201. For a solution containing a non-volatile solute \(B\) in a volatile solvent \(A\), Raoult's law for the solvent is:
ⓐ. \(p_A=x_Bp_A^\circ\)
ⓑ. \(p_A=x_Ap_B^\circ\)
ⓒ. \(p_A=x_Ap_A^\circ\)
ⓓ. \(p_A=p_A^\circ+x_B\)
Correct Answer: \(p_A=x_Ap_A^\circ\)
Explanation: The vapour pressure considered here is that of solvent \(A\). Raoult's law states that \(p_A=x_Ap_A^\circ\), where \(x_A\) is the mole fraction of solvent in the liquid phase. The pure vapour pressure must also belong to solvent \(A\), so \(p_A^\circ\) is used. The solute is non-volatile, so it does not add a separate vapour pressure contribution. Pairing \(x_B\) with \(p_A^\circ\) would describe lowering, not the solvent vapour pressure itself.
202. Assuming Raoult's law applies, a pure solvent has vapour pressure \(80\,kPa\) at a given temperature. A non-volatile solute is added so that \(x_A=0.75\) for the solvent. The vapour pressure of the solvent over the solution is:
ⓐ. \(20\,kPa\)
ⓑ. \(40\,kPa\)
ⓒ. \(60\,kPa\)
ⓓ. \(80\,kPa\)
Correct Answer: \(60\,kPa\)
Explanation: \( \textbf{Known data:} \) Pure solvent vapour pressure \(p_A^\circ=80\,kPa\), solvent mole fraction \(x_A=0.75\).
\( \textbf{Required quantity:} \) Vapour pressure of solvent over solution, \(p_A\).
\( \textbf{Raoult's law for solvent:} \)
\[
p_A=x_Ap_A^\circ
\]
\( \textbf{Substitution:} \)
\[
p_A=0.75\times80
\]
\( \textbf{Calculation:} \)
\[
p_A=60\,kPa
\]
\( \textbf{Final answer:} \) The solvent vapour pressure over the solution is \(60\,kPa\). The pure solvent pressure is not retained because the solvent mole fraction has fallen below \(1\).
203. If \(p_A^\circ\) is the vapour pressure of pure solvent and \(p_A\) is the vapour pressure of the solvent over a solution, the lowering of vapour pressure is:
ⓐ. \(p_A+p_A^\circ\)
ⓑ. \(p_A-p_A^\circ\)
ⓒ. \(\frac{p_A}{p_A^\circ}\)
ⓓ. \(p_A^\circ-p_A\)
Correct Answer: \(p_A^\circ-p_A\)
Explanation: Lowering of vapour pressure means the decrease from the pure solvent value to the solution value. The pure solvent has vapour pressure \(p_A^\circ\), while the solution has solvent vapour pressure \(p_A\). Since a non-volatile solute lowers the vapour pressure, \(p_A\lt p_A^\circ\). Therefore the lowering is \(p_A^\circ-p_A\). Writing \(p_A-p_A^\circ\) would give a negative value for a quantity normally reported as a positive lowering.
204. For an ideal binary solution with a non-volatile solute \(B\) in solvent \(A\), the relation between relative lowering of vapour pressure and solute mole fraction is:
ⓐ. \(\frac{p_A^\circ-p_A}{p_A^\circ}=x_A\)
ⓑ. \(\frac{p_A}{p_A^\circ}=x_B\)
ⓒ. \(\frac{p_A^\circ}{p_A}=x_B\)
ⓓ. \(\frac{p_A^\circ-p_A}{p_A^\circ}=x_B\)
Correct Answer: \(\frac{p_A^\circ-p_A}{p_A^\circ}=x_B\)
Explanation: For a non-volatile solute, Raoult's law gives \(p_A=x_Ap_A^\circ\). The lowering is \(p_A^\circ-p_A\). Substituting \(p_A=x_Ap_A^\circ\) gives \(p_A^\circ-p_A=p_A^\circ(1-x_A)\). In a binary solution, \(1-x_A=x_B\). Therefore \(\frac{p_A^\circ-p_A}{p_A^\circ}=x_B\), so the relative lowering equals the mole fraction of solute.
205. A non-volatile solute lowers the vapour pressure of water by \(2.0\%\). If the solution is dilute and ideal, the solute mole fraction is:
ⓐ. \(0.002\)
ⓑ. \(0.020\)
ⓒ. \(0.200\)
ⓓ. \(2.00\)
Correct Answer: \(0.020\)
Explanation: \( \textbf{Convert percentage to a fraction:} \)
\[
2.0\%=\frac{2.0}{100}=0.020
\]
\( \textbf{Use the relative-lowering relation:} \)
\[
\frac{p_A^\circ-p_A}{p_A^\circ}=x_B
\]
\( \textbf{Substitution:} \)
\[
x_B=0.020
\]
\( \textbf{Final answer:} \) The solute mole fraction is \(0.020\). Using \(2.0\) directly would treat a percentage as a fraction and overstate the mole fraction by a factor of \(100\).
206. Assuming Raoult's law applies, a pure solvent has vapour pressure \(50\,kPa\). After adding a non-volatile solute, the solute mole fraction is \(0.10\). The lowering of vapour pressure is:
ⓐ. \(2.5\,kPa\)
ⓑ. \(5.0\,kPa\)
ⓒ. \(45\,kPa\)
ⓓ. \(55\,kPa\)
Correct Answer: \(5.0\,kPa\)
Explanation: \( \textbf{Known data:} \) \(p_A^\circ=50\,kPa\), solute mole fraction \(x_B=0.10\).
\( \textbf{Relative lowering relation:} \)
\[
\frac{p_A^\circ-p_A}{p_A^\circ}=x_B
\]
\( \textbf{Lowering from this relation:} \)
\[
p_A^\circ-p_A=x_Bp_A^\circ
\]
\( \textbf{Substitution:} \)
\[
p_A^\circ-p_A=0.10\times50
\]
\( \textbf{Calculation:} \)
\[
p_A^\circ-p_A=5.0\,kPa
\]
\( \textbf{Final answer:} \) The lowering of vapour pressure is \(5.0\,kPa\). The final solvent vapour pressure would be \(45\,kPa\), not the lowering itself.
207. A student calculates relative lowering as \(\frac{p_A^\circ-p_A}{p_A}\). The correction is to divide by:
ⓐ. \(x_B\), liquid-phase solute mole fraction
ⓑ. \(p_B^\circ\), pure-solute vapour pressure
ⓒ. \(x_A\), liquid-phase solvent mole fraction
ⓓ. \(p_A^\circ\), pure-solvent vapour pressure
Correct Answer: \(p_A^\circ\), pure-solvent vapour pressure
Explanation: Relative lowering compares the lowering of vapour pressure with the original pure solvent vapour pressure. The numerator is \(p_A^\circ-p_A\). The denominator must be \(p_A^\circ\), not \(p_A\). This gives \(\frac{p_A^\circ-p_A}{p_A^\circ}\). Dividing by the solution vapour pressure changes the meaning of the ratio and will not equal the solute mole fraction in the standard relation.
208. For an ideal binary solution containing a non-volatile solute in a volatile solvent, the solvent mole fraction is \(0.88\). The relative lowering of vapour pressure is:
ⓐ. \(0.88\)
ⓑ. \(1.88\)
ⓒ. \(0.12\)
ⓓ. \(7.33\)
Correct Answer: \(0.12\)
Explanation: \( \textbf{Given data:} \) Solvent mole fraction \(x_A=0.88\).
\( \textbf{Binary mole-fraction relation:} \)
\[
x_A+x_B=1
\]
\( \textbf{Find solute mole fraction:} \)
\[
x_B=1-0.88=0.12
\]
\( \textbf{Relative lowering relation:} \)
\[
\frac{p_A^\circ-p_A}{p_A^\circ}=x_B
\]
\( \textbf{Substitution:} \)
\[
\frac{p_A^\circ-p_A}{p_A^\circ}=0.12
\]
\( \textbf{Final answer:} \) The relative lowering is \(0.12\). The solvent mole fraction gives the remaining fraction, while the lowering is governed by the solute mole fraction.
209. A table about non-volatile solute and vapour pressure is shown below.
| Row | Statement |
|---|
| P | For solvent \(A\), \(p_A=x_Ap_A^\circ\) |
| Q | Lowering of vapour pressure is \(p_A^\circ-p_A\) |
| R | Relative lowering is \(\frac{p_A^\circ-p_A}{p_A^\circ}\) |
| S | Relative lowering equals solvent mole fraction \(x_A\) |
The row that needs correction is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: Row P correctly gives Raoult's law for the solvent. Row Q correctly defines the lowering of vapour pressure. Row R correctly defines relative lowering as lowering divided by pure solvent vapour pressure. Row S is wrong because relative lowering equals the solute mole fraction \(x_B\), not the solvent mole fraction \(x_A\). In a binary solution, \(x_A\) and \(x_B\) add to \(1\), so interchanging them changes the meaning.
210. A non-volatile solute is dissolved in a volatile solvent at fixed temperature. If more solute particles are added without changing the solvent amount, the vapour pressure of the solvent generally:
ⓐ. increases as the solution becomes more concentrated
ⓑ. decreases as the solvent mole fraction decreases
ⓒ. remains equal to the pure-solvent value \(p_A^\circ\)
ⓓ. becomes equal to the vapour pressure of the solute
Correct Answer: decreases as the solvent mole fraction decreases
Explanation: Adding more non-volatile solute increases the solute mole fraction. In a binary solution, this lowers the solvent mole fraction \(x_A\). Since \(p_A=x_Ap_A^\circ\), a lower \(x_A\) gives a lower solvent vapour pressure. The solute does not supply its own appreciable vapour pressure. The trend follows composition, not crowding as a separate pressure source.
211. Assertion: Relative lowering of vapour pressure is a colligative property.
Reason: It depends on the number of solute particles through solute mole fraction, not on the chemical identity of the solute in ideal dilute behaviour.
ⓐ. 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 relative lowering of vapour pressure belongs to colligative properties. The Reason is also true because, for ideal dilute solutions, the effect depends on how many solute particles are present. The relation \(\frac{p_A^\circ-p_A}{p_A^\circ}=x_B\) shows the dependence on solute mole fraction. Chemical identity matters only through factors such as association or dissociation when particle count changes. The Reason directly explains why the property is classified as colligative.
212. A pure solvent has vapour pressure \(120\,kPa\). A solution has relative lowering of vapour pressure \(0.25\). The vapour pressure of the solvent over the solution is:
ⓐ. \(30\,kPa\)
ⓑ. \(60\,kPa\)
ⓒ. \(150\,kPa\)
ⓓ. \(90\,kPa\)
Correct Answer: \(90\,kPa\)
Explanation: \( \textbf{Known data:} \) \(p_A^\circ=120\,kPa\), relative lowering \(=0.25\).
\( \textbf{Relative lowering relation:} \)
\[
\frac{p_A^\circ-p_A}{p_A^\circ}=0.25
\]
\( \textbf{Find lowering:} \)
\[
p_A^\circ-p_A=0.25\times120=30\,kPa
\]
\( \textbf{Find solution vapour pressure:} \)
\[
p_A=120-30=90\,kPa
\]
\( \textbf{Final answer:} \) The solvent vapour pressure over the solution is \(90\,kPa\). The \(30\,kPa\) value is only the lowering, not the final vapour pressure.
213. An ideal solution has \(p_A^\circ=80\,kPa\), \(p_A=72\,kPa\), and contains a non-volatile solute. The solute mole fraction is:
ⓐ. \(0.10\)
ⓑ. \(0.20\)
ⓒ. \(0.72\)
ⓓ. \(0.90\)
Correct Answer: \(0.10\)
Explanation: \( \textbf{Relative-lowering relation:} \)
\[
x_B=\frac{p_A^\circ-p_A}{p_A^\circ}
\]
\( \textbf{Substitute the vapour pressures:} \)
\[
x_B=\frac{80-72}{80}
\]
\( \textbf{Vapour-pressure lowering:} \)
\[
80-72=8\,kPa
\]
\( \textbf{Calculate the mole fraction:} \)
\[
x_B=\frac{8}{80}=0.10
\]
\( \textbf{Final answer:} \) The solute mole fraction is \(0.10\). The value \(72\,kPa\) is the solvent vapour pressure over the solution, not a composition fraction.
214. For an ideal binary solution with non-volatile solute \(B\), \(x_B=0.20\). The ratio \(\frac{p_A}{p_A^\circ}\) is:
ⓐ. \(0.20\)
ⓑ. \(0.80\)
ⓒ. \(1.20\)
ⓓ. \(5.00\)
Correct Answer: \(0.80\)
Explanation: \( \textbf{Given data:} \) \(x_B=0.20\).
\( \textbf{Binary relation:} \)
\[
x_A=1-x_B=1-0.20=0.80
\]
\( \textbf{Raoult's law:} \)
\[
p_A=x_Ap_A^\circ
\]
\( \textbf{Divide by }p_A^\circ\textbf{:} \)
\[
\frac{p_A}{p_A^\circ}=x_A
\]
\( \textbf{Substitution:} \)
\[
\frac{p_A}{p_A^\circ}=0.80
\]
\( \textbf{Final answer:} \) The ratio is \(0.80\). The solute mole fraction gives the relative lowering, while the solvent mole fraction gives the pressure ratio.
215. Consider these statements for an ideal binary solution containing a non-volatile solute in volatile solvent \(A\).
Statement I: \(p_A=x_Ap_A^\circ\).
Statement II: \(\frac{p_A^\circ-p_A}{p_A^\circ}=x_B\).
Statement III: \(p_A^\circ-p_A=x_Ap_A^\circ\).
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Statement I is correct because Raoult's law for solvent \(A\) is \(p_A=x_Ap_A^\circ\). Statement II is correct because relative lowering equals the mole fraction of non-volatile solute \(B\). Statement III is false because the lowering is \(p_A^\circ-p_A=x_Bp_A^\circ\), not \(x_Ap_A^\circ\). The solvent mole fraction gives the remaining vapour pressure fraction. The solute mole fraction gives the lowered fraction.
216. In an ideal solution, \(6.0\,g\) of a non-volatile solute is dissolved in \(90\,g\) of water, giving a relative lowering of vapour pressure of \(0.02\). The approximate molar mass of the solute is:
ⓐ. \(30\,g\,mol^{-1}\)
ⓑ. \(60\,g\,mol^{-1}\)
ⓒ. \(120\,g\,mol^{-1}\)
ⓓ. \(300\,g\,mol^{-1}\)
Correct Answer: \(60\,g\,mol^{-1}\)
Explanation: \( \textbf{Moles of water:} \)
\[
n_A=\frac{90\,g}{18\,g\,mol^{-1}}=5.0\,mol
\]
\( \textbf{Use relative lowering:} \)
\[
x_B=\frac{p_A^\circ-p_A}{p_A^\circ}=0.02
\]
\( \textbf{Apply the exact mole-fraction relation:} \)
\[
0.02=\frac{n_B}{5.0+n_B}
\]
\( \textbf{Solve for solute moles:} \)
\[
0.02(5.0+n_B)=n_B
\]
\[
0.10=0.98n_B
\]
\[
n_B=0.102\,mol
\]
\( \textbf{Molar mass:} \)
\[
M_B=\frac{6.0\,g}{0.102\,mol}=58.8\,g\,mol^{-1}
\]
\( \textbf{Reporting to the nearest option:} \) \(58.8\,g\,mol^{-1}\approx60\,g\,mol^{-1}\).
\( \textbf{Final answer:} \) The approximate molar mass is \(60\,g\,mol^{-1}\). The dilute approximation \(x_B\approx\frac{n_B}{n_A}\) gives the same listed option.
217. Colligative properties of dilute solutions are mainly governed by:
ⓐ. the molar mass of each solute particle
ⓑ. the chemical identity of the dissolved solute
ⓒ. the number of dissolved solute particles
ⓓ. the size and shape of individual solute particles
Correct Answer: the number of dissolved solute particles
Explanation: Colligative properties depend primarily on the number of solute particles present in a given amount of solvent. They do not depend on the chemical identity of the solute when the solute behaves ideally and does not associate or dissociate unexpectedly. Relative lowering of vapour pressure is one such property because it depends on solute mole fraction. Elevation in boiling point, depression in freezing point, and osmotic pressure also follow the particle-count idea. This particle basis is why electrolytes may show abnormal values if they dissociate into more particles.
218. The set containing only colligative properties is:
ⓐ. vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure
ⓑ. solution density, coefficient of viscosity, surface tension, and refractive index of the liquid
ⓒ. optical rotation, electrical conductivity, solution colour, and boiling point of the pure solvent
ⓓ. molar mass, molecular formula, oxidation state, and atomic number of the dissolved solute
Correct Answer: vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure
Explanation: Colligative properties are solution properties controlled mainly by the number of solute particles. The four common colligative properties are relative lowering of vapour pressure, elevation in boiling point, depression in freezing point, and osmotic pressure. Density and viscosity are physical properties but are not classified as colligative properties in this context. Colour and odour depend strongly on chemical identity. Molar mass and molecular formula describe substances rather than colligative behaviour of solutions.
219. Two dilute aqueous solutions contain equal numbers of dissolved solute particles in equal amounts of water. If both solutes are non-volatile and nonelectrolytes, their boiling-point elevations and freezing-point depressions should be:
ⓐ. different because the solutes have different molar masses
ⓑ. zero because neither solute dissociates into ions
ⓒ. similar because the particle numbers are equal
ⓓ. larger for the solute with the greater molar mass
Correct Answer: similar because the particle numbers are equal
Explanation: For ideal dilute solutions of non-volatile nonelectrolytes, boiling-point elevation and freezing-point depression depend on the number of dissolved particles. If two solutions contain equal numbers of solute particles in equal amounts of solvent, these molality-based temperature changes should be similar. Different molar masses do not alter the result once the actual particle numbers are specified. Nonelectrolyte behaviour means each solute molecule remains one particle; it does not make the colligative effect zero. The comparison assumes no association, dissociation, or chemical reaction changes the particle count.
220. A solution shows larger elevation in boiling point than expected from the formula using the molecular solute amount. A likely reason is:
ⓐ. the solute dissociates to produce more particles
ⓑ. the solute becomes completely non-existent
ⓒ. the solvent mole fraction becomes greater than \(1\)
ⓓ. the solution has no vapour pressure at all
Correct Answer: the solute dissociates to produce more particles
Explanation: Colligative properties depend on the number of solute particles present in solution. If a solute dissociates, one formula unit can produce two or more particles. This increases the effective particle count compared with the calculated molecular amount. A higher particle count gives a larger colligative effect, such as greater boiling point elevation. This is one reason electrolytes can show abnormal molar mass values in colligative property calculations.
