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Solutions — NEET & JEE MCQs with answers
This chapter covers the formation and properties of solutions — homogeneous mixtures of two or more substances. It explains concentration units (mole fraction, molarity, molality, mass percentage), Henry's law for gas solubility, Raoult's law for vapour pressure of liquid solutions, and colligative properties (relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure) and their use in determining molar masses. The concept of van't Hoff factor accounts for electrolyte dissociation and association producing abnormal molar masses.
Topics in this chapter
- 1.1 Types of Solutions
- 1.2 Expressing Concentration of Solutions
- 1.3 Solubility
- 1.4 Vapour Pressure of Liquid Solutions
- 1.5 Ideal and Non-Ideal Solutions
- 1.6 Colligative Properties and Determination of Molar Mass
- 1.7 Abnormal Molar Masses
Free practice questions
17.1 g of sucrose ($M = 342$ g mol$^{-1}$) is dissolved in 250 g of water. Using $K_b = 0.52$ K kg mol$^{-1}$, the boiling point elevation of this solution is
- A. $0.026$ K
- B. $0.520$ K
- C. $0.104$ K
- D. $0.208$ K
Show answer
Answer: C — Moles of sucrose $= 17.1/342 = 0.0500$ mol; solvent mass $= 0.250$ kg, so $m = 0.0500/0.250 = 0.200$ mol kg$^{-1}$. $\Delta T_b = 0.52 \times 0.200 = 0.104$ K. Taking the moles of sucrose as the molality, i.e. treating the solvent as $1$ kg instead of $0.250$ kg, gives the small value $0.026$ K.
NCERT Class 12 Chemistry Ch 1 ("mass of solute in a known mass of the solvent is taken and DTb is")
In the Henry's law equation $p = K_H x$ (equation 1.11), what does the symbol $p$ denote?
- A. The total pressure of the gaseous mixture above the solution
- B. The vapour pressure of the pure solvent
- C. The partial pressure of the gas in the vapour phase
- D. The pressure at which the gas liquefies at that temperature
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Answer: C — The most commonly used form of Henry's law is stated as: the partial pressure of the gas in vapour phase ($p$) is proportional to the mole fraction of the gas ($x$) in the solution. Hence $p$ is the partial pressure of that gas in the vapour phase above the solution, not the total pressure of a gaseous mixture, not the vapour pressure of the pure solvent, and not a liquefaction pressure.
NCERT Class 12 Chemistry Ch 1 ("vapour phase (p) is proportional to the mole fraction")
When a solution does not obey Raoult's law over the entire range of concentration, it is called a
- A. Non-ideal solution
- B. Ideal solution
- C. Colligative solution
- D. Saturated solution
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Answer: A — This is NCERT's definition of a non-ideal solution. An ideal solution is the one that **does** obey Raoult's law throughout, so option B is the exact opposite. 'Colligative solution' is not a category at all, and saturation refers to the maximum solute a solvent can dissolve, not to vapour-pressure behaviour.
NCERT Class 12 Chemistry Ch 1 ("does not obey Raoult's law over the entire range of")
2.5 g of sodium chloride is dissolved in water and the solution is made up to a total volume of 500 mL. The mass by volume percentage of the solution is:
- A. $0.50\%$ (w/V)
- B. $5.0\%$ (w/V)
- C. $0.25\%$ (w/V)
- D. $1.25\%$ (w/V)
Show answer
Answer: A — Mass by volume percentage is the mass of solute present in 100 mL of solution. Here 500 mL contains 2.5 g, so 100 mL contains $2.5\times\dfrac{100}{500}=0.50\ \mathrm{g}$, giving $0.50\%$ (w/V). Option B ($5.0\%$) results from dropping a factor of ten, and option C ($0.25\%$) from using 1000 mL instead of 500 mL.
NCERT Class 12 Chemistry Ch 1 ("100 mL of the solution")
Which observation about the substances met in everyday life is made in the very first line of the Unit?
- A. Pure substances are rarely met with in normal life; most things are mixtures
- B. Pure substances are the commonest form of matter around us
- C. Every everyday substance turns out to be a binary solution
- D. Everyday substances are mostly heterogeneous suspensions
Show answer
Answer: A — The Unit opens by saying that in normal life we rarely come across pure substances and that most of these are mixtures containing two or more pure substances. That is the opposite of claiming pure substances are commonest, and the text does not claim everyday substances are binary solutions or heterogeneous suspensions.
NCERT Class 12 Chemistry Ch 1 ("In normal life we rarely come across pure substances")
One mole of KCl dissolved in one kg of water is expected to raise the boiling point of water by (take $K_b = 0.52$ K kg mol$^{-1}$ and ignore interionic attractions):
- A. 2.08 K
- B. 0.26 K
- C. 1.04 K
- D. 0.52 K
Show answer
Answer: C — One mole of KCl releases one mole each of $\mathrm{K^+}$ and $\mathrm{Cl^-}$, giving two moles of particles, so $\Delta T_b = i K_b m = 2 \times 0.52 \times 1 = 1.04$ K, exactly as NCERT computes. The distractor 0.52 K is what you get by forgetting dissociation ($i = 1$); 2.08 K would require $i = 4$.
NCERT Class 12 Chemistry Ch 1 ("would be expected to increase the boiling point by")
In a binary solution of two volatile liquids, component 1 is MORE volatile than component 2: $p_1^{\circ} = 119$ torr and $p_2^{\circ} = 37$ torr. The total vapour pressure over a solution in which $x_2 = 0.60$ is:
- A. $86.2$ torr
- B. $141.2$ torr
- C. $78.0$ torr
- D. $69.8$ torr
Show answer
Answer: D — Equation (1.16), $p_{total} = p_1^{\circ} + (p_2^{\circ} - p_1^{\circ})x_2$, holds whichever component is more volatile; here the bracket is negative. $p_{total} = 119 + (37 - 119) \times 0.60 = 119 - 49.2 = 69.8$ torr, i.e. $p_{total}$ DECREASES as $x_2$ rises because component 2 is the less volatile one. The distractor $141.2$ torr comes from $p_1^{\circ} + p_2^{\circ}x_2$; $86.2$ torr comes from using $x_2 = 0.40$; $78.0$ torr is the equimolar value.
NCERT Class 12 Chemistry Ch 1 ("solution decreases or increases with")
The mole fraction of $\mathrm{N_2}$ dissolved in water at 293 K under a partial pressure of $0.987$ bar is $1.29 \times 10^{-5}$. Given that 1 litre of water contains $55.5$ mol of water, how many millimoles of $\mathrm{N_2}$ dissolve in 1 litre of water?
- A. $1.29$ mmol
- B. $7.16$ mmol
- C. $0.358$ mmol
- D. $0.716$ mmol
Show answer
Answer: D — With $n$ = moles of dissolved $\mathrm{N_2}$, $x = n/(n + 55.5) \approx n/55.5$ because $n \ll 55.5$. Hence $n = 1.29 \times 10^{-5} \times 55.5 = 7.16 \times 10^{-4}$ mol $= 0.716$ mmol. Choosing $0.358$ mmol corresponds to using only $27.75$ mol of water, and $7.16$ mmol is a factor-of-ten slip.
NCERT Class 12 Chemistry Ch 1 ("As 1 litre of water contains 55.5 mol of it")
According to NCERT, the freezing point of a substance may be defined as the temperature at which
- A. the densities of the solid and the liquid phases become equal
- B. the vapour pressure of the liquid phase equals the atmospheric pressure
- C. the vapour pressure of the substance in its liquid phase equals its vapour pressure in the solid phase
- D. the vapour pressure of the substance falls to zero
Show answer
Answer: C — NCERT defines the freezing point as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase — the point at which solid and liquid are in dynamic equilibrium. Equality of vapour pressure with the atmospheric pressure defines the boiling point instead, and vanishing vapour pressure or equal densities are not the criterion.
NCERT Class 12 Chemistry Ch 1 ("the substance in its liquid phase is equal to its")
A concentration of exactly 1 part per million is the same as a percentage of:
- A. $10^{-4}\ \%$
- B. $10^{-6}\ \%$
- C. $10^{-3}\ \%$
- D. $10^{-2}\ \%$
Show answer
Answer: A — 1 ppm corresponds to a fraction of $\dfrac{1}{10^{6}}$. Since a percentage is that fraction multiplied by 100, 1 ppm $=\dfrac{100}{10^{6}}=10^{-4}\ \%$. Option B ($10^{-6}\ \%$) mistakes the fraction itself for a percentage, i.e. it omits the factor of 100 that converts one to the other.
NCERT Class 12 Chemistry Ch 1 ("Number of parts of the component")
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