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CHEMISTRY(KCET)

285375 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 10^{- 6}

2

1 \times 10^{- 5}

3

1 \times 10^{- 3}

4

1 \times 10^{- 4}

Explanation:

(c) Solubility of a gas $\propto$ mole fraction
$n_{N_{2}} = \frac{5.5 \times 10^{- 3}}{28} = 1.96 \times 10^{- 4}$ at 1 atm
Now, $p \propto$ mole fraction then at 5 atm , the mole fraction will increase 5 times.
Hence, $5 \times 1.96 \times 10^{- 4} = 9.8 \times 10^{- 4}$
$\approx 10 \times 10^{- 4} = 1.0 \times 10^{- 3}$

Karnataka CET 2014

CHEMISTRY(KCET)

285376 A solution of 1.25 g of $P$ in 50 g of water lowers freezing point by $^{-} {.3}^{\circ} C$ . Molar mass of P is 94 . $K_{f (water)} = 1.86 K kg {mol}^{- 1}$ . The degree of association of P in water is

1

80 %

2

60 %

3

65 %

4

75 %

Explanation:

(a) Given: $K_{f} = 1.86 K kg {mol}^{- 1}, w = 1.25 g$ ,
$W = 50 g, Δ T = {0.3}^{\circ} C, M =$ ?
As $P$ undergoes association,
$2 P ⇌ (P)_{2}; n = 2$
Using equation, $M = \frac{1000 \times K_{f} \times w}{W \times Δ T}$
$M = \frac{1000 \times 1.86 \times 1.25}{50 \times 0.3} = 155$
Now, $i = \frac{Normal mol. mass}{Observed mol. mass} = \frac{94}{155} = 0.606$
Then, the degree of association of $P$ is
$α = \frac{1 - i}{1 - \frac{1}{n}}$
$α = \frac{1 - 0.606}{1 - \frac{1}{2}} = 0.788$ or $78.8 % \sim 80 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285377 For an ideal binary liquid mixture

1

Δ S_{(mix)} = 0; Δ G_{(mix)} = 0

2

Δ H_{(mix)} = 0; Δ S_{(mix)} < 0

3

Δ V_{(mix)} = 0; Δ G_{(mix)} > 0

4

Δ S_{(mix)} > 0; Δ G_{(mix)} < 0

Explanation:

(d) For an ideal solution; $Δ H = 0$ and $Δ V = 0$ From, $Δ G = Δ H - T Δ S$
$Δ G = - ve$
Hence, $Δ G < 0$ and $Δ S > 0$

Karnataka CET 2014

CHEMISTRY(KCET)

285378 How many Coulombs are required to oxidise 0.1 mole of $H_{2} O$ to oxygen?

1

1.93 \times 10^{5} C

2

1.93 \times 10^{4} C

3

3.86 \times 10^{4} C

4

9.65 \times 10^{3} C

Explanation:

(b) $H_{2} O \to 2 H^{\oplus} + 2 e^{-} + \frac{1}{2} O_{2}$
$2 F = 1 mol$ of $H_{2} O = \frac{1}{2} mol$ of $O_{2}$
1 mol of $H_{2} O = 2 F$
0.1 mol of $H_{2} O = 2 \times 96500 \times 0.1 = 1.93 \times 10^{4} C$

Karnataka CET 2024

CHEMISTRY(KCET)

285375 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 10^{- 6}

2

1 \times 10^{- 5}

3

1 \times 10^{- 3}

4

1 \times 10^{- 4}

Explanation:

(c) Solubility of a gas $\propto$ mole fraction
$n_{N_{2}} = \frac{5.5 \times 10^{- 3}}{28} = 1.96 \times 10^{- 4}$ at 1 atm
Now, $p \propto$ mole fraction then at 5 atm , the mole fraction will increase 5 times.
Hence, $5 \times 1.96 \times 10^{- 4} = 9.8 \times 10^{- 4}$
$\approx 10 \times 10^{- 4} = 1.0 \times 10^{- 3}$

Karnataka CET 2014

CHEMISTRY(KCET)

285376 A solution of 1.25 g of $P$ in 50 g of water lowers freezing point by $^{-} {.3}^{\circ} C$ . Molar mass of P is 94 . $K_{f (water)} = 1.86 K kg {mol}^{- 1}$ . The degree of association of P in water is

1

80 %

2

60 %

3

65 %

4

75 %

Explanation:

(a) Given: $K_{f} = 1.86 K kg {mol}^{- 1}, w = 1.25 g$ ,
$W = 50 g, Δ T = {0.3}^{\circ} C, M =$ ?
As $P$ undergoes association,
$2 P ⇌ (P)_{2}; n = 2$
Using equation, $M = \frac{1000 \times K_{f} \times w}{W \times Δ T}$
$M = \frac{1000 \times 1.86 \times 1.25}{50 \times 0.3} = 155$
Now, $i = \frac{Normal mol. mass}{Observed mol. mass} = \frac{94}{155} = 0.606$
Then, the degree of association of $P$ is
$α = \frac{1 - i}{1 - \frac{1}{n}}$
$α = \frac{1 - 0.606}{1 - \frac{1}{2}} = 0.788$ or $78.8 % \sim 80 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285377 For an ideal binary liquid mixture

1

Δ S_{(mix)} = 0; Δ G_{(mix)} = 0

2

Δ H_{(mix)} = 0; Δ S_{(mix)} < 0

3

Δ V_{(mix)} = 0; Δ G_{(mix)} > 0

4

Δ S_{(mix)} > 0; Δ G_{(mix)} < 0

Explanation:

(d) For an ideal solution; $Δ H = 0$ and $Δ V = 0$ From, $Δ G = Δ H - T Δ S$
$Δ G = - ve$
Hence, $Δ G < 0$ and $Δ S > 0$

Karnataka CET 2014

CHEMISTRY(KCET)

285378 How many Coulombs are required to oxidise 0.1 mole of $H_{2} O$ to oxygen?

1

1.93 \times 10^{5} C

2

1.93 \times 10^{4} C

3

3.86 \times 10^{4} C

4

9.65 \times 10^{3} C

Explanation:

(b) $H_{2} O \to 2 H^{\oplus} + 2 e^{-} + \frac{1}{2} O_{2}$
$2 F = 1 mol$ of $H_{2} O = \frac{1}{2} mol$ of $O_{2}$
1 mol of $H_{2} O = 2 F$
0.1 mol of $H_{2} O = 2 \times 96500 \times 0.1 = 1.93 \times 10^{4} C$

Karnataka CET 2024

CHEMISTRY(KCET)

285375 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 10^{- 6}

2

1 \times 10^{- 5}

3

1 \times 10^{- 3}

4

1 \times 10^{- 4}

Explanation:

(c) Solubility of a gas $\propto$ mole fraction
$n_{N_{2}} = \frac{5.5 \times 10^{- 3}}{28} = 1.96 \times 10^{- 4}$ at 1 atm
Now, $p \propto$ mole fraction then at 5 atm , the mole fraction will increase 5 times.
Hence, $5 \times 1.96 \times 10^{- 4} = 9.8 \times 10^{- 4}$
$\approx 10 \times 10^{- 4} = 1.0 \times 10^{- 3}$

Karnataka CET 2014

CHEMISTRY(KCET)

285376 A solution of 1.25 g of $P$ in 50 g of water lowers freezing point by $^{-} {.3}^{\circ} C$ . Molar mass of P is 94 . $K_{f (water)} = 1.86 K kg {mol}^{- 1}$ . The degree of association of P in water is

1

80 %

2

60 %

3

65 %

4

75 %

Explanation:

(a) Given: $K_{f} = 1.86 K kg {mol}^{- 1}, w = 1.25 g$ ,
$W = 50 g, Δ T = {0.3}^{\circ} C, M =$ ?
As $P$ undergoes association,
$2 P ⇌ (P)_{2}; n = 2$
Using equation, $M = \frac{1000 \times K_{f} \times w}{W \times Δ T}$
$M = \frac{1000 \times 1.86 \times 1.25}{50 \times 0.3} = 155$
Now, $i = \frac{Normal mol. mass}{Observed mol. mass} = \frac{94}{155} = 0.606$
Then, the degree of association of $P$ is
$α = \frac{1 - i}{1 - \frac{1}{n}}$
$α = \frac{1 - 0.606}{1 - \frac{1}{2}} = 0.788$ or $78.8 % \sim 80 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285377 For an ideal binary liquid mixture

1

Δ S_{(mix)} = 0; Δ G_{(mix)} = 0

2

Δ H_{(mix)} = 0; Δ S_{(mix)} < 0

3

Δ V_{(mix)} = 0; Δ G_{(mix)} > 0

4

Δ S_{(mix)} > 0; Δ G_{(mix)} < 0

Explanation:

(d) For an ideal solution; $Δ H = 0$ and $Δ V = 0$ From, $Δ G = Δ H - T Δ S$
$Δ G = - ve$
Hence, $Δ G < 0$ and $Δ S > 0$

Karnataka CET 2014

CHEMISTRY(KCET)

285378 How many Coulombs are required to oxidise 0.1 mole of $H_{2} O$ to oxygen?

1

1.93 \times 10^{5} C

2

1.93 \times 10^{4} C

3

3.86 \times 10^{4} C

4

9.65 \times 10^{3} C

Explanation:

(b) $H_{2} O \to 2 H^{\oplus} + 2 e^{-} + \frac{1}{2} O_{2}$
$2 F = 1 mol$ of $H_{2} O = \frac{1}{2} mol$ of $O_{2}$
1 mol of $H_{2} O = 2 F$
0.1 mol of $H_{2} O = 2 \times 96500 \times 0.1 = 1.93 \times 10^{4} C$

Karnataka CET 2024

CHEMISTRY(KCET)

285375 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 10^{- 6}

2

1 \times 10^{- 5}

3

1 \times 10^{- 3}

4

1 \times 10^{- 4}

Explanation:

(c) Solubility of a gas $\propto$ mole fraction
$n_{N_{2}} = \frac{5.5 \times 10^{- 3}}{28} = 1.96 \times 10^{- 4}$ at 1 atm
Now, $p \propto$ mole fraction then at 5 atm , the mole fraction will increase 5 times.
Hence, $5 \times 1.96 \times 10^{- 4} = 9.8 \times 10^{- 4}$
$\approx 10 \times 10^{- 4} = 1.0 \times 10^{- 3}$

Karnataka CET 2014

CHEMISTRY(KCET)

285376 A solution of 1.25 g of $P$ in 50 g of water lowers freezing point by $^{-} {.3}^{\circ} C$ . Molar mass of P is 94 . $K_{f (water)} = 1.86 K kg {mol}^{- 1}$ . The degree of association of P in water is

1

80 %

2

60 %

3

65 %

4

75 %

Explanation:

(a) Given: $K_{f} = 1.86 K kg {mol}^{- 1}, w = 1.25 g$ ,
$W = 50 g, Δ T = {0.3}^{\circ} C, M =$ ?
As $P$ undergoes association,
$2 P ⇌ (P)_{2}; n = 2$
Using equation, $M = \frac{1000 \times K_{f} \times w}{W \times Δ T}$
$M = \frac{1000 \times 1.86 \times 1.25}{50 \times 0.3} = 155$
Now, $i = \frac{Normal mol. mass}{Observed mol. mass} = \frac{94}{155} = 0.606$
Then, the degree of association of $P$ is
$α = \frac{1 - i}{1 - \frac{1}{n}}$
$α = \frac{1 - 0.606}{1 - \frac{1}{2}} = 0.788$ or $78.8 % \sim 80 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285377 For an ideal binary liquid mixture

1

Δ S_{(mix)} = 0; Δ G_{(mix)} = 0

2

Δ H_{(mix)} = 0; Δ S_{(mix)} < 0

3

Δ V_{(mix)} = 0; Δ G_{(mix)} > 0

4

Δ S_{(mix)} > 0; Δ G_{(mix)} < 0

Explanation:

(d) For an ideal solution; $Δ H = 0$ and $Δ V = 0$ From, $Δ G = Δ H - T Δ S$
$Δ G = - ve$
Hence, $Δ G < 0$ and $Δ S > 0$

Karnataka CET 2014

CHEMISTRY(KCET)

285378 How many Coulombs are required to oxidise 0.1 mole of $H_{2} O$ to oxygen?

1

1.93 \times 10^{5} C

2

1.93 \times 10^{4} C

3

3.86 \times 10^{4} C

4

9.65 \times 10^{3} C

Explanation:

(b) $H_{2} O \to 2 H^{\oplus} + 2 e^{-} + \frac{1}{2} O_{2}$
$2 F = 1 mol$ of $H_{2} O = \frac{1}{2} mol$ of $O_{2}$
1 mol of $H_{2} O = 2 F$
0.1 mol of $H_{2} O = 2 \times 96500 \times 0.1 = 1.93 \times 10^{4} C$

Karnataka CET 2024