Chemistry Test

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Chemistry Test - 8

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following is correct regarding CH₂F₂?

A
∠H-C-H > ∠F-C-F

B
∠H-C-H < ∠F-C-F

C
∠H-C-H = ∠F-C-F

D
None of these

Solution

The electronegativity of C atom is high as compared to H atom, while the electronegativity of C atom is less as compared to F atom. So, the electron density around the central atom and hence the repulsions are higher in H-C-H as compared to F-C-F.

Therefore, option (1) is correct.
Question 2
4 / -1

Which of the following equations correctly represents the standard heat of formation (ΔH_f⁰) of methane?

A
C (diamond) + 4H(g)→CH4(g)

B
C (diamond) + 2H2(g)→CH4(g)

C
C (graphite) + 2H2(g)→CH4(g)

D
C (graphite) + 4H(g)→CH4(g)

Solution

Standard heat of formation (ΔH_f⁰)of methane is represented by

C (graphite) + 2H2(g)→CH4(g)
Question 3
4 / -1

If C₆H₆(I) + 15/2 O₂(g)→3H₂O(I) + 6CO₂(g); ΔH= – 3264.6 kJ mol^–1, then the energy obtained by burning 3.9 g of benzene in the air is

A
163.23 kJ

B
326.4 kJ

C
32.64 kJ

D
16.32 kJ

Solution

78 g of C₆H₆ gives heat

= 3264.6 kJ

∴3.9 g of C₆H₆ will give heat

= (3264.6 / 78) x 3.9= 163.23 kJ
Question 4
4 / -1

A 0.1 M aqueous solution of a weak acid is 2% ionised. If the ionic product of water is 1 x 10^-14 mol² dm^-6, then [OH^-] is

A
5 x 10^-12 M

B
2 x 10^-3 M

C
1 x 10^-14 M

D
None of these

Solution

since, it is a weak acid, the equation to calculate \(\left[\mathrm{H}^{\prime}\right]\) is
\[
\begin{array}{l}
\mathrm{C} \times \alpha(\alpha=\% \text { of ionisation }) \\
=0.1 \times 0.02=2 \times 10^{-3} \mathrm{M} \\
\mathrm{K}_{w}=\left[\mathrm{H}^{\prime}\right][\mathrm{OH}] \\
{\left[\mathrm{OH}^{-}\right]=\frac{\mathrm{K}_{w}}{\left[\mathrm{H}^{+}\right]}=\frac{1 \times 10^{-14}}{2 \times 10^{-3}}=5 \times 10^{-12} \mathrm{M}}
\end{array}
\]
Question 5
4 / -1

In which of the following reactions does H₂O₂ act as a reducing agent?

A
2FeCl₂ + 2HCl + H₂O₂ → 2FeCl₃ + 2H₂O

B
Cl₂ + H₂O₂ → 2HCl + O₂

C
2Hl + H₂O₂ → 2H₂O + l₂

D
H₂SO₃ + H₂O₂ → H₂SO₄ + H₂O

Solution

In the reaction Cl₂ + H₂O₂ → 2HCl + O₂, the oxidation number of oxygen changes from -1 to 0.

Hence, H₂O₂ is oxidised and acts as a reducing agent.
Question 6
4 / -1

Directions: In the following question, a statement of assertion is given, followed by a corresponding statement of the reason below it. Choose the correct option.

Assertion: Cu²⁺ and Cd²⁺are separated by first adding KCN solution and then, passing H₂S gas.

Reason: KCN reduces Cu²⁺ to Cu⁺and forms a complex with it.

A
Both assertion and reason are true and reason is the correct explanation of assertion.

B
Both assertion and reason are true, but reason is not the correct explanation of assertion.

C
Assertion is true, but reason is false.

D
Both assertion and the reason are false.

Solution

Both assertion and reason are true, but the reason is not a correct explanation for the assertion.

Cu⁺² makes a more stable complex with CN^- which is soluble and therefore the solution, now contains, the complex Cu⁺² ion and the free Cd⁺² ion. Hence only the sulphide of Cadmium will precipitate out on the passage of H₂S gas.
Question 7
4 / -1

A reaction was found to be in second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled with everything else kept the same, the rate of reaction will

A
be doubled

B
remain unchanged

C
be tripled

D
ncrease by a factor of 4

Solution

Rate \(\alpha[\mathrm{CO}]^{2} \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\left[\left(\mathrm{C}_{\mathrm{CO}}\right)_{2}\right]^{2}}{\left[\left(\mathrm{C}_{\mathrm{CO}}\right)_{1}\right]^{2}}\)
\(=\left[\frac{2 \times\left(\mathrm{C}_{\mathrm{CO}_{1}}\right)}{\left(\mathrm{C}_{\mathrm{CO}}\right)_{1}}\right]^{2}=4\)
Question 8
4 / -1

What volume of 0.1 M ethanol is required for the preparation of 5.6 g of C₂H₄ using concentrated H₂SO₄?

A
2000 mL

B
400 mL

C
200 mL

D
1 L

Solution

\(C H_{3}-C H_{2}-O H \stackrel{H_{2} S O_{4}}{\longrightarrow} C H_{2}=C H_{2}+H_{2} O\)

46 g 28 g

It is clear from the above equation that number of moles of ethene formed = Number of moles of ethanol used

Number of moles of ethene formed = 5.6/28 = 0.2

M = n/V

V = n/M = 0.2/0.1 = 2 L
Question 9
4 / -1

For gases at a constant volume, the correct relation for heat is

A
dQ=dE

B
dQ=dH

C
dQ=0

D
dE=0

Solution

For gases,

dQ = dE + W

W = P. V

At a constant volume, W = 0

⇒ dQ = dE
Question 10
4 / -1

The gas-phase reaction of nitric oxide and bromine yields nitrosyl bromide as 2NO(g) + Br₂(g) →2NOBr(g). What is the order of the reaction?

A
1

B
2

C
3

D
4

Solution

Rate law is ,

rate=k[NO]²[Br₂]

hence, wrt NO, order=2

wrt Br₂, order=1

∴overall order = 2+1=3

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Chemistry Test - 8

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Question 1

∠H-C-H > ∠F-C-F

∠H-C-H < ∠F-C-F

∠H-C-H = ∠F-C-F

None of these

Solution

Question 2

C (diamond) + 4H(g)→CH4(g)

C (diamond) + 2H2(g)→CH4(g)

C (graphite) + 2H2(g)→CH4(g)

C (graphite) + 4H(g)→CH4(g)

Solution

Question 3

163.23 kJ

326.4 kJ

32.64 kJ

16.32 kJ

Solution

Question 4

5 x 10-12 M

2 x 10-3 M

1 x 10-14 M

None of these

Solution

Question 5

2FeCl2 + 2HCl + H2O2 → 2FeCl3 + 2H2O

Cl2 + H2O2 → 2HCl + O2

2Hl + H2O2 → 2H2O + l2

H2SO3 + H2O2 → H2SO4 + H2O

Solution

Question 6

Both assertion and reason are true and reason is the correct explanation of assertion.

Both assertion and reason are true, but reason is not the correct explanation of assertion.

Assertion is true, but reason is false.

Both assertion and the reason are false.

Solution

Question 7

be doubled

remain unchanged

be tripled

ncrease by a factor of 4

Solution

Question 8

2000 mL

400 mL

200 mL

1 L

Solution

Question 9

dQ=dE

dQ=dH

dQ=0

dE=0

Solution

Question 10

1

2

3

4

Solution

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5 x 10^-12 M

2 x 10^-3 M

1 x 10^-14 M

2FeCl₂ + 2HCl + H₂O₂ → 2FeCl₃ + 2H₂O

Cl₂ + H₂O₂ → 2HCl + O₂

2Hl + H₂O₂ → 2H₂O + l₂

H₂SO₃ + H₂O₂ → H₂SO₄ + H₂O