 0.37
(d) 0.66
Answer: (c) 0.37
In simple words: Using the mole fraction and the number of moles of water, we get 0.37 millimoles of gas.

Exam Tip: Keep in mind that 1 L of water contains approximately 55.5 moles of water.

Question (v) Which of the following statements is correct?
(a) \( K_H \) increases with increase of temperature.
(b) \( K_H \) decreases with increase of temperature.
(c) \( K_H \) remains constant with increase of temperature.
(d) \( K_H \) first increases then decreases, with increase of temperature.
Answer: (a) \( K_H \) increases with increase of temperature.
In simple words: Gases become less soluble when water gets warmer, which corresponds to an increase in Henry's constant.

Exam Tip: Since dissolution of most gases is exothermic, higher temperature shifts the equilibrium to decrease solubility, raising \( K_H \).

Case 6. Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase.
A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapor pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapor pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. (Source: Jayawardena, J. A. E. C., Vanniarachchi, M. P. G., & Wansapala, M. A. J. (2017). Freezing point depression of different Sucrose solutions and coconut water.)

Question (i) When a non volatile solid is added to pure water it will
(a) boil above 100ºC and freeze above 0ºC.
(b) boil below 100ºC and freeze above 0ºC.
(c) boil above 100ºC and freeze below 0ºC.
(d) boil below 100ºC and freeze below 0ºC.
Answer: (c) boil above 100ºC and freeze below 0ºC.
In simple words: Adding salt or sugar to water makes it boil at a higher temperature and freeze at a lower temperature.

Exam Tip: This describes both boiling point elevation and freezing point depression simultaneously.

Question (ii) Colligative properties are
(a) dependent only on the concentration of the solute and independent of the solvent’s and solute’s identity.
(b) dependent only on the identity of the solute and the concentration of the solute and independent of the solvent’s identity.
(c) dependent on the identity of the solvent and solute and thus on the concentration of the solute.
(d) dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity.
Answer: (d) dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity.
In simple words: These properties depend on the amount of solute particles and the solvent, but not on the kind of solute.

Exam Tip: Colligative properties are defined by solute particle concentration, completely ignoring solute identity.

Question (iii) Assume three samples of juices A, B and C have glucose as the only sugar present in them. The concentration of sample A, B and C are 0.1 M, 0.5 M and 0.2 M respectively. Freezing point will be highest for the fruit juice
(a) A
(b) B
(c) C
(d) All have same freezing point
Answer: (a) A
In simple words: The juice with the lowest concentration freezes at the highest temperature because it has the least drop.

Exam Tip: Remember that the highest freezing point corresponds to the least depression (lowest concentration).

Question (iv) Identify which of the following is a colligative property?
(a) Freezing point
(b) Boiling point
(c) Osmotic pressure
(d) All of the options
Answer: (c) Osmotic pressure
In simple words: Osmotic pressure is the only direct colligative property listed here.

Exam Tip: Be careful: 'boiling point' is not a colligative property, but 'elevation in boiling point' is.