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Test: Henry's Law - NEET MCQ


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29 Questions MCQ Test - Test: Henry's Law

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Test: Henry's Law - 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 (KH = 0.034 mol/kg bar) 

Detailed Solution for Test: Henry's Law - Question 1

Test: Henry's Law - Question 2
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CO(g)is dissolved in H2Oat 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

Detailed Solution for Test: Henry's Law - Question 2

The mole fraction of a gas in solution is related to partial pressure of the gas above the solution by Henry’s law.

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Test: Henry's Law - Question 3
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 H2S 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 H2S is

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Test: Henry's Law - Question 4
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 Variation of Kh (Henry’s law constant) with temperature T is shown by following graphs I - IV.


Correct representation is
Detailed Solution for Test: Henry's Law - Question 4

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.

Test: Henry's Law - 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)
Detailed Solution for Test: Henry's Law - Question 5

Note: Kh (Henry’s law constant) is in pressure unit, hence we use relation, Concentration xKH = Pressure

Test: Henry's Law - 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, KH (Henry’s law constant) of 02 = 1.3 x 10-3 mol/kg bar at 25° C.
Detailed Solution for Test: Henry's Law - Question 6

In this case, KH has been given in the unit of mol/kg bar. Thus, solubility of 02 (or concentration) = KH Pgas

Test: Henry's Law - Question 7
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Following data has been given for C02 for the concentration in H20.

If solution of C02 in H20 is heated from 273 to 333 K, pressure ( p2 ) needed to keep C02 in the solution is
Detailed Solution for Test: Henry's Law - Question 7

Test: Henry's Law - 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 (nt = total moles, K1and K2 are Henry’s constants)
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Test: Henry's Law - Question 9
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Which is correct about Henry's law
Detailed Solution for Test: Henry's Law - Question 9

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 = KHx
KH is the henry's law constant. 

Test: Henry's Law - 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
Detailed Solution for Test: Henry's Law - Question 10

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)

Test: Henry's Law - 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 H20. What should be the pressure of CO(g) above the solution to obtain 0.010 M CO solution?
Detailed Solution for Test: Henry's Law - Question 11

Volume of CO = 0 .0354 mL = 0 .0354 x 10- 3 L

p = 1 atm, T = 273 K

*Multiple options can be correct
Test: Henry's Law - 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).
Detailed Solution for Test: Henry's Law - Question 12

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 KH, smaller the solubility (x) of the gas at a given pressure thus, correct.

*Multiple options can be correct
Test: Henry's Law - Question 13
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Select the correct statement(s).
Detailed Solution for Test: Henry's Law - Question 13

(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 KH value increases then     (solubility) decreases. Thus, correct.
(c) KH increases with increase in temperature thus, correct.
(d) KH is independent of pressure thus, incorrect.

*Multiple options can be correct
Test: Henry's Law - 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) + H20 ( / ) ----- » H2CO3(aq ); ΔH = - ve
Detailed Solution for Test: Henry's Law - Question 14

(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.

Test: Henry's Law - 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.
Detailed Solution for Test: Henry's Law - Question 15

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.

Test: Henry's Law - Question 16
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Assertion: Soft drinks and soda water bottles are sealed under high pressure.
Reason: High pressure maintains the taste and texture of soft drinks.
Detailed Solution for Test: Henry's Law - Question 16

The correct option is A

Assertion is true, but the reason is false.
Explanation for Assertion:-

  • The soda drinks and soda waters are sealed under high pressure.
  • This increases, the solubility of the Carbon dioxide(CO2).

Explanation for Reason:-

  • According to Henry's law, the partial pressure of the gas is directly proportional to the solubility of the gas.
  • That is more the pressure of the gas more will be the solubility of the gas.
  • Thus, due to this, to increase the solubility of (CO2), the bottles are sealed under high pressure.
  • It is not done to improve or maintain the taste or texture of the sodas or drinks.

Therefore, option (A) Assertion is true, but the reason is false is the correct answer.

Test: Henry's Law - Question 17
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Statement I Scuba (instrum ent used for breathing) divers must cope with high concentration of dissolved gases while breathing at high pressure under water.

Statement II Formation of bubbles in blood blocks capillaries and creates bends which are painful and dangerous to life.
Detailed Solution for Test: Henry's Law - Question 17

(a) When divers breathe under high pressure, solubility of atmospheric gases in blood increases. When divers come at the surface, pressure gradually decreases. Gases (N2 and 02) are released in the form of bubbles which blocks capillaries and creates a medical symptoms called bends which are painful and dangerous. To avoid bends as well as toxic effects of high concentration of nitrogen in the blood, the tanks used by divers are filled with air diluted with helium (11.7% He, 56.2% N2 and 32.1% Og). Thus, both Statements I and II are correct and Statement II is the correct explanation of Statement I.

Test: Henry's Law - Question 18
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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
 

The major components of air are 02 and N2 with approximate proportion of 21% and 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K, KH(N2 ) = 6.5 1 x 107 mm, KH( 02 ) = 3.30 x 107 mm Match the parameters in Column I with their values in Column II and select the answer from the codes given below.

Codes:
Detailed Solution for Test: Henry's Law - Question 18

Test: Henry's Law - Question 19
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This section contains a graph describing theory, experiments, data, etc. Two 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

Q. Henry’s law constant for HCI gas is (in torr)
Detailed Solution for Test: Henry's Law - Question 19

Based on given figure, pressure of  

Test: Henry's Law - Question 20
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The value for Henry’s constant for helium, hydrogen, nitrogen and oxygen at 293 K are 144.97 kbar, 69.16 kbar, 76.48 kbar and 34.86 kbar respectively. Which of the gas will be having maximum solubility?
Detailed Solution for Test: Henry's Law - Question 20

The correct answer is option B.

At a constant pressure KH is inversely proportional to x therefore higher the KH lower the solubility.
The increasing solubility order is:
He < N2 < H2 < O2
 

Test: Henry's Law - Question 21
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A solution in which no more solute can be dissolved at a given temperature and pressure is called a saturated solution. At saturated solution stage equilibrium gets established among which two processes?
Detailed Solution for Test: Henry's Law - Question 21

At saturated solution stage equilibrium gets established among dissolution and crystallisation.

Test: Henry's Law - Question 22
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A solution in which no more solute can be dissolved at the given temperature and pressure is called a
Detailed Solution for Test: Henry's Law - Question 22

The correct answer is option  D
In a saturated solution, more solute cannot be dissolved at a given temperature.
This is because, the solute dissolves in a solvent because of space between particles of solvent but on continuous addition of solute, the space between the solvent particles gets fulfilled. Thus no more solute particle can dissolve in a solvent.

Test: Henry's Law - Question 23
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Which expression correctly represents Henry’s Law?
Detailed Solution for Test: Henry's Law - Question 23

The correct answer is option A
Henry’s law is a gas law which states that the amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid when the temperature is kept constant. The constant of proportionality for this relationship is called Henry’s law constant (usually denoted by ‘kH‘). The mathematical formula of Henry’s law is given by:
P ∝ C (or) P = kH.C
Where,
 

  • ‘P’ denotes the partial pressure of the gas in the atmosphere above the liquid.

  • ‘C’ denotes the concentration of the dissolved gas.

  • ‘kH’ is the Henry’s law constant of the gas.

This law was formulated in the early 19th century by the English chemist William Henry. It can be noted that Henry's law constant can be expressed in two different ways. If the constant is defined in terms of solubility/pressure, it is referred to as the Henry’s law solubility constant (denoted by ‘H’). On the other hand, if the proportionality constant is defined in terms of pressure/solubility, it is called the Henry’s law volatility constant (denoted by ‘kH’).

Test: Henry's Law - Question 24
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Value of Henry’s constant KH is ______.
Detailed Solution for Test: Henry's Law - Question 24

The correct answer is option A

According to Henry's law, solubility of a gas in a liquid is directly proportional to pressure of gas.
PαX
P = KHX
P = Partial pressure of gas
X = solubility of gas in liquid
KH = Henry's constant
KH depends only on the nature of gas, nature of liquid and temperature (T). As temperature increases, 'KH' increases and 'X' decreases.

Test: Henry's Law - Question 25
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Some solute particles in solution collide with other solid solute particles present and get separated out of solution. This process is known as
Detailed Solution for Test: Henry's Law - Question 25
Crystallization Process

  • Definition: Crystallization is the process in which solute particles in a solution collide with other solid solute particles present and get separated out of the solution.

  • Explanation: During the process of crystallization, the solute particles come together and form a crystalline solid structure, separating from the liquid solvent.

  • Factors affecting crystallization: Various factors such as temperature, concentration of the solution, and presence of impurities can affect the process of crystallization.

  • Applications: Crystallization is used in various industries such as pharmaceuticals, food processing, and chemical manufacturing to purify substances and obtain pure crystals.

  • Steps involved: The process of crystallization typically involves heating the solution to dissolve the solute, cooling the solution to allow crystals to form, and then isolating the crystals from the remaining liquid.

Test: Henry's Law - Question 26
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The value for Henry’s constant for argon, carbon dioxide, methane and vinyl chloride at 298 K are 40.3 kbar, 1.67 kbar, 0.413 kbar and 0.611 kbar respectively. Which of the gas will be having least solubility?
Detailed Solution for Test: Henry's Law - Question 26

The correct answer is Option A
The higher the value of KH, the lower is the solubility of the gas in the liquid. Hence the order of increasing solubility of the gases will be Ar <  CO2<CH4< HCHO

Test: Henry's Law - Question 27
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Which of the following gases will be least soluble in water?
Detailed Solution for Test: Henry's Law - Question 27
Least Soluble Gas in Water

  • Hydrogen chloride: Hydrogen chloride is highly soluble in water due to its polar nature.

  • Sulphur dioxide: Sulphur dioxide is soluble in water, forming sulfurous acid.

  • Ammonia: Ammonia is also soluble in water, forming ammonium hydroxide.

  • Nitrogen: Nitrogen is the least soluble gas in water among the options provided. Nitrogen is a nonpolar gas and does not readily dissolve in water. Its solubility in water is very low compared to the other gases listed.


Therefore, nitrogen will be the least soluble gas in water among the given options.

Test: Henry's Law - Question 28
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According to Henry’s Law at a constant temperature the solubility of gas in a liquid is directly proportional to the
Detailed Solution for Test: Henry's Law - Question 28

The correct answer is option D
Henry's law states that at a constant temperature, the solubility of a gas is directly proportional to the pressure of the 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=KH​ x
Here, KH​ is Henry's law constant.
Hence, the given statement is Henry's law.

Test: Henry's Law - Question 29
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When a large amount of solute dissolves in a given amount of solvent it forms
Detailed Solution for Test: Henry's Law - Question 29
Concentrated Solution

  • Definition: A concentrated solution is formed when a large amount of solute is dissolved in a given amount of solvent.

  • Characteristics:

    • It contains a high amount of solute compared to the solvent.

    • The solution is stable at a specific temperature.

    • It may reach a point where no more solute can be dissolved, leading to a saturated solution.



  • Examples:

    • Sugar dissolved in water beyond its saturation point to form a concentrated sugar solution.

    • Adding more salt to water until it can no longer dissolve, resulting in a concentrated salt solution.



  • Uses:

    • Concentrated solutions are commonly used in laboratory experiments and industrial processes.

    • They are essential in creating specific chemical reactions or for producing concentrated products.



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