Chemistry Test

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Chemistry Test - 18

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Weekly Quiz Competition

Question 1
4 / -1

In which of the following cases is rate of reaction not affected by pressure?

A
PCl₃ (g) + Cl₂ (g) ⇌ PCl₅ (g)

B
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

C
N₂ (g) + O₂ (g) ⇌ 2NO (g)

D
2SO₂ (g) + O₂ (g) ⇌ 2SO₂(g)

Solution

Since the number of moles of reactants and products are the same, there will not be any effect of pressure on the reaction.
Question 2
4 / -1

The favourable conditions for Haber's process are:

A
Low temperature and low pressure

B
Low temperature and high pressure

C
High temperature and low pressure

D
High temperature and high pressure

Solution

The favourable conditions for Haber's process are low temperature and high pressure, as:

(1) According to Le Chatelier's principle, low temperature will shift the equilibrium to the right because the reaction is exothermic.

(2) High pressure on the reaction at equilibrium favours the shift of the equilibrium to the right because the forward reaction proceeds with a decrease in the number of gaseous moles.
Question 3
4 / -1

If equivalent conductance of 1 M benzoic acid is 12.8 ohm^-1 cm², and the values of conductance of benzoate ion and H⁺ ion respectively are 42 and 288.42 ohm^-1 cm², then its degree of dissociation is:

A
39%

B
3.9%

C
0.35%

D
0.039%

Solution

\(\begin{aligned} \alpha &=\frac{\lambda_{v}}{\lambda_{\infty}} \\ \text { Given, } \lambda_{C_{6} \mathrm{H}_{5}}^{\infty} \mathrm{COO}^{-}=& 42 \mathrm{ohm}^{-1} \mathrm{cm}^{2} \\ \lambda_{\mathrm{H}^{+}}^{\infty}=& 288.42 \mathrm{ohm}^{-1} \mathrm{cm}^{2} \\ \lambda^{\infty} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} &=\lambda^{\infty} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}+\lambda^{\infty} \mathrm{H}^{+} \\ &=42.0+288.42 \\ &=330.42 \mathrm{ohm}^{-1} \mathrm{cm}^{2} \\ \lambda_{v} &=12.8 \mathrm{ohm}^{-1} \mathrm{cm}^{2} \\ \alpha=\frac{\lambda_{v}}{\lambda_{\infty}} &=\frac{12.8}{330.42} \\ &=0.039 \\ &=0.039 \times 100 \\ &=3.9 \% \end{aligned}\)
Question 4
4 / -1

\(2 A+B \rightarrow 3 C+D\)

Which of the following does not express the reaction rate for the above reaction?

A
\(A=\frac{-d[A]}{2 d t}\)

B
\(\mathrm{B}=\frac{-d[B]}{d t}\)

C
\(\mathrm{C}=\frac{+d[C]}{3 \mathrm{d} t}\)

D
\(\mathrm{D}=\frac{-d[D]}{2 d t}\)

Solution

rate of reaction for different stoicometric cofficients of reactant and products is given as:
\(-\frac{d[A]}{2 d t}=-\frac{d[B]}{d t}=\frac{d[C]}{3 d t}=\frac{d[D]}{d t}\)
Question 5
4 / -1

Directions: The following question has four choices out of which ONLY ONE is correct.

Which of the following biomolecules is insoluble in water?

A
α-Keratin

B
Haemoglobin

C
Ribonuclease

D
Adenine

Solution

α- Keratin which are associated the most with structural proteins. α- Keratin contains a lot of cysteine amino acids (C). These Cysteine amino acids forms disulphide bonds or bridges. These disulphide bridges are very strong covalent bonds and are not easily broken by any means. That's why disulphide bridges are so crucial to structural proteins. Also these fibrous proteins have intra-chain H-bonding. Thus, the disulphide bridges (Covalent bonds) and H - bonding cause insolubility.
Question 6
4 / -1

Which of the following statements is correct about the compounds I, II and III?

A
I and II are identical.

B
II and III are diastereomers.

C
I and III are enantiomers.

D
II and III are enantiomers.

Solution

Enantiomers are non-superimposable mirror images of each other.

Diastereomers have the same configuration at one chiral carbon atom and a different configuration at other carbon atoms.

from given three compounds,

I and II are enantiomers.

I and III are diastereomers.

II and III are diastereomers.
Question 7
4 / -1

The vapour pressure of two volatile liquid mixtures is p_t = 5.3 + 2x_B (in mm of Hg), where X_B is mole fraction of B in mixture. What is the ratio of mole fractions of A and B in liquid phase?

(P_A and P_Bare vapour pressures of A and B)

A
\(\frac{73}{53}\)

B
\(\frac{73}{53} \times \frac{P_{A}}{P_{B}}\)

C
\(\frac{53}{73}\)

D
\(\frac{53}{73} \times \frac{P_{A}}{P_{s}}\)

Solution

\(p_{t}=5.3+2 x_{\mathrm{B}}\)
If \(x_{\mathrm{B}}=0, \mathrm{p}_{\mathrm{t}}=\mathrm{p}_{\mathrm{A}}^{\circ}=5.3\)
and if \(\mathrm{x}_{\mathrm{B}}=1, \mathrm{p}_{\mathrm{t}}=\mathrm{p}_{\mathrm{e}}^{\circ}=7.3\)
But \(\mathrm{p}_{\mathrm{A}}=\mathrm{x}_{\mathrm{A}} \mathrm{p}_{\mathrm{A}}^{\circ} \quad\) and \(\mathrm{p}_{\mathrm{e}}=\mathrm{x}_{\mathrm{B}} \mathrm{p}_{\theta}^{\circ}\)
\(\frac{\mathrm{x}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{B}}}=\frac{\mathrm{p}_{\mathrm{A}}}{\mathrm{p}_{\mathrm{B}}} \times \frac{\mathrm{p}_{\mathrm{e}}^{\circ}}{\mathrm{p}_{\mathrm{A}}^{\circ}}=\frac{73}{53} \frac{\mathrm{p}_{\mathrm{A}}}{\mathrm{p}_{\mathrm{B}}}\)
Question 8
4 / -1

The decomposition of a certain mass of CaCO₃ gave 11.2 dm³ of CO₂ gas at STP. The mass of KOH required to completely neutralize the gas is

A
56 g

B
28 g

C
42 g

D
20 g

Solution

Weight of \(11.2 \mathrm{dm}^{3}\) of \(\mathrm{CO}_{2}\) gas at \(\mathrm{STP}\) is \(44 / 2=22 \mathrm{g}\)
\(\mathrm{KOH}+\mathrm{CO}_{2} \rightarrow \mathrm{KHCO}_{3}\)
\(56 \mathrm{g} \quad 44 \mathrm{g}\)
\(\mathrm{KOH}\) required for complete neutralization of \(22 \mathrm{g}\) \(\mathrm{CO}_{2}=\frac{56}{44} \times 22=28 \mathrm{g}\)
Question 9
4 / -1

If at any temperature, Kp is the equilibrium constant for the reaction N₂ (g) + 3H₂ (g) ⇌ 2 NH₃(g), then what will be the effect of increasing the pressure from 5 atm to 50 atm?

A
Increase in the yield of ammonia as well as the value of K_p

B
Increase in the yield of NH₃, but no change in the value of K_p

C
Decrease in the yield of NH₃ as well as the value of K_p

D
Decrease in the yield of ammonia as well as the value of K_c

Solution

Since the number of moles of reactants is more than that of products, increasing pressure will increase the yield of NH₃ as well as the value of K_p.
Question 10
4 / -1

The vapour pressure of a pure liquid ‘A’ is 70 torr. at 27°C. It forms an ideal solution with another liquid B.The mole fraction of B is 0.2 and total pressure of the solution is 84 torr at 27°C.The vapour pressure of pure liquid B at 27°C is

A
14 torr

B
56 torr

C
140 torr

D
70 torr

Solution

\(\mathrm{p}_{\mathrm{A}}=(1-0.2) \times 70=0.8 \times 70=56 \mathrm{torr}\)
\(\mathrm{p}_{\mathrm{B}}=84-56=28 \mathrm{torr}\)
\(28=0.2 \mathrm{p}_{\mathrm{B}}^{\circ}\)
\(p_{\mathrm{\theta}}^{\circ}=140 \mathrm{torr}\)

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Chemistry Test - 18

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Chemistry Test - 18

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Question 1

PCl3 (g) + Cl2 (g) ⇌ PCl5 (g)

N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

N2 (g) + O2 (g) ⇌ 2NO (g)

2SO2 (g) + O2 (g) ⇌ 2SO2 (g)

Solution

Question 2

Low temperature and low pressure

Low temperature and high pressure

High temperature and low pressure

High temperature and high pressure

Solution

Question 3

39%

3.9%

0.35%

0.039%

Solution

Question 4

\(A=\frac{-d[A]}{2 d t}\)

\(\mathrm{B}=\frac{-d[B]}{d t}\)

\(\mathrm{C}=\frac{+d[C]}{3 \mathrm{d} t}\)

\(\mathrm{D}=\frac{-d[D]}{2 d t}\)

Solution

Question 5

α-Keratin

Haemoglobin

Ribonuclease

Adenine

Solution

Question 6

I and II are identical.

II and III are diastereomers.

I and III are enantiomers.

II and III are enantiomers.

Solution

Question 7

\(\frac{73}{53}\)

\(\frac{73}{53} \times \frac{P_{A}}{P_{B}}\)

\(\frac{53}{73}\)

\(\frac{53}{73} \times \frac{P_{A}}{P_{s}}\)

Solution

Question 8

56 g

28 g

42 g

20 g

Solution

Question 9

Increase in the yield of ammonia as well as the value of Kp

Increase in the yield of NH3, but no change in the value of Kp

Decrease in the yield of NH3 as well as the value of Kp

Decrease in the yield of ammonia as well as the value of Kc

Solution

Question 10

14 torr

56 torr

140 torr

70 torr

Solution

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PCl₃ (g) + Cl₂ (g) ⇌ PCl₅ (g)

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

N₂ (g) + O₂ (g) ⇌ 2NO (g)

2SO₂ (g) + O₂ (g) ⇌ 2SO₂(g)

Increase in the yield of ammonia as well as the value of K_p

Increase in the yield of NH₃, but no change in the value of K_p

Decrease in the yield of NH₃ as well as the value of K_p

Decrease in the yield of ammonia as well as the value of K_c