Chemistry Test

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Chemistry Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following can act only as a reducing agent?

A
K₂Cr₂O₇

B
CaOCl₂

C
PbO₂

D
H₂S

Solution

\(S\) can have oxidation states in the
range -2 to \(+6 .\) In \(H_{2} S\), it has oxidation state of -2 . Thus, it can only lose electrons. Hence, \(H_{2} S\) can act only as reducing agent. \(K_{2} C r_{2} O_{7}, C a O C l_{2}\) and \(P b O_{2}\) are
oxidizing agents while \(H_{2} S\) is a reducing agent.
Question 2
1 / -0

12 g N₂, 4 gm H₂ and 9 gm O₂ are put into a one litre container at 27^∘C. What is the total pressure.

A
33.37 atm

B
66.74 atm

C
133.48

D
None of these

Solution

12 \(\mathrm{g} N_{2}\) corresponds to \(\frac{12}{28}=0.429 \mathrm{mol}, 4 \mathrm{gm} H_{2}\) corresponds to \(\frac{4}{2}=2 \mathrm{mol}\) and
\(9 \mathrm{gm}\)
\(O_{2}\) corresponds to \(\frac{9}{32}=0.28 .\) The total number of moles are \(0.429 \mathrm{mol}+2 \mathrm{mol}+0.28 \mathrm{mol}=2.71 \mathrm{mol}\) The total pressure is
\(\frac{n R T}{V}=\frac{2.71 \times 0.0821 \times 300}{1}=66.74 a t m\)
Question 3
1 / -0

Two containers x and y contains the same gas.If the P,V and T of the gas in x are 2 times compared to those in y and if the mass of gas in y is A, the mass of the gas in x will be:

A
A

B
A / 2

C
2A

D
4A

Solution

\(P V=n R T=\frac{w}{M} R T\)
\(w=M \frac{P V}{R T}\)
For the gas in the container \(x\) \(w_{x}=M \frac{P_{x} V_{x}}{R T_{x}} \ldots \ldots .(1)\)
Eor the gas in container \(y\) \(w_{y}=M \frac{P_{y} V_{y}}{R T_{y}} \ldots \ldots .(2)\)
Divide equation ( 1 ) by equation ( 2 ) \(\frac{w_{x}}{w_{y}}=\frac{M \frac{P_{z} V_{x}}{R T_{z}}}{M \frac{P_{y} V_{y}}{R T_{y}}}\)
The \(P, V\) and \(T\) of the gas in \(x\) are 2 times compared to those in \(y\) and the mass of gas in \(y\) is \(A\) \(\frac{w_{x}}{A}=\frac{M \frac{2 P_{y} 2 V_{y}}{R 2 T_{y}}}{M \frac{P_{y} V_{y}}{R T_{y}}}\)
\(w_{x}=2 A\)
Hence, the mass of the gas in \(x\) will be \(2 A\).
Question 4
1 / -0

In the following reactions:

Identify the correct option which represents the oxidised substances?

A
K,SO₂,I₂

B
K,H₂S,I₂

C
AlCl₃,H₂S,I₂

D
AlCl₃,SO₂,I₂

Solution

(i) \(A l^{3+} \rightarrow A l^{0}\) and \(K^{0} \rightarrow K^{+}\)
Thus, \(A l^{3+}\) is reduced and \(K\) is oxidised. Note the atom having higher oxidation number in a conjugate redox pair is oxidant and is therefore reduced.
(ii) Similarly, \(S O_{2}\) is reduced and \(H_{2} S\) is oxidised.
(iii) Similarly, neither one is reduced nor one is oxidised (". No change in oxidation number during reaction).
(iv) \(I_{2}\) is itself oxidised as well as reduced. \(I_{2}^{0} \rightarrow 2 I^{5+}+10 e\)
\(2 e+I_{2}^{0} \rightarrow 2 I^{-}\)
Question 5
1 / -0

A
84.76

B
63.5

C
127.0

D
158.75

Solution
Question 6
1 / -0

Calculate relative rate of effusion of SO₂ to CH₄ under given condition

( i ) Under similar condition of pressure & temperature

( ii ) Through a container containing SO₂ and CH₄ in 3 : 2 mass ratio

( iii ) If the mixture obtained by effusing out a mixture (ⁿSO₂/ⁿCH₄=8/1) for three effusing steps.

A
( i ) 1/2 ( ii ) 3/16 ( iii ) 1/2

B
( i ) 1/4 ( ii ) 3/8 ( iii ) 1/4

C
5( i ) 1/6 ( ii ) 5/16 ( iii ) 1/2

D
None fo these

Solution

The relative rate of effusion is inversely proportional to the square root of molar mass.It is directly proportional to the number of moles. The molar masses of sulphur dioxide and methane are \(64 \mathrm{g} / \mathrm{mol}\) and \(16 \mathrm{g} / \mathrm{mol}\) respectively.
(i) Under similar condition of pressure \& temperature \(\frac{r_{S O_{2}}}{r_{C H_{4}}}=\sqrt{\frac{M_{C H_{4}}}{M_{S O_{2}}}}=\sqrt{\frac{16}{64}}=\frac{1}{2}\)
(ii) Through a container containing \(S O_{2}\) and \(C H_{4}\) in 3: 2 mass ratio \(\frac{r_{S O_{2}}}{r_{C H_{4}}}=\sqrt{\frac{n_{S O_{2}} M_{C H_{4}}}{n_{C H_{4}} M_{S O_{2}}}}=\sqrt{\frac{\frac{3}{64} \times 16}{\frac{2}{16} \times 64}}=\frac{3}{16}\)
(iii) If the mixture obtained by effusing out a mixture \(\left(n_{S O_{2}} / n_{C H_{4}}=8 / 1\right)\) for three effusing steps. \(\frac{r_{S O_{2}}}{r_{C H_{4}}}=\sqrt{\frac{M_{C H_{4}}}{M_{S O_{2}}}}=\sqrt{\frac{16}{64}}=\frac{1}{2}\)
Question 7
1 / -0

Out of all reactions, in which reaction hydrogen is acting as oxidizing agent?

A
Reaction with iodine to give hydrogen iodide

B
Reaction with lithium to give lithium hydride

C
Reaction with nitrogen to give ammonia

D
Reaction with sulphur to give hydrogen sulphide

Solution

When hydrogen acts as oxidizing agent, it itself gets reduced. Hydrogen atom accepts an electron to form hydride ion. Thus, it reacts with lithium to form lithium hydride. (Lithium gets easily oxidised as it highly electropositive)

Hence, lithium is oxidized and therefore, hydrogen acts as oxidizing agent.
Question 8
1 / -0

dN_u / N vs u graphs for different gases are identical if ___________ ratio is same for the gases concerned.

A
T / M

B
P / V

C
T / V

D
T / R

Solution

\(\frac{d N_{u}}{N}\) vs \(u\) graphs for different gases are identical if \(T / M\) ratio is same for the gases concerned. Note according to Maxwell-Boltzmann law for molecular velocities, the probability (p) of finding molecules with velocity u is given by the expression \(p=4 \pi\left(\frac{M}{2 \pi R T}\right)^{3 / 2} \cdot e^{-\frac{M u^{2}}{2 R T}} \cdot u^{2}\)
Question 9
1 / -0

What is the oxidation number of chlorine in ClO⁻³?

A
+5

B
+3

C
+4

D
+2

Solution

Let \(x\) be the oxidation number of chlorine in \(C l O_{3}\). The oxidation number of oxygen is -2 The sum of the oxidation numbers of chlorine and oxygen is -1 , which is equal to the charge on ion. Hence, \(x+3(-2)=-1\) or, \(x=+5\)
Question 10
1 / -0

\(1 mol\) of ferric oxalate is oxidized by \(xmol\) of \(MnO _{4}^{-}\) and also \(1 mol\) of ferrous oxalate is oxidised by y mol of \(M n O_{4}^{-}\) in acidic medium. The ration \(\frac{x}{y}\) is:

A
2 : 1

B
1 : 2

C
3 : 1

D
1 : 3

Solution

\(Fe _{2}(204) 3+ MnO _{4}^{-} \rightarrow Mn ^{2+}+ Co _{2}\)
\(1 \times 6=x \times 5\)
\(x=\frac{6}{5}\)
\(M n O_{4}^{-}+F e C_{2} O_{4} \rightarrow M n^{2+}+F e^{3+}+C o_{2}\)
4 -factor \(=2+1=3\)
\(y \times 5=1 \times 3\)
\(y=\frac{3}{5}\)
\(\frac{x}{y}=\frac{6}{5} / \frac{3}{5}=2: 1\)

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Re-Attempt Weekly Quiz Competition

Chemistry Test - 17

Result

Chemistry Test - 17

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SHARING IS CARING

Question 1

K2Cr2O7

CaOCl2

PbO2

H2S

Solution

Question 2

33.37 atm

66.74 atm

133.48

None of these

Solution

Question 3

A

A / 2

2A

4A

Solution

Question 4

K,SO2,I2

K,H2S,I2

AlCl3,H2S,I2

AlCl3,SO2,I2

Solution

Question 5

84.76

63.5

127.0

158.75

Solution

Question 6

( i ) 1/2 ( ii ) 3/16 ( iii ) 1/2

( i ) 1/4 ( ii ) 3/8 ( iii ) 1/4

5( i ) 1/6 ( ii ) 5/16 ( iii ) 1/2

None fo these

Solution

Question 7

Reaction with iodine to give hydrogen iodide

Reaction with lithium to give lithium hydride

Reaction with nitrogen to give ammonia

Reaction with sulphur to give hydrogen sulphide

Solution

Question 8

T / M

P / V

T / V

T / R

Solution

Question 9

+5

+3

+4

+2

Solution

Question 10

2 : 1

1 : 2

3 : 1

1 : 3

Solution

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K₂Cr₂O₇

CaOCl₂

PbO₂

H₂S

K,SO₂,I₂

K,H₂S,I₂

AlCl₃,H₂S,I₂

AlCl₃,SO₂,I₂