Solutions and Phase Equilibria Questions and Answers

Q: August 21, 1986, a cloud of CO2 gas suddenly erupted from a lake in cameroon. Which of the following account for this incident ?(I) Over the years CO2 has saturated in the upper layers of lake water.(II) Over the years large quantities CO2 has dissolved in the bottom layers of water.(III) Heavy winds could have overturned the lake water.(IV) This event is a natural phenomenon explained by Raoult's law.

The incident is accounted for by (II) and (III).The 1986 disaster at Lake Nyos in Cameroon was a limnic eruption. Volcanic CO2 seeping into the bottom of the deep, stratified lake dissolved in the cold, high-pressure bottom water over many years (II); by Henry's law, high pressure at depth allows large amounts of gas to stay dissolved. A disturbance, possibly wind, a landslide or cooling of surface water, overturned the layers (III). As deep water rose, the pressure fell, the dissolved CO2 came out of solution explosively, and the dense gas cloud flowed down the valleys, killing about 1,700 people.The gas accumulated in the bottom layers, not the upper ones, so (I) is wrong, and the relevant law is Henry's law for gas solubility, not Raoult's law for vapour pressure, so (IV) is wrong.

Q: In the following phase diagram, the process corresponding to sublimation is :

Sublimation corresponds to I → III.In this pressure-temperature diagram region I, at low temperature beside the steep fusion line, is the solid; region II, above the vaporisation curve, is the liquid; and region III, at high temperature and low pressure, is the gas. The three lines meet at the triple point.Sublimation is the direct change from solid to gas, crossing the short sublimation line below the triple point, so it is I → III. I → II is melting, II → III is vaporisation, and III → II is condensation.

Q: Both the vapour pressure at 25°C and normal boiling point of toluene are higher than that of water, which of the following explains these observations ?

The explanation is that toluene has a lower heat of vaporization than water.The Clausius-Clapeyron equation links how fast vapour pressure rises with temperature to the enthalpy of vaporization:\[ \ln P = -\frac{\Delta H_{vap}}{RT} + C \]A small ΔHvap means a shallow curve. Toluene starts with a higher vapour pressure than water at 25 °C, yet reaches 1 atm (its boiling point) at a higher temperature, 110.6 °C against 100 °C. Its vapour pressure must therefore rise more slowly with temperature, which requires the smaller ΔHvap: about 33 kJ/mol for toluene against about 41 kJ/mol for hydrogen-bonded water.The first option states the opposite of the usual trend (a higher vapour pressure normally goes with a lower boiling point). A higher molar mass would lower the vapour pressure, not raise it, and the density of the vapour does not govern evaporation.

Q: Consider an ideal solution of two liquids A and B. If partial vapour pressures of both liquids are same, then total pressure of the solution is :

The total pressure is 2PA°PB°/(PA° + PB°).By Raoult's law, pA = xAPA° and pB = xBPB°. Setting them equal, with xB = 1 − xA:\[ \begin{aligned} x_{A}P_{A}^{o} &= (1-x_{A})P_{B}^{o} \\ x_{A} &= \frac{P_{B}^{o}}{P_{A}^{o}+P_{B}^{o}} \\ P &= 2x_{A}P_{A}^{o} = \frac{2P_{A}^{o}P_{B}^{o}}{P_{A}^{o}+P_{B}^{o}} \end{aligned} \]The first option is the partial pressure of one component, half the total. The last two are ratios of pressures, dimensionless numbers, so they cannot be a pressure at all.

Q: The normal boiling point of a liquid is 200 K. If the liquid behaves as ideal liquid at 76 mm, then its boiling point will be close to : (Given : \(\dfrac{\Delta H}{R} \approx 2303\,\text{K}\))

The boiling point at 76 mm is close to 167 K.Use the Clausius-Clapeyron equation between the normal boiling point (760 mm, 200 K) and the new pressure (76 mm, T2):\[ \begin{aligned} \ln\frac{p_{2}}{p_{1}} &= -\frac{\Delta H}{R} \\ &\quad \times \left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right) \end{aligned} \]\[ \begin{aligned} \ln 0.1 &= -2.303 \\ -2.303 &= -2303\left(\frac{1}{T_{2}}-\frac{1}{200}\right) \\ \frac{1}{T_{2}} &= 0.005 + 0.001 = 0.006 \\ T_{2} &= 166.7\ \text{K} \end{aligned} \]Lower pressure must lower the boiling point, so 200 K (unchanged) and 440 K (higher) are impossible, and 20 K is what you get by scaling the temperature in proportion to the pressure, which the equation does not allow.

Q: According to the phase diagram given below, the composition of the layer 'a' relative to the layer 'b' at point 'c' is :

The ratio is bc/ac (option B), by the lever rule.On a temperature-composition diagram, when the overall composition c lies in the two-phase region, the mixture splits into the two layers whose compositions are a and b. The amounts of the two layers are in inverse ratio to the distances of their compositions from c:\[ \frac{n_a}{n_b} = \frac{bc}{ac} \]The nearer the point is to a, the more of layer a there is, so the quantity of layer a relative to layer b is the longer lever bc over the shorter ac. The stem words this as the composition of layer a relative to layer b, and the key takes it as this ratio of amounts. ac/bc is the inverse, and ab, the distance between the two layers, does not enter the ratio, so C and D are wrong. The diagram is hand drawn and marks a and b on the composition axis, so the distances are read as composition differences.

41.

August 21, 1986, a cloud of CO₂ gas suddenly erupted from a lake in cameroon. Which of the following account for this incident ?

(I) Over the years CO₂ has saturated in the upper layers of lake water.
(II) Over the years large quantities CO₂ has dissolved in the bottom layers of water.
(III) Heavy winds could have overturned the lake water.
(IV) This event is a natural phenomenon explained by Raoult's law.

A.(II), (III) and (IV)
B.(I), (III)
C.(I), (II) and (IV)
D.(II), (III)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The incident is accounted for by (II) and (III).

The 1986 disaster at Lake Nyos in Cameroon was a limnic eruption. Volcanic CO₂ seeping into the bottom of the deep, stratified lake dissolved in the cold, high-pressure bottom water over many years (II); by Henry's law, high pressure at depth allows large amounts of gas to stay dissolved. A disturbance, possibly wind, a landslide or cooling of surface water, overturned the layers (III). As deep water rose, the pressure fell, the dissolved CO₂ came out of solution explosively, and the dense gas cloud flowed down the valleys, killing about 1,700 people.

The gas accumulated in the bottom layers, not the upper ones, so (I) is wrong, and the relevant law is Henry's law for gas solubility, not Raoult's law for vapour pressure, so (IV) is wrong.

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42.

In the following phase diagram, the process corresponding to sublimation is :

Pressure-temperature phase diagram with three regions: I at low temperature to the left of a steep line, II above a curve rising to the right, and III at high temperature and low pressure below the curve; the lines meet at a triple point

A.I → II
B.I → III
C.II → III
D.III → II

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

Sublimation corresponds to I → III.

In this pressure-temperature diagram region I, at low temperature beside the steep fusion line, is the solid; region II, above the vaporisation curve, is the liquid; and region III, at high temperature and low pressure, is the gas. The three lines meet at the triple point.

Sublimation is the direct change from solid to gas, crossing the short sublimation line below the triple point, so it is I → III. I → II is melting, II → III is vaporisation, and III → II is condensation.

Discuss in Forum

43.

Both the vapour pressure at 25°C and normal boiling point of toluene are higher than that of water, which of the following explains these observations ?

A.Liquids with higher vapour pressures typically have higher boiling points
B.Toluene has a higher molar mass than water
C.Toluene has a lower heat of vaporization than water
D.The density of toluene vapour is greater than that of water vapour

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The explanation is that toluene has a lower heat of vaporization than water.

The Clausius-Clapeyron equation links how fast vapour pressure rises with temperature to the enthalpy of vaporization:

\[ \ln P = -\frac{\Delta H_{vap}}{RT} + C \]

A small ΔH_vap means a shallow curve. Toluene starts with a higher vapour pressure than water at 25 °C, yet reaches 1 atm (its boiling point) at a higher temperature, 110.6 °C against 100 °C. Its vapour pressure must therefore rise more slowly with temperature, which requires the smaller ΔH_vap: about 33 kJ/mol for toluene against about 41 kJ/mol for hydrogen-bonded water.

The first option states the opposite of the usual trend (a higher vapour pressure normally goes with a lower boiling point). A higher molar mass would lower the vapour pressure, not raise it, and the density of the vapour does not govern evaporation.

Discuss in Forum

44.

Consider an ideal solution of two liquids A and B. If partial vapour pressures of both liquids are same, then total pressure of the solution is :

A.\(\frac{P_{A}^{o}P_{B}^{o}}{P_{A}^{o}+P_{B}^{o}}\)
B.\(\frac{2P_{A}^{o}P_{B}^{o}}{P_{A}^{o}+P_{B}^{o}}\)
C.\(\frac{P_{A}^{o}}{P_{A}^{o}+P_{B}^{o}}\)
D.\(\frac{2P_{B}^{o}}{P_{A}^{o}+P_{B}^{o}}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The total pressure is 2P_A°P_B°/(P_A° + P_B°).

By Raoult's law, p_A = x_AP_A° and p_B = x_BP_B°. Setting them equal, with x_B = 1 − x_A:

\[ \begin{aligned} x_{A}P_{A}^{o} &= (1-x_{A})P_{B}^{o} \\ x_{A} &= \frac{P_{B}^{o}}{P_{A}^{o}+P_{B}^{o}} \\ P &= 2x_{A}P_{A}^{o} = \frac{2P_{A}^{o}P_{B}^{o}}{P_{A}^{o}+P_{B}^{o}} \end{aligned} \]

The first option is the partial pressure of one component, half the total. The last two are ratios of pressures, dimensionless numbers, so they cannot be a pressure at all.

Discuss in Forum

45.

The normal boiling point of a liquid is 200 K. If the liquid behaves as ideal liquid at 76 mm, then its boiling point will be close to : (Given : \(\dfrac{\Delta H}{R} \approx 2303\,\text{K}\))

A.167 K
B.20 K
C.200 K
D.440 K

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The boiling point at 76 mm is close to 167 K.

Use the Clausius-Clapeyron equation between the normal boiling point (760 mm, 200 K) and the new pressure (76 mm, T₂):

\[ \begin{aligned} \ln\frac{p_{2}}{p_{1}} &= -\frac{\Delta H}{R} \\ &\quad \times \left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right) \end{aligned} \]

\[ \begin{aligned} \ln 0.1 &= -2.303 \\ -2.303 &= -2303\left(\frac{1}{T_{2}}-\frac{1}{200}\right) \\ \frac{1}{T_{2}} &= 0.005 + 0.001 = 0.006 \\ T_{2} &= 166.7\ \text{K} \end{aligned} \]

Lower pressure must lower the boiling point, so 200 K (unchanged) and 440 K (higher) are impossible, and 20 K is what you get by scaling the temperature in proportion to the pressure, which the equation does not allow.

Discuss in Forum

46.

According to the phase diagram given below, the composition of the layer 'a' relative to the layer 'b' at point 'c' is :

A hand-drawn temperature against x2 diagram with a dome-shaped curve; a and b are marked on the composition axis at the two feet of the dome and the point c lies inside the dome

A.\(\frac{ac}{bc}\)
B.\(\frac{bc}{ac}\)
C.\(\frac{ac}{ab}\)
D.\(\frac{bc}{ab}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The ratio is bc/ac (option B), by the lever rule.

On a temperature-composition diagram, when the overall composition c lies in the two-phase region, the mixture splits into the two layers whose compositions are a and b. The amounts of the two layers are in inverse ratio to the distances of their compositions from c:

\[ \frac{n_a}{n_b} = \frac{bc}{ac} \]

The nearer the point is to a, the more of layer a there is, so the quantity of layer a relative to layer b is the longer lever bc over the shorter ac. The stem words this as the composition of layer a relative to layer b, and the key takes it as this ratio of amounts. ac/bc is the inverse, and ab, the distance between the two layers, does not enter the ratio, so C and D are wrong. The diagram is hand drawn and marks a and b on the composition axis, so the distances are read as composition differences.

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