Solutions Test

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Solutions Test - 8

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Weekly Quiz Competition

Question 1
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Which combination of following terms is matched correctly?
I. Vapour pressure

II. Intermolecular forces

III. Latent heat of vaporisation

A
a

B
b

C
c

D
d

Solution

(a) Liquid Vapour

(a) If vapour pressure is high, it means equilibrium is in forward side. This is attained when intermolecular forces holding the molecules are of weak nature. Low heat is required to convert liquid into vapour. Thus, (a) is correct.

(b) Thus, incorrect.

(c) When vapour pressure is low, it means intermolecular forces are strong hence (c) is incorrect.

(d) When vapour pressure is low, intermolecular forces are strong, hence large heat is required thus, (d) is incorrect,
Question 2
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The solubility of urea in methanol is 17 g urea/100 mL methanol. Density of methanol is 0.792 g/mL at 293 K at which vapour pressure of methanol is 96.0 mm of Hg. Thus, vapour pressure of saturated solution of urea in methanol is

A
86.11 mm hg

B
90.27 mm hg

C
88.13 mm hg

D
80.12 mm hg

Solution
Question 3
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One or More than One Options Correct Type
This section contains 3 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is/are [JEE Advanced 2013]

A
(ΔS)_system >0

B
(ΔS)_surrounding =0

C
ΔH =0

D
All are correct

Solution

When binary mixture is ideal, then by Raoult ’s law

ΔV = 0, ΔH_mix = 0, no enthalpy change

Mixing is spontaneous, thus ΔG <0 By thermodynamic property,

Thus, on mixing, entropy increases
Question 4
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Total vapour pressure of mixture of 1 mole of a volatile component A (p °A = 100 mm Hg) and 3 moles of volatile component B(p °B = 60 mm Hg) is 75 mm Hg. For such case,

A
Positive deviation from Raoult ’s law

B
Boiling point is lowered

C
Force of attraction between A and B is less than between A and A,A and B

D
All are correct

Solution

But, actual vapour pressure = 75 mm Hg. Vapour pressure is increased. Force of attraction between A and B is smaller than between A and A and between B and B. Thus, more vaporisation due to decrease in boiling point, a case of positive deviation.
Question 5
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At 35 °C, the vapour pressure of carbon disulphide (CS₂ )is 512 mm Hg and that of acetone (CH₃ COCH₃ )is 344 mm Hg. A solution of CS₂ and CH₃ COCH₃ in which mole fraction of CS₂ is 0.25 has a total vapour pressure of 600 mm Hg. Which of the following statement(s) is/are correct?

A
A mixture of 100 mL of CH₃ COCH₃ and 100 mL of CS₂ has a volume of 200 mL

B
When CH₃ COCH₃ and CS₂ are mixed at 35 °C, heat must be absorbed in order to produce a solution at 35 °C

C
Process of mixing is exothermic

D
Entropy of mixing is zero

Solution

(A for acetone, C for carbon disulphide)

But, actual p_total = 600 mm Hg

Thus, there is positive deviation. Boiling point is lowered.
Thus, ΔH = + ve (endothermic), i.e. heat is absorbed.
Question 6
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Comprehension Type
This section contains 2 paragraphs , each describing theory, experiments, data, etc. four questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

Passage I
At 300 K, vapour pressure of pure liquid A = 575torr
and that of pure liquid B = 390torr
In binary mixture of liquids A and B with ideal behaviour, mole fraction of A in vapour phase =0.35
Q. Binary mixture contains

A
27 mol% A

B
73 mol % A

C
50 mol% A

D
47 mol% A

Solution

Let moles of A = n_A and moles of B = n_B

Mole fraction of A in vapour state

Thus, A is 27 mole %.
Total vapour pressure of binary mixture
Question 7
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Passage I

At 300 K, vapour pressure of pure liquid A = 575torr and that of pure liquid B = 390torr In binary mixture of liquids A and B with ideal behaviour, mole fraction of A in vapour phase =0.35

Q. Total vapour pressure of the binary mixture is

A
440 torr

B
455 torr

C
140 torr

D
483 torr

Solution

Let moles of A = n_A and moles of B = n_B

Mole fraction of A in vapour state

Thus, A is 27 mole %.
Total vapour pressure of binary mixture
Question 8
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Passage II

At 40 °C, the vapour pressure (in torr) of methanol-ethanol solution is represented by

where, is the mole fraction of methanol.

Q. Difference of vapour pressures of volatile liquids (methanol-ethanol) in pure state at 40 °C is

A
135 torr

B
119 torr

C
254 torr

D
16 torr

Solution
Question 9
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Passage II

At 40 °C, the vapour pressure (in torr) of methanol-ethanol solution is represented by

where, is the mole fraction of methanol.

Q.

What is the mole fraction of methanol in vapour phase in a binary mixture with mole fraction of ethanol 0.50?

A
0.50

B
0.35

C
0.65

D
0.25

Solution

(C)
Question 10
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Matching List Type
Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

100 g of binary mixture of A and B contains 60% of A. Mixture is subjected to the following operation.

Original mixture I II containing vapours of A and B containing vapours of A and B

Match the content of Column I with their values in Column II and select the answer from the codes given below.

A
a

B
b

C
c

D
d

Solution