Board Previous Year Questions

How to Score Full Marks in the Board Exam

This section is a complete bank of board-style questions with model answers for Solutions. Each answer is written the way you should write it in the exam — stating the definition or formula first, then substituting with units, and boxing the final answer. Examiners award method marks, so always show the formula and steps even if you are unsure of the arithmetic.

The questions are arranged from 1-mark definitions through 2-3 mark reasoning and numericals to 5-mark derivations. Practise writing each answer in full sentences with correct units.

1-Mark Questions (Definitions & Direct)

Q1. Define molarity. Answer: Molarity (M) is the number of moles of solute dissolved per litre of solution: $M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$. Its unit is mol L$^{-1}$.

Q2. Why is molarity temperature-dependent? Answer: Molarity is defined per litre of solution. Volume changes (expands or contracts) with temperature, so the same number of moles occupies a different volume — hence molarity changes with temperature.

Q3. Define mole fraction. Answer: The mole fraction of a component is the ratio of its number of moles to the total number of moles of all components: $x_A = \dfrac{n_A}{n_A + n_B}$. It is dimensionless and temperature-independent.

Q4. State Henry's law. Answer: The partial pressure of a gas over a solution is directly proportional to the mole fraction of the gas in the solution: $p = K_H x$, where $K_H$ is the Henry's law constant.

1-Mark Questions (continued)

Q5. State Raoult's law for a solution of a non-volatile solute. Answer: The vapour pressure of the solution equals the product of the vapour pressure of the pure solvent and the mole fraction of the solvent: $p = p_A^0\, x_A$.

Q6. What are isotonic solutions? Answer: Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions; there is no net osmosis between them.

Q7. Define an ideal solution. Answer: An ideal solution is one that obeys Raoult's law over the entire range of concentration, with $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$.

Q8. What is the van't Hoff factor? Answer: The van't Hoff factor $i$ is the ratio of the observed colligative property to the calculated colligative property for the same amount of solute. Equivalently, it is the ratio of the actual number of particles in solution to the number expected if no association or dissociation occurred.

2-Mark Reasoning Questions

Q9. Why does the vapour pressure of a solvent decrease when a non-volatile solute is added? Answer: The non-volatile solute particles reduce the mole fraction of the solvent in the liquid phase. By Raoult's law, $p = p_A^0 x_A$, and since $x_A < 1$ when solute is present, the vapour pressure of the solution is always less than that of the pure solvent.

Q10. Why is the freezing point depression of 0.1 m NaCl about twice that of 0.1 m glucose? Answer: NaCl is a strong electrolyte that dissociates completely into Na$^+$ and Cl$^-$, giving about two particles per formula unit (van't Hoff factor $i \approx 2$). Glucose is a non-electrolyte ($i = 1$). Since colligative properties depend on the number of particles, the depression for NaCl is roughly double that of glucose.

2-Mark Reasoning Questions (continued)

Q11. Why does a soft-drink bottle fizz when opened? Answer: The bottle is sealed under a high partial pressure of CO$_2$, which (by Henry's law) keeps a large amount of CO$_2$ dissolved. On opening, the CO$_2$ pressure above the liquid falls suddenly, so the solubility of CO$_2$ decreases and the excess gas escapes as effervescence (fizz).

Q12. Why is the molality of a solution preferred over molarity for expressing concentration in colligative-property studies? Answer: Molality is defined per kilogram of solvent (a mass), which does not change with temperature. Since colligative-property experiments often involve temperature changes (boiling, freezing), molality gives a temperature-independent and therefore reliable measure of concentration.

Q13. Define azeotropes and give one example of each type. Answer: Azeotropes are binary mixtures having a constant boiling point whose vapour has the same composition as the liquid, so they cannot be separated by fractional distillation. Minimum-boiling azeotrope: ethanol-water (~95% ethanol by mass). Maximum-boiling azeotrope: nitric acid-water (~68% HNO$_3$ by mass).

2-3 Mark Reasoning Questions

Q14. Distinguish between an ideal and a non-ideal solution (any three points). Answer: (i) An ideal solution obeys Raoult's law over the whole composition range; a non-ideal solution does not. (ii) For an ideal solution $\Delta_{mix}H = 0$; for a non-ideal solution $\Delta_{mix}H \ne 0$. (iii) For an ideal solution $\Delta_{mix}V = 0$; for a non-ideal solution $\Delta_{mix}V \ne 0$. Example of ideal: benzene + toluene; non-ideal: ethanol + acetone.

Q15. What is meant by positive and negative deviation from Raoult's law? Give one example of each. Answer: A solution shows positive deviation when its vapour pressure is higher than predicted by Raoult's law — the A-B forces are weaker than A-A and B-B, and $\Delta_{mix}H > 0$ (e.g. ethanol + acetone). It shows negative deviation when its vapour pressure is lower than predicted — the A-B forces are stronger (often new hydrogen bonds), and $\Delta_{mix}H < 0$ (e.g. chloroform + acetone).

Q16. Why is the osmotic-pressure method preferred for determining the molar masses of macromolecules such as proteins? Answer: Osmotic pressure gives a measurable value even for very dilute solutions of high-molar-mass solutes, where the boiling-point elevation and freezing-point depression would be too small to measure accurately. It is also measured at room temperature, avoiding heat damage to delicate biological molecules.

3-Mark Numericals

Q17. Calculate the molarity of a solution containing 5 g of NaOH in 450 mL of solution. (Molar mass of NaOH = 40 g mol$^{-1}$) Answer: Moles of NaOH $= \dfrac{5}{40} = 0.125$ mol. Volume $= 0.450$ L. Molarity $= \dfrac{0.125}{0.450} = \mathbf{0.278\ mol\ L^{-1}}$.

Q18. A solution of glucose (molar mass 180 g mol$^{-1}$) contains 18 g of glucose in 250 g of water. Calculate the molality. Answer: Moles of glucose $= \dfrac{18}{180} = 0.10$ mol. Mass of solvent $= 0.250$ kg. Molality $= \dfrac{0.10}{0.250} = \mathbf{0.40\ mol\ kg^{-1}}$.

Q19. 15 g of an unknown non-electrolyte dissolved in 450 g of water gave a freezing-point depression of 0.34 K. Calculate the molar mass. ($K_f = 1.86$ K kg mol$^{-1}$) Answer: $M_B = \dfrac{K_f \times w_B \times 1000}{\Delta T_f \times w_A} = \dfrac{1.86 \times 15 \times 1000}{0.34 \times 450} = \dfrac{27900}{153} = \mathbf{182.4\ g\ mol^{-1}}$.

3-Mark Numericals (continued)

Q20. Calculate the osmotic pressure of a solution containing 3.0 g of urea (molar mass 60 g mol$^{-1}$) in 250 mL of solution at 300 K. ($R = 0.0821$ L atm K$^{-1}$ mol$^{-1}$) Answer: Concentration $C = \dfrac{3.0/60}{0.250} = 0.20$ M. Osmotic pressure $\Pi = CRT = 0.20 \times 0.0821 \times 300 = \mathbf{4.93\ atm}$.

Q21. The vapour pressure of pure water at 298 K is 23.8 mm Hg. Calculate the vapour pressure of a solution in which the mole fraction of water is 0.95. Answer: By Raoult's law $p = p^0_{water}\, x_{water} = 23.8 \times 0.95 = \mathbf{22.61\ mm\ Hg}$.

Q22. Calculate the boiling point of a solution containing 6.0 g of urea (molar mass 60) in 200 g of water. ($K_b = 0.52$ K kg mol$^{-1}$, normal b.p. = 100 °C) Answer: Molality $= \dfrac{6.0/60}{0.200} = 0.50$ m. $\Delta T_b = K_b \times m = 0.52 \times 0.50 = 0.26$ K. Boiling point $= 100 + 0.26 = \mathbf{100.26\ ^\circ C}$.

3-Mark Numericals (continued)

Q23. A 0.20 molal aqueous KCl solution shows a freezing-point depression of 0.68 K. Calculate the van't Hoff factor. ($K_f = 1.86$ K kg mol$^{-1}$) Answer: Calculated (normal) $\Delta T_f = K_f \times m = 1.86 \times 0.20 = 0.372$ K. $i = \dfrac{\text{observed}}{\text{calculated}} = \dfrac{0.68}{0.372} = \mathbf{1.83}$. (KCl is about 83% dissociated.)

Q24. Calculate the mass of ascorbic acid (molar mass 176 g mol$^{-1}$) to be dissolved in 75 g of acetic acid to lower its freezing point by 1.5 K. ($K_f$ of acetic acid = 3.9 K kg mol$^{-1}$) Answer: Molality $= \dfrac{\Delta T_f}{K_f} = \dfrac{1.5}{3.9} = 0.3846$ m. Moles $= 0.3846 \times 0.075 = 0.02885$ mol. Mass $= 0.02885 \times 176 = \mathbf{5.08\ g}$.

Q25. The Henry's law constant for N$_2$ at 293 K is 76.48 kbar. If N$_2$ exerts a partial pressure of 0.987 bar, find the mole fraction of N$_2$ dissolved in water. Answer: $x = \dfrac{p}{K_H} = \dfrac{0.987}{76.48 \times 10^3} = \mathbf{1.29 \times 10^{-5}}$.

3-Mark Numericals & Interconversion

Q26. A 20% (by mass) aqueous solution of KI has density 1.202 g mL$^{-1}$. Calculate the molality of the solution. (Molar mass of KI = 166 g mol$^{-1}$) Answer: In 100 g of solution: KI = 20 g, water = 80 g. Moles of KI $= \dfrac{20}{166} = 0.1205$ mol. Mass of solvent $= 0.080$ kg. Molality $= \dfrac{0.1205}{0.080} = \mathbf{1.51\ mol\ kg^{-1}}$.

Q27. Calculate the molarity of the KI solution in Q26. Answer: Volume of solution $= \dfrac{100}{1.202} = 83.2$ mL $= 0.0832$ L. Molarity $= \dfrac{0.1205}{0.0832} = \mathbf{1.45\ mol\ L^{-1}}$.

Q28. Calculate the mass percentage of benzene when 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Answer: Total mass $= 22 + 122 = 144$ g. Mass % of benzene $= \dfrac{22}{144} \times 100 = \mathbf{15.28\%}$ (and CCl$_4$ = 84.72%).

3-Mark Numericals (continued)

Q29. A first-glance estimate: 18 g of glucose (molar mass 180) is dissolved in 1 kg of water. What is the molality, and what is the relative lowering of vapour pressure? Answer: Molality $= \dfrac{18/180}{1} = 0.10$ m. Moles glucose = 0.10, moles water $= \dfrac{1000}{18} = 55.55$. Relative lowering $= x_{solute} = \dfrac{0.10}{0.10 + 55.55} = \mathbf{1.80 \times 10^{-3}}$.

Q30. Calculate the freezing point of a solution containing 60 g of glucose (molar mass 180) in 250 g of water. ($K_f = 1.86$ K kg mol$^{-1}$) Answer: Molality $= \dfrac{60/180}{0.250} = \dfrac{0.333}{0.250} = 1.333$ m. $\Delta T_f = 1.86 \times 1.333 = 2.48$ K. Freezing point $= 0 - 2.48 = \mathbf{-2.48\ ^\circ C}$.

Q31. The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point rose to 354.11 K. Calculate the molar mass. ($K_b$ for benzene = 2.53 K kg mol$^{-1}$) Answer: $\Delta T_b = 354.11 - 353.23 = 0.88$ K. $M_B = \dfrac{K_b w_B \times 1000}{\Delta T_b w_A} = \dfrac{2.53 \times 1.80 \times 1000}{0.88 \times 90} = \mathbf{57.5\ g\ mol^{-1}}$.

5-Mark / Long-Answer Questions

Q32. Derive the relation between the relative lowering of vapour pressure and the mole fraction of the solute. Answer: For a solution of a non-volatile solute in a volatile solvent, by Raoult's law the vapour pressure of the solution is $p_A = p_A^0 x_A$. The lowering of vapour pressure is $\Delta p = p_A^0 - p_A = p_A^0 - p_A^0 x_A = p_A^0(1 - x_A)$. Since $x_A + x_B = 1$, we have $1 - x_A = x_B$. Therefore the relative lowering of vapour pressure $\dfrac{p_A^0 - p_A}{p_A^0} = x_B$, the mole fraction of the solute. This shows the relative lowering equals the mole fraction of solute — a colligative property.

Q33. Explain abnormal molar mass with the help of the van't Hoff factor. How is it related to the degree of dissociation? Answer: When a solute dissociates (electrolytes) or associates (e.g. benzoic acid in benzene), the number of particles in solution differs from the number of formula units dissolved, so the molar mass calculated from a colligative property is abnormal. The van't Hoff factor $i$ corrects this: $i = \dfrac{\text{normal molar mass}}{\text{observed molar mass}}$. For dissociation into $n$ ions with degree $\alpha$: $i = 1 + (n-1)\alpha$ (so $i > 1$). For association of $n$ molecules into one with degree $\alpha$: $i = 1 - \alpha\left(1 - \dfrac{1}{n}\right)$ (so $i < 1$). Each colligative equation is then multiplied by $i$.

5-Mark / Long-Answer Questions (continued)

Q34. State and explain Raoult's law for a binary solution of two volatile liquids. How is the composition of the vapour related to it? Answer: For a binary solution of two volatile liquids A and B, Raoult's law states that the partial vapour pressure of each component is proportional to its mole fraction in the solution: $p_A = p_A^0 x_A$ and $p_B = p_B^0 x_B$. By Dalton's law, the total vapour pressure is $p_{total} = p_A^0 x_A + p_B^0 x_B$. The composition of the vapour is given by the mole fractions $y_A = \dfrac{p_A}{p_{total}}$ and $y_B = \dfrac{p_B}{p_{total}}$. The vapour is always richer in the more volatile component (the one with higher $p^0$), which is the basis of fractional distillation.

Q35. Two solutions A and B are isotonic. Solution A contains 6 g of a substance X in 1 L, and solution B contains 9 g of glucose (molar mass 180) in 1 L, both at the same temperature. Find the molar mass of X. Answer: Isotonic solutions have equal osmotic pressure, so for solutions at the same temperature and in the same volume, their molar concentrations are equal. For glucose, $C = \dfrac{9/180}{1} = 0.05$ M. Therefore $\dfrac{6/M_X}{1} = 0.05$, giving $M_X = \dfrac{6}{0.05} = \mathbf{120\ g\ mol^{-1}}$.

Q36. Why do gases always tend to be less soluble in liquids as the temperature is raised? Explain using Henry's law and give one consequence. Answer: Dissolution of a gas in a liquid is usually exothermic. By Le Chatelier's principle, raising the temperature shifts the dissolution equilibrium backwards, so less gas remains dissolved. In terms of Henry's law, the constant $K_H$ increases with temperature, and since $x = p/K_H$, the mole fraction dissolved falls. A consequence: aquatic life is endangered in warm water because warm water holds less dissolved oxygen (this also explains thermal pollution).