Chemistry Test

Result

Chemistry Test - 38

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Ammonium carbamate dissociates as

$\mathrm{NH}_{2} \mathrm{COONH}_{4}(s) 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})$

In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that partial pressure of NH₃ now equals to the original total pressure. Calculate the ratio of partial pressure of CO₂ now to the original partial pressure of CO₂ :

A
4

B
9

C
$\frac{4}{9}$

D
$\frac{2}{9}$

Solution

(i) $\mathrm{NH}_{2} \mathrm{COONH}_{4}(\mathrm{s}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{NH}_{3}^{(\mathrm{g})}$
$2 \mathbf{T}$
$\mathrm{K}_{\mathrm{p}}=(\mathrm{p})^{1}(2 \mathrm{p})^{2}=4 \mathrm{p}^{3}$
Total pressure $=3 p=P_{e q}$
$\mathrm{NH}_{2} \mathrm{COONH}_{4}(\mathrm{s}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{NH}_{3}^{(\mathrm{g})}$
$\mathrm{P}_{1}$
$\left(P_{1}\right)^{1}(3 P)^{2}=4 P^{3}$
$\mathrm{p}_{1}=\frac{4 \mathrm{p}^{3}}{9 \mathrm{p}^{2}}=\frac{4}{9} \mathrm{p}$
$\frac{p_{1}}{p}=\frac{4}{9}$
Hence, the correct option is (C)
Question 2
1 / -0

The number of electrons lost in the following change is :

Fe+H₂O⟶Fe₃O₄+H₂

A
2

B
4

C
6

D
8

Solution

$3 F e+8 e^{-} \longrightarrow\left(F e^{8 / 3+}\right)_{3}$
$2 H^{\oplus} \longrightarrow H_{2}+2 e^{-} \times 4$
$3 F e+4 H_{2} O \longrightarrow F e_{3} O_{4}+4 H_{2} O$
Total number of electrons lost or gained is $8 .$
Hence, the correct option is (D)
Question 3
1 / -0

In the reaction C(s)+CO₂⇌2CO(g), the equilibrium pressure is 12 atm. If 50% CO₂ reacts, calculate K_P.

A
K_P=0.25atm

B
K_P=4atm

C
K_P=16atm

D
K_P=8atm

Solution

$C(s)+C O_{2} \rightleftharpoons 2 C O(g)$
(Gaseous moles before dissociation) $\left(1-\frac{50}{100}\right) \frac{2 \times 100}{50}$ (Gaseous moles after
dissociation) Total moles $=1.5$ and $\triangle n=1$
Total pressure given at equilibrium $=12$ atm
$K_{P}=\frac{\left(n_{C O}\right)^{2}}{n_{C O_{2}}} \times\left[\frac{P}{\sum n}\right]^{\sum n}=\frac{(1)^{2}}{0.5} \times\left(\frac{12}{1.5}\right)^{1}$
$K_{P}=\frac{12}{1.5 \times 0.5}=16 \mathrm{atm}$
Hence, the correct option is (C)
Question 4
1 / -0

Addition of HI on double bond of propene yields isopropyl iodide and not n-propyl iodide, because addition proceeds through :

A
a more stable carbanion

B
a more stable carbonium ion

C
a more stable free radical

D
none

Solution

The reaction proceeds through a more stable carbonium ion as shown below:
$C H_{3}-C H=C H_{2} \stackrel{H^{+}}{\longrightarrow} C H_{3}-C H-C H_{3} \stackrel{I}{\longrightarrow} C H_{3}-C H I-C H_{3}$
Hence, the correct option is (B)
Question 5
1 / -0

In an alkaline medium, KMnO₄ reacts as follows (At wt. of K=39.09, Mn=54.94, O=16.0): 2KMnO4+2KOH→2K2MnO4+H2O+[O]

i) The equivalent weight of KMnO₄ in alkaline medium is 158.

ii) The number of electrons gained by one molecule of KMnO₄ in the alkaline medium is 1.

iii) In the alkaline medium, KMnO₄ acts as reducing agent.

Select correct option.

A
Only i is correct

B
i and ii are correct

C
Only iii is correct

D
Only ii is correct

Solution

$\mathrm{KMnO}_{4} \rightarrow \mathrm{Mn}^{+7}$
$K_{2} M n O_{4} \rightarrow M n^{+6}$
So, one $e^{-}$ is gained by a single molecule of $K M n O_{4}$ in the alkaline medium and thus, the valency factor is
1.
So, equivalent weight $=$ molecular weight $=39 \times 1+54 \times 94+4 \times 16$
$=158$
So, i and ii are correct statement. Oxidation number is reduced and therefore, potassium has reduced itself which makes it an oxidizing agent.
Hence, the correct option is (B)
Question 6
1 / -0

How many d-orbital(s) is/are involved in hybridization of [AuCl₄]⁻ ?

A
4

B
3

C
1

D
2

Solution

Option (D) is correct.

Tetrachloroaurate [AuCl₄]⁻ is dsp² (or sp²d) hybridized. One 4s, two 4p orbitals (on the x and y axes), and one 3d orbital from the Au(III) become hybridized. There are no unpaired electrons in this polyatomic ion.

This hybridization is usually not discussed much in general chemistry.
Hence, the correct option is (C)
Question 7
1 / -0

In which of the following, the highest oxidation state is not possible?

A
[XeO₆]⁴⁻

B
XeF₈

C
OsO₄

D
RuO₄

Solution

Xe shows +8 oxidation state in XeF₈ but it does not exist because of steric hindrance of 8F atoms.
Hence, the correct option is (B)
Question 8
1 / -0

How many maximum atom(s) is/are present in same plane of Cr(CO)₆ ?

A
8

B
13

C
9

D
5

Solution

Hence, the correct option is (C)
Question 9
1 / -0

Which of the following is used as freezing mixture?

A
Mixture of ether and liquid CO₂

B
Mixture of ether and ethyl alcohol

C
Mixture of ether and dry ice

D
Mixture of ethyl alcohol and dry ice

Solution

A mixture of dry ice and solvents such as acetone, alcohol or ether acts as a freezing mixture.
Hence, the correct option is (C)
Question 10
1 / -0

In the formation of a chloride ion from an isolated gaseous chlorine atom, 3.8 eV energy is released, which would be equal to:

A
electron affinity of Cl⁻

B
ionisation potential of Cl

C
electronegativity of Cl

D
ionisation potential of Cl⁻

Solution

In the formation of a chloride ion from an isolated gaseous chlorine atom, 3.8 eV energy is released which would be equal to Ionisation potential of Cl⁻.

The electron affinity or electron gain enthalpy of the chlorine atom is equal in magnitude and opposite in sign to the ionization potential of chloride ion.
Hence, the correct option is (D)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemistry Test - 38

Result

Chemistry Test - 38

Score

-

Rank

-

SHARING IS CARING

Question 1

4

9

49

29

Solution

Question 2

2

4

6

8

Solution

Question 3

KP=0.25atm

KP=4atm

KP=16atm

KP=8atm

Solution

Question 4

a more stable carbanion

a more stable carbonium ion

a more stable free radical

none

Solution

Question 5

Only i is correct

i and ii are correct

Only iii is correct

Only ii is correct

Solution

Question 6

4

3

1

2

Solution

Question 7

[XeO6]4−

XeF8

OsO4

RuO4

Solution

Question 8

8

13

9

5

Solution

Question 9

Mixture of ether and liquid CO2

Mixture of ether and ethyl alcohol

Mixture of ether and dry ice

Mixture of ethyl alcohol and dry ice

Solution

Question 10

electron affinity of Cl−

ionisation potential of Cl

electronegativity of Cl

ionisation potential of Cl−

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$\frac{4}{9}$

$\frac{2}{9}$

K_P=0.25atm

K_P=4atm

K_P=16atm

K_P=8atm

[XeO₆]⁴⁻

XeF₈

OsO₄

RuO₄

Mixture of ether and liquid CO₂

electron affinity of Cl⁻

ionisation potential of Cl⁻