301. If a solute associates in solution, the observed colligative property is generally:
ⓐ. smaller because association reduces the particle count
ⓑ. larger because association increases the particle count
ⓒ. unchanged because colligative properties ignore particle count
ⓓ. negative for every concentration of the associated solute
Correct Answer: smaller because association reduces the particle count
Explanation: Association means two or more solute particles combine to form a larger aggregate. This reduces the total number of independent particles in solution. Since colligative properties depend on particle number, the observed effect becomes smaller than expected from the formula-unit amount. A smaller colligative effect can make the apparent molar mass larger than the normal molar mass. The key idea is reduction in particle count, not disappearance of solute mass.
302. If a solute dissociates in solution, the apparent molar mass calculated from a colligative property is generally:
ⓐ. higher than the normal molar mass
ⓑ. equal to the normal molar mass
ⓒ. twice the normal molar mass for every dissociation
ⓓ. lower than the normal molar mass
Correct Answer: lower than the normal molar mass
Explanation: Dissociation breaks one formula unit into two or more particles. This increases the effective number of solute particles in solution. A greater particle count gives a larger colligative effect than expected for the original molecular amount. When the usual formula is applied without allowing for dissociation, the calculated apparent molar mass comes out smaller. This is why dissociation is linked with lower abnormal molar mass.
303. A solute has normal molar mass \(60\,g\,mol^{-1}\), but a colligative-property experiment gives apparent molar mass \(120\,g\,mol^{-1}\). The best explanation is:
ⓐ. complete dissociation into two particles
ⓑ. association into larger particles
ⓒ. no solute dissolved in the solvent
ⓓ. conversion of solvent into solute
Correct Answer: association into larger particles
Explanation: The apparent molar mass is larger than the normal molar mass. A larger apparent molar mass means the experiment suggests fewer particles than expected for the given solute mass. Fewer particles are produced when solute molecules associate into larger units. Dissociation would do the opposite and make the apparent molar mass smaller. The observation therefore supports association in solution.
304. Assertion: Association of solute particles can give a higher apparent molar mass.
Reason: Association decreases the number of independent solute particles in solution.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true because association often makes the apparent molar mass greater than the normal molar mass. The Reason is also true because association combines particles and reduces the number of independent solute particles. Colligative properties become smaller when particle count decreases. If the smaller effect is interpreted using the usual formula, fewer moles are calculated for the same mass. Fewer moles for the same mass means a larger apparent molar mass.
305. A table summarizes particle changes and apparent molar mass.
| Row | Process | Particle count | Apparent molar mass |
|---|
| P | Association | Decreases | Higher than normal |
| Q | Dissociation | Increases | Lower than normal |
| R | No association or dissociation | As expected | Normal value |
| S | Association | Increases | Lower than normal |
The row that needs correction is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: Row P is correct because association reduces the number of independent particles and gives a higher apparent molar mass. Row Q is correct because dissociation increases particle count and gives a lower apparent molar mass. Row R is correct for a solute that remains as expected in solution. Row S reverses the association effect. Association does not increase particle count; it decreases it by combining particles.
306. A weak electrolyte partially dissociates in water. Compared with a nonelectrolyte solution of the same formal concentration at the same temperature, its osmotic pressure is expected to be:
ⓐ. smaller because dissociation reduces the particle count
ⓑ. larger because dissociation increases particle count
ⓒ. unchanged because the formal concentration is the same
ⓓ. smaller because ions have greater molar masses
Correct Answer: larger because dissociation increases particle count
Explanation: Osmotic pressure depends on the effective number concentration of solute particles. A nonelectrolyte molecule usually remains as one particle. A weak electrolyte partially dissociates into ions, increasing the number of particles above the formal solute-unit concentration. Therefore its osmotic pressure is higher than that of a nonelectrolyte with the same formal concentration. The amount of increase depends on the degree of dissociation.
307. The van't Hoff factor \(i\) is used in colligative properties mainly to account for:
ⓐ. thermal expansion of the solvent
ⓑ. the molar mass used in the concentration calculation
ⓒ. association or dissociation of solute particles
ⓓ. the pure-solvent vapour pressure at the measurement temperature
Correct Answer: association or dissociation of solute particles
Explanation: Colligative properties depend on the number of solute particles present in solution. If a solute associates, the number of particles decreases; if it dissociates, the number increases. The van't Hoff factor \(i\) modifies the ideal colligative-property relation for this change in effective particle count. It does not merely account for solute molar mass, solvent expansion, or pure-solvent vapour pressure. Its purpose is to connect the observed effect with the actual number of dissolved particles.
308. For a nonelectrolyte solute that neither associates nor dissociates in solution, the van't Hoff factor is:
ⓐ. \(0\)
ⓑ. \(0.5\)
ⓒ. \(2\)
ⓓ. \(1\)
Correct Answer: \(1\)
Explanation: If a solute neither associates nor dissociates, each solute molecule remains as one independent particle in solution. The effective number of particles is then the same as the expected number calculated from the molecular amount. In such a case, the observed colligative property equals the normal calculated value. Therefore \(i=1\). Values different from \(1\) indicate association, dissociation, or another particle-count change.
309. The correct comparison for van't Hoff factor is:
ⓐ. \(i\gt1\) for dissociation and \(i\lt1\) for association
ⓑ. \(i\lt1\) for dissociation and \(i\gt1\) for association
ⓒ. \(i=1\) for both dissociation and association
ⓓ. \(i\gt1\) for both dissociation and association
Correct Answer: \(i\gt1\) for dissociation and \(i\lt1\) for association
Explanation: Dissociation increases the number of solute particles in solution. Since colligative properties increase with particle number, dissociation gives \(i\gt1\). Association combines solute particles and reduces the number of independent particles. This makes the observed colligative effect smaller than expected, so \(i\lt1\). The value of \(i\) is therefore a direct clue to whether particle count has increased or decreased.
310. The relation between van't Hoff factor, normal molar mass, and abnormal molar mass is:
ⓐ. \(i=\frac{M_{\text{abnormal}}}{M_{\text{normal}}}\)
ⓑ. \(i=M_{\text{normal}}+M_{\text{abnormal}}\)
ⓒ. \(i=\frac{M_{\text{normal}}}{M_{\text{abnormal}}}\)
ⓓ. \(i=M_{\text{normal}}-M_{\text{abnormal}}\)
Correct Answer: \(i=\frac{M_{\text{normal}}}{M_{\text{abnormal}}}\)
Explanation: The apparent or abnormal molar mass is obtained from a colligative property measurement without properly accounting for association or dissociation. If dissociation occurs, the colligative effect is larger and the abnormal molar mass becomes smaller. If association occurs, the colligative effect is smaller and the abnormal molar mass becomes larger. The relation that captures both cases is \(i=\frac{M_{\text{normal}}}{M_{\text{abnormal}}}\). This ratio connects particle-count correction with molar-mass abnormality.
311. A solute has normal molar mass \(60\,g\,mol^{-1}\), but its apparent molar mass from a colligative property is \(30\,g\,mol^{-1}\). The van't Hoff factor is:
ⓐ. \(0.5\)
ⓑ. \(1.0\)
ⓒ. \(2.0\)
ⓓ. \(3.0\)
Correct Answer: \(2.0\)
Explanation: \( \textbf{Known data:} \) Normal molar mass \(M_{\text{normal}}=60\,g\,mol^{-1}\), abnormal molar mass \(M_{\text{abnormal}}=30\,g\,mol^{-1}\).
\( \textbf{Relation:} \)
\[
i=\frac{M_{\text{normal}}}{M_{\text{abnormal}}}
\]
\( \textbf{Substitution:} \)
\[
i=\frac{60}{30}
\]
\( \textbf{Calculation:} \)
\[
i=2.0
\]
\( \textbf{Final answer:} \) The van't Hoff factor is \(2.0\). Since \(i\gt1\), the result suggests dissociation or an increase in effective particle number.
312. A table about van't Hoff factor is shown below.
| Row | Process | Particle count | Expected \(i\) |
|---|
| P | No association or dissociation | Unchanged | \(1\) |
| Q | Dissociation | Increases | \(i\gt1\) |
| R | Association | Decreases | \(i\lt1\) |
| S | Association | Increases | \(i\gt1\) |
The row that needs correction is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: Row P is correct because unchanged particle count gives \(i=1\). Row Q is correct because dissociation increases the number of particles and gives \(i\gt1\). Row R is correct because association decreases particle count and gives \(i\lt1\). Row S reverses the association effect. Association combines particles, so it cannot be described as increasing particle count in this context.
313. The boiling point elevation formula corrected using van't Hoff factor is:
ⓐ. \(\Delta T_b=iK_bm\)
ⓑ. \(\Delta T_b=\frac{K_bm}{i}\)
ⓒ. \(\Delta T_b=i+K_b+m\)
ⓓ. \(\Delta T_b=\frac{i}{K_bm}\)
Correct Answer: \(\Delta T_b=iK_bm\)
Explanation: The normal boiling point elevation relation is \(\Delta T_b=K_bm\). When association or dissociation changes the effective particle count, the van't Hoff factor \(i\) is included. The corrected formula becomes \(\Delta T_b=iK_bm\). If \(i\gt1\), the elevation is larger than the nonelectrolyte value. If \(i\lt1\), the elevation is smaller because fewer effective particles are present.
314. The freezing point depression for an electrolyte solution is commonly written as:
ⓐ. \(\Delta T_f=K_bm\)
ⓑ. \(\Delta T_f=iK_fm\)
ⓒ. \(\Delta T_f=\frac{K_f}{im}\)
ⓓ. \(\Delta T_f=iK_bm\)
Correct Answer: \(\Delta T_f=iK_fm\)
Explanation: The ordinary freezing point depression relation is \(\Delta T_f=K_fm\). For a solute that dissociates or associates, the effective particle count differs from the formula-unit count. The van't Hoff factor \(i\) corrects this difference, giving \(\Delta T_f=iK_fm\). The constant \(K_f\) must be used for freezing point depression. Using \(K_b\) would mix the boiling point constant with the freezing point property.
315. The osmotic pressure equation modified for association or dissociation is:
ⓐ. \(\pi=iCRT\)
ⓑ. \(\pi=\frac{CRT}{i}\)
ⓒ. \(\pi=i+C+R+T\)
ⓓ. \(\pi=\frac{i}{CRT}\)
Correct Answer: \(\pi=iCRT\)
Explanation: For an ideal dilute nonelectrolyte solution, osmotic pressure is \(\pi=CRT\). If the solute dissociates, more particles are present and osmotic pressure increases. If the solute associates, fewer particles are present and osmotic pressure decreases. The van't Hoff factor \(i\) adjusts the formula to \(\pi=iCRT\). The factor multiplies the particle concentration effect because osmotic pressure is directly proportional to effective particle number.
316. A \(0.10\,mol\,kg^{-1}\) solution of a solute has \(i=2\). If \(K_f=1.86\,K\,kg\,mol^{-1}\), the freezing point depression is:
ⓐ. \(0.186\,K\)
ⓑ. \(0.372\,K\)
ⓒ. \(1.86\,K\)
ⓓ. \(3.72\,K\)
Correct Answer: \(0.372\,K\)
Explanation: \( \textbf{Known data:} \) \(i=2\), \(m=0.10\,mol\,kg^{-1}\), \(K_f=1.86\,K\,kg\,mol^{-1}\).
\( \textbf{Corrected freezing point relation:} \)
\[
\Delta T_f=iK_fm
\]
\( \textbf{Substitution:} \)
\[
\Delta T_f=2\times1.86\times0.10
\]
\( \textbf{First multiply }1.86\times0.10\textbf{:} \)
\[
1.86\times0.10=0.186
\]
\( \textbf{Apply }i\textbf{:} \)
\[
\Delta T_f=2\times0.186=0.372\,K
\]
\( \textbf{Final answer:} \) The freezing point depression is \(0.372\,K\). The factor \(i=2\) doubles the nonelectrolyte value.
317. For a solute, the calculated normal boiling point elevation is \(0.20\,K\), but the observed elevation is \(0.30\,K\). The van't Hoff factor is:
ⓐ. \(0.50\)
ⓑ. \(0.67\)
ⓒ. \(2.00\)
ⓓ. \(1.50\)
Correct Answer: \(1.50\)
Explanation: \( \textbf{Normal calculated elevation:} \) \(0.20\,K\).
\( \textbf{Observed elevation:} \) \(0.30\,K\).
\( \textbf{Relation for van't Hoff factor:} \)
\[
i=\frac{\text{observed colligative property}}{\text{normal calculated colligative property}}
\]
\( \textbf{Substitution:} \)
\[
i=\frac{0.30}{0.20}
\]
\( \textbf{Calculation:} \)
\[
i=1.50
\]
\( \textbf{Final answer:} \) The van't Hoff factor is \(1.50\). Since \(i\gt1\), the observed effect is larger than the normal value because the effective number of particles has increased.
318. Observed boiling-point data for a \(0.50\,mol\,kg^{-1}\) solution give \(\Delta T_b=0.13\,K\) in a solvent with \(K_b=0.52\,K\,kg\,mol^{-1}\). The most suitable conclusion is:
ⓐ. the solute is completely dissociated into two ions
ⓑ. the solute is behaving as a nonelectrolyte with \(i=1\)
ⓒ. the solute is associated with \(i=0.5\)
ⓓ. the solvent constant has no role in the result
Correct Answer: the solute is associated with \(i=0.5\)
Explanation: \( \textbf{Expected nonelectrolyte elevation:} \)
\[
\Delta T_{b,\text{normal}}=K_bm
\]
\[
\Delta T_{b,\text{normal}}=0.52\times0.50=0.26\,K
\]
\( \textbf{Observed elevation:} \) \(0.13\,K\).
\( \textbf{Find the van't Hoff factor:} \)
\[
i=\frac{\Delta T_{b,\text{observed}}}{\Delta T_{b,\text{normal}}}
\]
\[
i=\frac{0.13}{0.26}=0.50
\]
\( \textbf{Interpretation:} \) Since \(i\lt1\), the effective number of solute particles is lower than expected, indicating association.
\( \textbf{Final answer:} \) The solute is associated with \(i=0.5\). Complete dimerisation is one common limiting case that gives this value.
319. A solute shows \(i=0.5\) in a solvent. If the normal calculated osmotic pressure is \(4.0\,bar\), the observed osmotic pressure is:
ⓐ. \(0.5\,bar\)
ⓑ. \(2.0\,bar\)
ⓒ. \(4.0\,bar\)
ⓓ. \(8.0\,bar\)
Correct Answer: \(2.0\,bar\)
Explanation: \( \textbf{Given data:} \) \(i=0.5\), normal osmotic pressure \(=4.0\,bar\).
\( \textbf{Relation:} \)
\[
\pi_{\text{observed}}=i\pi_{\text{normal}}
\]
\( \textbf{Substitution:} \)
\[
\pi_{\text{observed}}=0.5\times4.0
\]
\( \textbf{Calculation:} \)
\[
\pi_{\text{observed}}=2.0\,bar
\]
\( \textbf{Final answer:} \) The observed osmotic pressure is \(2.0\,bar\). A value of \(i=0.5\) indicates fewer effective particles than expected, as in complete dimerisation.
320. Complete dissociation occurs in a \(0.10\,M\) aqueous \(CaCl_2\) solution. At \(300\,K\), using \(R=0.083\,L\,bar\,K^{-1}\,mol^{-1}\), its osmotic pressure is:
ⓐ. \(2.49\,bar\)
ⓑ. \(4.98\,bar\)
ⓒ. \(7.47\,bar\)
ⓓ. \(24.9\,bar\)
Correct Answer: \(7.47\,bar\)
Explanation: \( \textbf{Dissociation pattern:} \)
\[
CaCl_2\rightarrow Ca^{2+}+2Cl^-
\]
Complete dissociation gives \(i=3\).
\( \textbf{Known data:} \) \(C=0.10\,mol\,L^{-1}\), \(T=300\,K\), and \(R=0.083\,L\,bar\,K^{-1}\,mol^{-1}\).
\( \textbf{Corrected osmotic-pressure relation:} \)
\[
\pi=iCRT
\]
\( \textbf{Substitution:} \)
\[
\pi=3\times0.10\times0.083\times300
\]
\( \textbf{Intermediate product:} \)
\[
0.083\times300=24.9
\]
\( \textbf{Calculation:} \)
\[
\pi=3\times0.10\times24.9=7.47\,bar
\]
\( \textbf{Final answer:} \) The osmotic pressure is \(7.47\,bar\). Complete dissociation triples the effective particle concentration relative to a nonelectrolyte at the same formal concentration.
