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CHXII02:SOLUTIONS

319385 Which mixture is used for respiration by deep sea divers?

1

N e + O_{2}

2

K r + O_{2}

3

A r + O_{2}

4

H e + O_{2}

MHTCET - 2019

CHXII02:SOLUTIONS

319386 Value of Henry’s constant, $K_{H}$

1 Increases with increase in temperature.

2 Decreases with increase in temperature.

3 Remains constant.

4 First increases then decreases.

Explanation:

Value of $K_{H}$ depends on temperature and nature of solvent.

CHXII02:SOLUTIONS

319387 Assertion :
The solubility of a gas in a liquid increases with increase of pressure.
Reason :
The solubility of a gas in a liquid is directly proportional to the pressure of the gas.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Henry's law states that the solubility of a gas in a liquid is directly proportional to the gas pressure above the liquid, expressed mathematically as $x = k P$ , where ' $x$ ' is solubility, ' $P$ ' is partial pressure, and ' $k$ ' is the Henry's law constant representing the proportionality between gas solubility and pressure. So option (1) is correct.

CHXII02:SOLUTIONS

319388 5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is approximately

1

1 \times 1 0^{- 6}

2

1 \times 1 0^{- 5}

3

1 \times 1 0^{- 3}

4

1 \times 1 0^{- 4}

Explanation:

Solubility of a gas $\propto$ mole fraction
$n_{N_{2}} = \frac{5 .5 \times 1 0^{- 3}}{28} = 1 .96 \times 1 0^{- 4} a t 1 a t m$
Now, $p \propto$ mole fraction then at 5 atm, the mole
fraction will increase 5 times.
Hence, $5 \times 1 .96 \times 1 0^{- 4} = 9 .8 \times 1 0^{- 4}$
$\approx 10 \times 1 0^{- 4} = 1 .0 \times 1 0^{- 3}$

KCET - 2014

CHXII02:SOLUTIONS

319385 Which mixture is used for respiration by deep sea divers?

1

N e + O_{2}

2

K r + O_{2}

3

A r + O_{2}

4

H e + O_{2}

MHTCET - 2019

CHXII02:SOLUTIONS

319386 Value of Henry’s constant, $K_{H}$

1 Increases with increase in temperature.

2 Decreases with increase in temperature.

3 Remains constant.

4 First increases then decreases.

Explanation:

Value of $K_{H}$ depends on temperature and nature of solvent.

CHXII02:SOLUTIONS

319387 Assertion :
The solubility of a gas in a liquid increases with increase of pressure.
Reason :
The solubility of a gas in a liquid is directly proportional to the pressure of the gas.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Henry's law states that the solubility of a gas in a liquid is directly proportional to the gas pressure above the liquid, expressed mathematically as $x = k P$ , where ' $x$ ' is solubility, ' $P$ ' is partial pressure, and ' $k$ ' is the Henry's law constant representing the proportionality between gas solubility and pressure. So option (1) is correct.

CHXII02:SOLUTIONS

319388 5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is approximately

1

1 \times 1 0^{- 6}

2

1 \times 1 0^{- 5}

3

1 \times 1 0^{- 3}

4

1 \times 1 0^{- 4}

Explanation:

Solubility of a gas $\propto$ mole fraction
$n_{N_{2}} = \frac{5 .5 \times 1 0^{- 3}}{28} = 1 .96 \times 1 0^{- 4} a t 1 a t m$
Now, $p \propto$ mole fraction then at 5 atm, the mole
fraction will increase 5 times.
Hence, $5 \times 1 .96 \times 1 0^{- 4} = 9 .8 \times 1 0^{- 4}$
$\approx 10 \times 1 0^{- 4} = 1 .0 \times 1 0^{- 3}$

KCET - 2014

CHXII02:SOLUTIONS

319385 Which mixture is used for respiration by deep sea divers?

1

N e + O_{2}

2

K r + O_{2}

3

A r + O_{2}

4

H e + O_{2}

MHTCET - 2019

CHXII02:SOLUTIONS

319386 Value of Henry’s constant, $K_{H}$

1 Increases with increase in temperature.

2 Decreases with increase in temperature.

3 Remains constant.

4 First increases then decreases.

Explanation:

Value of $K_{H}$ depends on temperature and nature of solvent.

CHXII02:SOLUTIONS

319387 Assertion :
The solubility of a gas in a liquid increases with increase of pressure.
Reason :
The solubility of a gas in a liquid is directly proportional to the pressure of the gas.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Henry's law states that the solubility of a gas in a liquid is directly proportional to the gas pressure above the liquid, expressed mathematically as $x = k P$ , where ' $x$ ' is solubility, ' $P$ ' is partial pressure, and ' $k$ ' is the Henry's law constant representing the proportionality between gas solubility and pressure. So option (1) is correct.

CHXII02:SOLUTIONS

319388 5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is approximately

1

1 \times 1 0^{- 6}

2

1 \times 1 0^{- 5}

3

1 \times 1 0^{- 3}

4

1 \times 1 0^{- 4}

Explanation:

Solubility of a gas $\propto$ mole fraction
$n_{N_{2}} = \frac{5 .5 \times 1 0^{- 3}}{28} = 1 .96 \times 1 0^{- 4} a t 1 a t m$
Now, $p \propto$ mole fraction then at 5 atm, the mole
fraction will increase 5 times.
Hence, $5 \times 1 .96 \times 1 0^{- 4} = 9 .8 \times 1 0^{- 4}$
$\approx 10 \times 1 0^{- 4} = 1 .0 \times 1 0^{- 3}$

KCET - 2014

CHXII02:SOLUTIONS

319385 Which mixture is used for respiration by deep sea divers?

1

N e + O_{2}

2

K r + O_{2}

3

A r + O_{2}

4

H e + O_{2}

MHTCET - 2019

CHXII02:SOLUTIONS

319386 Value of Henry’s constant, $K_{H}$

1 Increases with increase in temperature.

2 Decreases with increase in temperature.

3 Remains constant.

4 First increases then decreases.

Explanation:

Value of $K_{H}$ depends on temperature and nature of solvent.

CHXII02:SOLUTIONS

319387 Assertion :
The solubility of a gas in a liquid increases with increase of pressure.
Reason :
The solubility of a gas in a liquid is directly proportional to the pressure of the gas.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Henry's law states that the solubility of a gas in a liquid is directly proportional to the gas pressure above the liquid, expressed mathematically as $x = k P$ , where ' $x$ ' is solubility, ' $P$ ' is partial pressure, and ' $k$ ' is the Henry's law constant representing the proportionality between gas solubility and pressure. So option (1) is correct.

CHXII02:SOLUTIONS

319388 5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is approximately

1

1 \times 1 0^{- 6}

2

1 \times 1 0^{- 5}

3

1 \times 1 0^{- 3}

4

1 \times 1 0^{- 4}

Explanation:

Solubility of a gas $\propto$ mole fraction
$n_{N_{2}} = \frac{5 .5 \times 1 0^{- 3}}{28} = 1 .96 \times 1 0^{- 4} a t 1 a t m$
Now, $p \propto$ mole fraction then at 5 atm, the mole
fraction will increase 5 times.
Hence, $5 \times 1 .96 \times 1 0^{- 4} = 9 .8 \times 1 0^{- 4}$
$\approx 10 \times 1 0^{- 4} = 1 .0 \times 1 0^{- 3}$

KCET - 2014

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