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Chemistry Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

If \(0.2 mol\) of \(H_{2}(g)\) and \(2.0 mol\) of \(S ( s )\) are \(mixed\) in a 1.0 litre vessel at \(90^{\circ} C ,\) the partial pressure of \(H_{2} S(g)\) formed according to the reaction,

\(H_{2}(g)+S(s) \Longleftrightarrow H_{2} S(g) ; K_{p}=6.8 \times 10^{-2}\) would be

A
0.38 atm

B
0.19 atm

C
0.6 atm

D
none of the above

Solution

\(H_{2}(g)+S(s) \Leftrightarrow H_{2} S(g)\)
\(0.2\quad\quad\quad 2\quad \quad\quad 0\)
\(0.2-x \quad 2-x\quad\quad x\)
\(\because \Delta n=0, K_{p}=K_{c}\)
\(K_{c}=\frac{\left[H_{2} S\right]}{\left[H_{2}\right]}=\frac{x}{(0.2-x)}=6.8 \times 10^{-2}\)
On solving this, we get
\(=>x=1.27 \times 10^{-2}\)
\(P_{H_{2}S}=[x] R T=1.27 \times 10^{-2} \times 0.0821 \times(273+90)=0.38 a t m\)
Hence, the correct option is (A)
Question 2
1 / -0

Nitroethane can be obtained from ethane by which of the following method?

A
Action with HNO₃ (conc.) at 100⁰C

B
Action with diluted HNO₃at 200⁰C

C
Action with HNO₃(conc.) at 475⁰C

D
Action with HNO₃ (dil.) at 100⁰C

Solution

Nitroethane can be obtained from ethane by the action of \(H N O_{3}\) (concentrated) at \(475^{0} \mathrm{C}\) \(C H_{3}-C H_{3}+H O N O_{2} \stackrel{475^{\circ} C}{\longrightarrow} C H_{3} C H_{2} N O_{2}+H_{2} O\)
Hence, the correct option is (C)
Question 3
1 / -0

What is the minimum quantity of methyl iodide required for preparing one mole of ethane by Wurtz reaction ?(At.wt.of iodine=127)

A
142 gram

B
568 gram

C
326 gram

D
284 gram

Solution

Ethane formation by Wurtz reaction requires two moles of methyl iodide.

MW =127+15=142 gm

Hence, minimum quantity of methyl iodie required=2×142=284 gm.
Hence, the correct option is (D)
Question 4
1 / -0

On heating sodium propanoate with sodium hydroxide and quicklime, the gas evolved is :

A
C₂H₂

B
C₂H₆

C
CH₄

D
C₂H₄

Solution

\(C H_{3} C H_{2} C O O N a+N a O H \stackrel{C a O}{\longrightarrow} C_{2} H_{6}+N a_{2} C O_{3}\)
On heating sodium propanoate with sodium hydroxide and quicklime, the gas evolved is ethane.
Hence, the correct option is (B)
Question 5
1 / -0

What is \(X\) in the following sequence of reactions? \(X \stackrel{N a}{\longrightarrow H_{2}} Y \frac{N a O H}{C a O} C H_{4}\)

A
Methanoic acid

B
Ethanoic acid

C
Propane

D
Methane

Solution

\(X\) is ethanoic acid in the given sequence of reactions. The reactions are as follows:
\(C H_{3} C O O H+N a \rightarrow C H_{3} C O O N a+\frac{1}{2} H_{2}\)
\(\mathrm{CH}_{3} \mathrm{COONa}+\mathrm{NaOH} \stackrel{\mathrm{CaO}}{\longrightarrow} \mathrm{CH}_{4}+\mathrm{Na}_{2} \mathrm{CO}_{3}\)
Hence, \(X\) is \(C H_{3} C O O H .\)
Hence, the correct option is (B)
Question 6
1 / -0

N₂O₄ is 25% dissociated at 37^oCand one atmosphere pressure. Calculate KpKp and the percentage dissociation at 0.1 atmosphere and 37^oC.

A
0.166 atm and 37% respecively

B
0.266 atm and 87% respectively

C
0.316 atm and 47% respectively

D
0.316 atm and 37% respectively

Solution

Let initially 1 mole of \(N_{2} O_{4}\) be present. Total pressure is 1 atm.
\(N_{2} O_{4} \rightleftharpoons N O_{2}\)
Initial moles
10
Equilibrium moles0.75 0.5 The expression for the equilibrium constant is:
\(K_{p}=\frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}}=\frac{0.4^{2}}{0.6}=0.266\)
Now total pressure is 0.1 atm. \(N_{2} O_{4} \rightleftharpoons N O_{2}\)
Equilibrium moles are \(1-x\) and \(2 x\) Mole fraction are \(\frac{1-x}{1+x}, \frac{2 x}{1+x}\) The expression for the equilibrium constant is \(K_{p}=\frac{P_{N O_{2}}^{2}}{P_{N_{2}} o_{4}}\) \(0.266=\frac{\left(\frac{2 x}{1+x}\right)^{2}}{\left(\frac{1-x}{1+x}\right) \times 0.1} .\) Thus \(x=0.87 .\) Hence, the degree of dissociation is \(87 \%\)
Hence, the correct option is (B)
Question 7
1 / -0

Ethylene is converted to ethane in the presence of NiNi at 300⁰C. In this reaction the hybridization of carbon changes from :

A
sp to sp²

B
sp² to sp³

C
sp³ to sp

D
sp to sp³

Solution

In ethylene C₂H₄, C atom is sp²hybridised

In ethane C₂H₆, CC atom is sp³hybridised as it does not contain double bonds
Hence, the correct option is (B)
Question 8
1 / -0

For the decomposition, \(N H_{2} C O O N H_{4}(s) \Longleftrightarrow 2 N H_{3}(g)+C O_{2}(g)\)
\(K_{p}=2.9 \times 10^{-5} a t m^{3}\)
The total pressure of gases at equilibrium when 1.0 mol of \(N H_{2} C O O N H_{4}(s)\) was taken
to start with would be :

A
0.0194 atm

B
0.0388 atm

C
0.0582 atm

D
0.0776 atm

Solution

\(N H_{2} C O O N H_{4} \rightarrow 2 N H_{3}+C O_{2}\)
Initial moles \(\quad 1 \quad 0 \quad 0\)
Moles at equilibriumi-x \(\quad 2 \mathrm{x} \quad \mathrm{x}\) Mole fraction
Partial pressure \(0.66 \mathrm{P}\) o.34P
The expression for the equilibrium constant is \(K_{p}=P_{N H_{3}}^{2} P_{C O_{2}}\) Substitute values in the above expression \(2 \times 10^{-5}=(0.66 P)^{2} \times 0.34 P\)
Hence, \(P=0.0582 a t m\)
Hence, the correct option is (C)
Question 9
1 / -0

The correct order of activation of benzene ring towards an electrophile is :

A
(IV)>(III)>(II)>(I)

B
(III)>(I)>(IV)>(II)

C
(III)>(IV)>(I)>(II)

D
(IV)>(III)>(II)>(I)

Solution

\((I I I)>(I V)>(I)>(I I)\)
The trend is on the basis of resonance of functional groups with benzene ring. \(O C H_{3}, O=O,-O C O C H_{3}\) are electron donating groups and fluorine is an electron withdrawing group.
Hence, the correct option is (C)
Question 10
1 / -0

Solid ammonium carbamate dissociates to give ammonia and carbon dioxide as follows:

NH₂COONH₄(s)⇌2NH₃(g)+CO₂(g)

At equilibrium, ammonia is added such that partial pressures of NH₃now equals the original total pressure. Calculate the ratio of the total pressures now to the original total pressure.

A
31/37

B
60/40

C
31/9

D
62/27

Solution

\(N H_{2} C O O N H_{4}(s) \rightleftharpoons 2 N H_{3}(g)+C O_{2}(g)\)
\(2 P\)
Initial \(K_{P}=\left(P_{N H_{3}}\right)^{2}\left(P_{C O_{2}}\right)\)
\(K_{P}=(2 P)^{2}(P) \quad \ldots \ldots(i)\)
\(P_{T}(\)initial\()=3 P\)
\(\quad \quad N H_{2} C O O N H_{4}(s) \rightleftharpoons 2 N H_{3}(g)+C O_{2}(g)\)
Final \(\quad 3 P \quad P^{\prime}\)
\(K_{p}=(3 P)^{2}\left(P^{\prime}\right) \quad \ldots \ldots(i i)\)
From eq. (i) and (ii) \((2 P)^{2}(P)=(3 P)^{2}\left(P^{\prime}\right)\)
\(P^{\prime}=\frac{4 P}{9}\)
\(\frac{P_{T}(N e w)}{P_{T}(O l d)}=\frac{3 P+P^{\prime}}{3 P}=\frac{3 P+\frac{4 P}{9}}{3 P}=\frac{31}{27}\)
Hence, the correct option is (A)

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Chemistry Test - 22

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Question 1

0.38 atm

0.19 atm

0.6 atm

none of the above

Solution

Question 2

Action with HNO3 (conc.) at 1000C

Action with diluted HNO3at 2000C

Action with HNO3 (conc.) at 4750C

Action with HNO3 (dil.) at 1000C

Solution

Question 3

142 gram

568 gram

326 gram

284 gram

Solution

Question 4

C2H2

C2H6

CH4

C2H4

Solution

Question 5

Methanoic acid

Ethanoic acid

Propane

Methane

Solution

Question 6

0.166 atm and 37% respecively

0.266 atm and 87% respectively

0.316 atm and 47% respectively

0.316 atm and 37% respectively

Solution

Question 7

sp to sp2

sp2 to sp3

sp3 to sp

sp to sp3

Solution

Question 8

0.0194 atm

0.0388 atm

0.0582 atm

0.0776 atm

Solution

Question 9

(IV)>(III)>(II)>(I)

(III)>(I)>(IV)>(II)

(III)>(IV)>(I)>(II)

(IV)>(III)>(II)>(I)

Solution

Question 10

31/37

60/40

31/9

62/27

Solution

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Action with HNO₃ (conc.) at 100⁰C

Action with diluted HNO₃at 200⁰C

Action with HNO₃(conc.) at 475⁰C

Action with HNO₃ (dil.) at 100⁰C

C₂H₂

C₂H₆

CH₄

C₂H₄

sp to sp²

sp² to sp³

sp³ to sp

sp to sp³