Solutions Test

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Solutions Test - 3

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Question 1
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Only One Option Correct Type
This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct
An unopened soda has an aqueous concentration of C02 at 25 °C equal to 0.0506 molal. Thus, pressure of C0₂ gas in the can is (K_H = 0.034 mol/kg bar)

A
0.671 bar

B
1.49 bar

C
1.20 bar

D
1.71 bar

Solution
Question 2
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CO(g)is dissolved in H₂ Oat 25 °C and 0.010 atm. Henry ’s law constant for this system is 5.80 x10⁴ atm. Thus, mole fraction of CO(g)is

A
1.72 x 10^-7

B
8.28 x 10^-7

C
0.99

D
0.01

Solution

The mole fraction of a gas in solution is related to partial pressure of the gas above the solution by Henry ’s law.
Question 3
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H₂ S gas is used in qualitative analysis of inorganic cations. Its solubility in water at STP is 0.195 mol kg^-1 . Thus, Henry ’s law constant (in atm molar^-1 ) for H₂ S is

A
0.195

B
5.128

C
3.897 x 10³

D
2.565 x 10^-4

Solution
Question 4
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Variation of K_h (Henry ’s law constant) with temperature T is shown by following graphs I - IV.

Correct representation is

A
I

B
II

C
III

D
IV

Solution

Partial pressure
If temperature is lowered, solubility of gases increases at a given pressure. Thus, KH increases with increase in temperature. It is due to this reason that aquatic species are more comfortable in cold water rather than in warm water.
Question 5
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Concentration of C0₂ (in mole fraction) in fat when partial pressure of C0₂ is 55 kPa at 25 °C, is (Henry ’s law constant of C0₂ = 8.6 x 10⁴ torr)

A
1.6 x 10^-4

B
4.8 x 10^-3

C
8.6 x 10^-4

D
2.4 x 10^-3

Solution

Note: K_h (Henry ’s law constant) is in pressure unit, hence we use relation, Concentration xK_H = Pressure
Question 6
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What is the concentration of 0₂ in a fresh water stream in equilibrium with air at 25 °C and 1.0 bar? Given, K_H (Henry ’s law constant) of 0₂ = 1.3 x 10^-3 mol/kg bar at 25 °C.

A
8.736 x 10^-3 g/kg

B
4.16 x 10^-2 g/kg

C
24.04 g/kg

D
114.5 g/kg

Solution

In this case, K_H has been given in the unit of mol/kg bar. Thus, solubility of 0₂ (or concentration) = K_H P_gas
Question 7
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Following data has been given for C0₂ for the concentration in H₂ 0.
If solution of C0₂ in H₂ 0 is heated from 273 to 333 K, pressure ( p₂ ) needed to keep C0₂ in the solution is

A
0.108 atm

B
0.805 atm

C
0.212 atm

D
0.468 atm

Solution
Question 8
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Relation between the volume of gas (2) that dissolves in a fixed volume of solvent(1) and the partial pressure of gas (2) is (n_t = total moles, K₁ and K₂ are Henry ’s constants)

A
n_t RT/K₂

B
n₂ RT/K₂ p₂

C
n₁ RT/n₂

D
None of these

Solution
Question 9
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Which is correct about Henry 's law

A
The gas in contact with the liquid should behave as an ideal gas

B
There should not be any chemical interaction between the gas and liquid

C
The pressure applied should be high

D
All of these

Solution

Henry 's law states that at a constant temperature, the solubility of a gas is directly proportional to the pressure of a gas.
In other words, The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
p = K_H x
K_H is the henry 's law constant.
Question 10
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Henry ’s law constant for N₂ at 293 K is 76.48 kbar. N₂ exerts a partial pressure of 0.987 bar. If N₂ gas is bubbled through water at 293 K, then number of millimoles of N₂ that will dissolve in 1 L of water is

A
0.0716

B
1.29 x 10^-5

C
1.29 x 10^-2

D
5.55

Solution

Henry ’s law constant is in the unit of pressure, hence we use relation
Let number of moles of nitrogen = n
(Since, is very small)
Question 11
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A handbook lists the solubility of carbon monoxide in water at 0 °C and 1 atm pressure as 0.0354 mL CO per mL of H₂ 0. What should be the pressure of CO(g) above the solution to obtain 0.010 M CO solution?

A
0.633 atm

B
6.33 atm

C
1 atm

D
0.01 atm

Solution

Volume of CO = 0 .0354 mL = 0 .0354 x 10^- 3 L
p = 1 atm, T = 273 K
Question 12
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One or More than One Options Correct Type
This section contains 2 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct .

Select the correct statement(s).

A
Gases which form ions in the solvent are soluble in that solvent

B
Gases which are liquefied easily are more soluble in water than the other gases which are not liquefied easily

C
Higher the value of K_H at a given pressure, lower is the solubility of the gas in the liquid

D
All are correct

Solution

NH₃ , C0₂ , S0₃ ...... gases form ions in water, hence these gases are soluble in water. 0₂ , N₂ .......do not form ions in water hence, insoluble thus correct.

(b) NH₃ and C0₂ are liquefied easily hence, soluble in water thus, correct.

Higher the value of K_H , smaller the solubility (x) of the gas at a given pressure thus, correct.
Question 13
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Select the correct statement(s).

A
K_H (Henry ’s law constant) is a function of nature of gases

B
Higher the value of K_H at a given pressure, the lower is the solubility of the gas in the liquid

C
K_H increases with increase in temperature

D
All are correct

Solution

(a) At constant temperature, different gases have different KH values. Thus, KH is a function of nature of gases.
Thus, correct.
(b) p (partial pressure) =
If p is constant and K_H value increases then (solubility) decreases. Thus, correct.
(c) K_H increases with increase in temperature thus, correct.
(d) K_H is independent of pressure thus, incorrect.
Question 14
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Statement Type
This section is based on Statement I and Statement II. Select the correct anser from the codes given below
Statement I Solubility of gases in liquids decreases with rise in temperature.
Statement II CO₂ (g) + H₂ 0 ( / ) ----- »H₂ CO₃ (aq ); ΔH = - ve

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

(b) Dissolution of gas can be considered as the reverse of vaporisation, i.e. condensation in which heat is evolved. It is thus, exothermic ΔH <0. By Le-Chatelier ’s principle, increase in temperature shifts the dissolution equilibrium in the direction of form ation of C0₂ (g).
Thus, increase in temperature decreases solubility. Thus, both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I.
Question 15
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Statement I: People living at high altitudes have symptoms of a condition known as anoxia.
Statement II: At high altitudes, the partial pressure of oxygen is less than that at the ground level.

A
Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

At high altitudes, the partial pressure of oxygen is less than that at the ground level. This leads to the low concentration of O₂ in the blood and tissues of the people residing in hill areas. This produces weakness and loss of memory.
Symptoms of this conditions is known as ANOXIA. Thus, Statements I are correct but Statement II is the correct explanation of Statement I.