Chemistry Test

Result

Chemistry Test - 6

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
3 / -1

A non-volatile solute (A) is dissolved in a volatile solvent (B). The vapour pressure of the solution is P_s. The vapour pressure of the pure solvent is P⁰_B. If X is mole fraction then which of the following is correct?

A
p_s=p⁰_B.X_A

B
p⁰=p_BX_B

C
p_s=p⁰_B.X_B

D
p⁰_B=p_sX_A

Solution

By Raoult's law

P_S=P⁰_BX_B+P⁰_AX_A

But A is non-volatile solute

So P⁰_A=0

P_S=P⁰_BX_B

Hence option C is correct.
Question 2
3 / -1

Directions: The following question has four choices, out of which ONE or MORE are correct.

For the first order reaction ,

\(2 N_{2} O_{5}(g) \rightarrow 4 N O_{2}(g)+O_{2}(g)\)

A
the concentration of the reactant decreases exponentially with time

B
the half-life of the reaction decreases with increasing temperature

C
the reaction proceeds to 99.6% completion in eight half-life duration

D
All of the above

Solution

Option 3 is incorrect because half-life period of first order reaction is independent from initial concentration of reactants, while options 1, 2 and 4 are correct as the concentration of reactant which is following first order kinetics always decreases exponentially and becomes zero at infinity.

As the temperature increases, the rate constant increases and the half-life decreases since half-life is inversely dependent on rate constant.

\(C_{t}=C \cdot e^{-t t}\)
\(\frac{t_{1} \alpha}{2} \frac{1}{K} \cdot K\) increases on increasing \(T\) After eight half-lives. \(c=\frac{c_{0}}{2^{3}}\)
\(\%\) completion \(=\frac{C_{0}-\frac{o_{0}}{q^{3}}}{C_{0}} \times 100=99.6 \%\)

Hence option D is correct.
Question 3
3 / -1

The brown ring test for NO₂and NO₃ is due to the formation of complex ion with formula:

A
[Fe(H₂O)₆]²⁺

B
[Fe(NO)(CN)₅]²⁻

C
[Fe(H₂O)₅NO]²⁺

D
[Fe(H₂O)(NO)₅]²⁺

Solution

Option (C) is correct. The brown ring test for \(N O_{2}\) and \(N O_{3}\) is due to the formation of complex, \(\left[F e\left(H_{2} O\right)_{5} N O\right]^{2+}\) Brown ring test is the test to find out the presence of nitrate radical. For this equal volume of the freshly prepared \(F e S O_{4}\) is mixed with solution to be analysed. To this concentrated \(H_{2} S O_{4}\) is added along the sides of the test tube. A brown ring will forms at the junction of the two liquids. This is nitroso Ferrous Sulphate \(\left(F e S O_{4} \cdot N O\right)\) Nitric Acid (Nitrate) reduces Ferrous to Ferric and itself is reduced to nitrogen monoxide \(N O\). This reacts with FeSO \(_{4}\) and forms the brown ring.

Hence option C is correct.
Question 4
3 / -1

Reaction that takes place at graphite anode in dry cell is:

A
Zn²++2e⁻→Zn(s)

B
Zn(s)→Zn²++2e⁻

C
Mn²⁺+2e⁻→Mn(s)

D
Mn(s)→Mn⁺+e⁻+1.5V

Solution

In the dry cell, the oxidation of zinc occurs at graphite anode. Zinc loses two electrons to form Zn2+ ion. The electrode reaction is Zn→Zn²+2e⁻
Hence, correct option is B.
Question 5
3 / -1

Directions: The following question has four choices, out of which one or more is/are correct.

In the depression of freezing point experiment, it is found that the

A
vapour pressure of the solution is less than that of pure solvent

B
depression in the freezing point will be the same for a solution of glucose and that of NaCl, each 1 m in concentration

C
cryoscopic constant is the same for all substances at a constant temperature

D
none of these

Solution

Statement (1) is correct as the vapour pressure of a solvent decreases on addition of a non-volatile solute.

The colligative properties depend on the number of particles in the solution. Hence, the change will not be the same for the solution of an electrolyte and a non-electrolyte, even if they have equal concentrations. Hence, statement (2) is incorrect.

The cryoscopic constant is specific for each solvent and is a function of temperature. Hence, (3) is incorrect.

Hence option A is correct.
Question 6
3 / -1

The van't Hoff's factor for 0.1 M Ba(NO₃)2 solution is 2.74. The degree of dissociation is :

A
9103%

B
87%

C
100%

D
74%

Solution

The van't Hoffs factor for \(0.1 MBa \left( NO _{3}\right)_{2}\) solution is \(2.74 .\) The degree of dissociation is \(87 \% i=1+2 \alpha[\) for \(\left.B a\left(N O_{3}\right)_{3} \rightleftharpoons B a^{2+}+2 N O_{3}\right]\)
\(\therefore 2.74=1+2 \alpha\)
\(\therefore \alpha=\frac{1.74}{2}=0.87=87 \%\)
Hence option B is correct.
Question 7
3 / -1

Which of the following species is desirable substance in extraction of copper but not in extraction of iron?

A
CaSiO₃

B
FeSiO₃

C
SiO₂

D
coke

Solution

Haematite ore of iron contain Si as impurity where it reacts with O₂ to produce SiO₂ which is further used in formation of slag.
But in extraction of copper SiO₂ is needed to form slag.
Hence option C is correct.
Question 8
3 / -1

Directions: The following question has four choices, out of which ONE or MORE are correct.

The standard redox potentials E^o of the following systems are:

The oxidising powers of the various species are related as:

A
Ce⁴⁺ > Sn⁴⁺

B
Ce⁴⁺ > MnO^-₄

C
MnO^-₄ > Sn⁴⁺

D
All of the above

Solution

The higher the reduction potential, the higher the tendency to get reduced, and the higher the oxidising power.

Hence option D is correct.
Question 9
3 / -1

Galena (PbS) on heating in limited supply of air gives lead metal. This is known as:

A
smelting

B
calcination

C
self reduction

D
sulphatizing roasting

Solution

2PbS+3O₂→2PbO+2SO₂
PbS+2PbO→3Pb+SO₂(self reduction)
Pb,Cu and Hg are obtained by self reduction process.
Hence option C is correct.
Question 10
3 / -1

Directions: The following question has four choices out of which ONE or MORE is/are correct.

The aqueous solution of the following salts will be coloured in the case of

A
Zn(NO₃)₂

B
LiNO₃

C
Co(NO₃)₂

D
None of these

Solution

Co(NO₃)₂ have unpaired electron; hence, they are coloured. Zn(NO₃)₂, LiNO₃ and potash alum have no unpaired electron; hence, they are colorless.

Hence option C is correct.
Question 11
3 / -1

Directions: The following question has four choices, out of which one or more is/are correct.

Identify the binary mixture(s) that can be separated into individual compounds, by at least one of the reagents (NaOH or NaHCO3), as shown in the given scheme.

A
C₆H₅OH and C₆H₅COOH

B
C₆H₅CH₂OH and C₆H₅OH

C
C₆H₅CH₂OH and C₆H₅CH₂COOH

D
All of the above

Solution

(1) Both benzoic acid and phenol are soluble in NaOH, hence inseparable using this reagent.

However, only benzoic acid (C₆H₅COOH) is soluble in NaHCO3, while benzyl alcohol (C₆H₅CH₂OH) is not. Hence, the mixture can be separated using NaHCO₃.

(2) Both benzyl alcohol (C₆H₅CH₂OH) and phenol (C₆H₅OH) react with NaOH, but are not soluble in it. Both alcohols and phenol do not react with NaHCO₃ and cannot be separated using this reagent.

(3) -phenyl acetic acid (C₆H₅CH₂COOH) is soluble in NaOH and NaHCO₃. Benzyl alcohol (C₆H₅CH₂OH) is not. Hence, the mixture is separable with both the reagents.

Hence option D is correct.
Question 12
3 / -1

Rate of a reaction can be expressed by Arrhenius equation as, k=Ae^−E/RT . In this reaction, E represents _____________ .

A
the energy above which all the colliding molecules will react

B
the energy below which colliding molecules will not react

C
the total energy of the reacting molecules at a temperature T

D
the fraction of molecules with energy greater than the activation energy of the reaction

Solution

In the Arrhenius equation, k=Ae^−E/RT , E represents the activation energy. It is the minimum energy that the reacting molecules should possess so that the collision is effective. The colliding molecules will not react if their energy is lower than the activation energy.
Hence option B is correct.
Question 13
3 / -1

The hybridization of iodine in IF₅ is :

A
sp³d

B
sp³d²

C
sp³d³

D
sp³

Solution

Out of 7 valencies of iodine 5 is satisfied by 5 fluorine atoms. Thus, iodine has one lone pair of electrons remaining after the formation of IF₇.
Thus there are 7 electrons surrounding the central iodine out of which 5 are bond pairs and 1 is a lone pair. Thus IF₇shows sp³d² hybridization.
Hence option B is correct.
Question 14
3 / -1

Calamine is:

A
ZnS

B
PbCO₃

C
ZnCO₃

D
MgCO₃

Solution

(A) Zinc sulphide ZnS is Zinc blende.
(B) Lead carbonate (PbCO₃) is Cerrusite.
(C) Calamine is zinc carbonate (ZnCO₃)
(D) Magnesium carbonate M_gCO₃ is magnesite.
Hence option C is correct.
Question 15
3 / -1

During depression of freezing point in a solution, which of the following are in equilibrium?

A
Liquid solvent and solid solvent

B
Liquid solvent and solid solute

C
Liquid solute and solid solute

D
Liquid solute and solid solvent

Solution

During the depression of freezing point in a solution, liquid solvent and solid solvent are in equilibrium. During freezing of a solution, only the solvent freezes out and the equilibrium exists between solid and liquid forms of the solvent. Vapour pressure of the solid and liquid forms must be the same at freezing point, because otherwise, the system would not be at equilibrium, the lowering of the vapour pressure leads to the lowering of temperature at which the vapour pressures of the liquid and frozen forms of the solution will be equal.
Hence option A is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemistry Test - 6

Result

Chemistry Test - 6

Score

-

Rank

-

SHARING IS CARING

Question 1

ps=p0B.XA

p0=pBXB

ps=p0B.XB

p0B=psXA

Solution

Question 2

the concentration of the reactant decreases exponentially with time

the half-life of the reaction decreases with increasing temperature

the reaction proceeds to 99.6% completion in eight half-life duration

All of the above

Solution

Question 3

[Fe(H2O)6]2+

[Fe(NO)(CN)5]2−

[Fe(H2O)5NO]2+

[Fe(H2O)(NO)5]2+

Solution

Question 4

Zn2++2e−→Zn(s)

Zn(s)→Zn2++2e−

Mn2++2e−→Mn(s)

Mn(s)→Mn++e−+1.5V

Solution

Question 5

vapour pressure of the solution is less than that of pure solvent

depression in the freezing point will be the same for a solution of glucose and that of NaCl, each 1 m in concentration

cryoscopic constant is the same for all substances at a constant temperature

none of these

Solution

Question 6

9103%

87%

100%

74%

Solution

Question 7

CaSiO3

FeSiO3

SiO2

coke

Solution

Question 8

Ce4+ > Sn4+

Ce4+ > MnO-4

MnO-4 > Sn4+

All of the above

Solution

Question 9

smelting

calcination

self reduction

sulphatizing roasting

Solution

Question 10

Zn(NO3)2

LiNO3

Co(NO3)2

None of these

Solution

Question 11

C6H5OH and C6H5COOH

C6H5CH2OH and C6H5OH

C6H5CH2OH and C6H5CH2COOH

All of the above

Solution

Question 12

the energy above which all the colliding molecules will react

the energy below which colliding molecules will not react

the total energy of the reacting molecules at a temperature T

the fraction of molecules with energy greater than the activation energy of the reaction

Solution

p_s=p⁰_B.X_A

p⁰=p_BX_B

p_s=p⁰_B.X_B

p⁰_B=p_sX_A

[Fe(H₂O)₆]²⁺

[Fe(NO)(CN)₅]²⁻

[Fe(H₂O)₅NO]²⁺

[Fe(H₂O)(NO)₅]²⁺

Zn²++2e⁻→Zn(s)

Zn(s)→Zn²++2e⁻

Mn²⁺+2e⁻→Mn(s)

Mn(s)→Mn⁺+e⁻+1.5V

CaSiO₃

FeSiO₃

SiO₂

Ce⁴⁺ > Sn⁴⁺

Ce⁴⁺ > MnO^-₄

MnO^-₄ > Sn⁴⁺

Zn(NO₃)₂

LiNO₃

Co(NO₃)₂

C₆H₅OH and C₆H₅COOH

C₆H₅CH₂OH and C₆H₅OH

C₆H₅CH₂OH and C₆H₅CH₂COOH

sp³d

sp³d²

sp³d³

sp³

PbCO₃

ZnCO₃

MgCO₃