Chemistry Test

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Chemistry Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

A
3.62 x 10²

B
1.00

C
9.94 x 10^-1

D
2.80 x 10^-3

Solution
Question 2
1 / -0

If equal masses of oxygen and nitrogen are placed in separate containers of equal volume at the same temperature, which one of the following statements is true? (MW of N₂=28,O₂=32)

A
Both flasks contain the same number of molecules

B
The pressure in the nitrogen flask is greater than in the oxygen flask

C
More molecules are present in the oxygen flask

D
Molecules in the oxygen flask are moving faster on the average than the ones in the nitrogen flask

Solution

The ideal gas equation is PV=nRT=(m/M)RT.
Here, m is the mass and M is the molecular weight.
For oxygen and nitrogen, the volume, mass and temperature are same.
Hence their pressures are inversely proportional to their molecular weights.
The molecular weight of nitrogen (28) is smaller than that of oxygen (32).
Hence, the pressure in the nitrogen flask is greater than in the oxygen flask.
Hence, the correct option is (B)
Question 3
1 / -0

A proton is accelerated to one-tenth of the velocity of light. If its velocity can be measured with a precision ±1%, then its uncertainty in position is:

A
1.05×10⁻¹³m

B
1.13×10⁻¹³m

C
1.15×10⁻¹³m

D
1.25×10⁻¹³m

Solution

Uncertainty in position is given by \(\Delta x=\frac{h}{4 \pi m \Delta v}\)
\(\Delta v=\frac{c}{10}=3 \times 10^{7} m\)
\(\Delta x=\frac{6.62 \times 10^{-34}}{4 \pi \times 1.672 \times 10^{-27} \times 3 \times 10^{7}}=1.05 \times 10^{-13} \mathrm{m}\)
Hence, the correct option is (A)
Question 4
1 / -0

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5×10⁻¹mol L⁻¹.
Pcl₅(g)⇌PCl₃(g)+Cl2(g),KC=8.3×10⁻³ mol L⁻¹

A
0.08 M

B
0.04 M

C
0.02 M

D
0.02 M

Solution

Let \(x M\) be the equilibrium concentration of \(P C l_{3}\) The equilibrium reaction is shown below. \(P C l_{5}(g) \rightleftharpoons P C l_{3}(g)+C l_{2}(g)\)
The equilibrium concentrations of \(P C l_{5}, P C l_{3}\) and \(C l_{2}\) are \(0.5 \times 10^{-2} M, x\) and \(x\) respectively. The expression of the equilibrium constant is \(K_{c}=\frac{\left[P C l_{3}\right]\left[C l_{2}\right]}{\left[P C l_{3}\right]}\) Substitute values in the above expression. \(8.3 \times 10^{-3}=\frac{x^{2}}{0.5 \times 10^{-1}}\)
\(x^{2}=4.15 m \times 10^{-4}\)
\(x \simeq 0.02 M\)
Thus, option \(D\) is the correct answer.
Question 5
1 / -0

Chooe the correct amswer using given codes. Use T if statement is true and F if it is false.
(I) Sulphide ions reacts with Na₂[Fe(CN)₅(NO)] to form a purple coloured compound Na₄[Fe(CN)₅(NOS)]. In the reaction, the oxidation state of iron changes.
(II) Pt(IV) compounds are relatively more stable than Ni(IV) compounds.
(III) The welding of magnesium can be done in the atmosphere of helium.
(IV) LiAIH₄ on hydrolysis will give H₂.

A
FFTT

B
FTTT

C
TFTF

D
TFTT

Solution

(I) Oxidation state of iron in \(N a_{2}\left[F e(C N)_{5}(N O)\right]\) is +3 Oxidation state of iron in \(N a_{4}\left[F e(C N)_{5}(N O S)\right]\) is +3 In this reaction, the oxidation state of iron does not change. Thus, statement is false ( \(F\) ).
(II) Going down in the group, the ionization energy required to remove 4 electrons from \(N i\) is higher than required for formation of \(P t^{+4} .\) Thus \(P t(I V)\) compounds are more stable and \(N i(I V)\) compounds do not exist. This is due to the presence of shielding effect of \((n-1) d\) sub shell electrons. Thus, statement is true ( \(T\) ).
(III) The welding of magnesium can be done in the atmosphere of helium. Due to higher ionization potential of \(H e\) inert gas, it produces hotter arc at higher voltage, providing wide deep bead; this is an advantage for magnesium alloys. Thus, statement is true ( \(T\) ).
(IV) \(L i A I H_{4}\) on hydrolysis will give \(H_{2} .\) It reacts violently with water by producing hydrogen gas. Hence it should not be exposed to moisture and the reactions are performed in inert and dry atmosphere. \(L i A l H_{4}+4 H_{2} O \longrightarrow L i O H+A l(O H)_{3}+4 H_{2}\)
Thus, statement is true ( \(T\) ).
Hence, the correct option is (B)
Question 6
1 / -0

Of the following statements :
(P)C₆H₅N=CH−C₆H₅ is a Schiff's base.
(Q)A dye is obtained by the reaction of aniline and C₆H₅N=NCl.
(R)C₆H₅CH₂NH₂ on treatment with [NaNO₂+HCl] gives diazonium salt.
(S)p−Toluidine on treatment with [HNO₂+HCl] gives diazonium salt.

A
only (P)and(Q) are correct

B
only (P)and(R)(P)and(R) are correct

C
only (R)and(S) are correct

D
(P),(Q)and(S) are correct

Solution

A Schiff base is a substituted imine(R₂C=N−R′). It is used as antioxidant. Thus (P)C₆H₅N=CH−C₆H₅ is a Schiff's base. Thus statment (P) is correct.
Aniline reacts with C₆H₅N=NCl to form higly coloured azo dye. This reaction is known as azo coupling. Thus the statement (Q) is correct.
C₆H₅CH₂NH₂ is a primary aliphatic amine. When it is treated with [NaNO₂+HCl], it gives unstable diazonium salt which decomposes to form benzyl alcohol. Hence, statement (R) is incorrect.
p-Toluidne is an aromatic primary amine. Hence, it on treatment with [NaNO₂+HCl] gives diazonium salt. Hence, statement (S) is correct.
Hence, the correct option is (D)
Question 7
1 / -0

Chloride samples are prepared for analysis by using NaCl,KCl and NH₄Cl separately or as mixture. What minimum volume of 5% by weight AgNO₃ solution (specific gravity =1.02gmL−1) must be added to a sample of 0.321 g, in order to ensure complete precipitation of chloride in every possible case?

A
20 mL

B
30 mL

C
40 mL

D
50 mL

Solution

\(0.321 \mathrm{g}\) of a sample of \(\mathrm{NH}_{4} \mathrm{Cl}\) contains \(\frac{0.321}{53.5}=0.006 \mathrm{mol},\) so it will provide 0.006 moles of chloride ion. \(A g N O_{3}+N H_{4} C l \rightarrow A g C l\)
For complete precipitation, moles of \(A g N O_{3}\) required is 0.006 mol. Let, \(V\) mL of solution is required, so, \(V \times 1.02 \times \frac{0.05}{160}=0.006 \Longrightarrow V=20\) \(\mathrm{mL}\)
Hence, the correct option is (A)
Question 8
1 / -0

Bronsted-Lowry acid among the following is :

A
[Cu(NH₃)₄]²⁺

B
[FeCl₄]⁻

C
[Fe(H2O)₆]³⁺

D
[Zn(OH)₄]²⁻

Solution

Option (C) is correct. \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is an example of Bronsted-Lowry acid.
When a coordinate covalent bond is formed between a metal cation and a water molecule, the positive charge of the metal ion and its small size means that the electron of the \(H_{2} O \rightarrow M^{n+}\) bond are strongly attracted to the metal. As a result, the \(O-H\) bonds of the bound water molecules are polarized. The net effect is that a hydrogen atom of the coordinated water molecule is removed as \(H^{+}\) more readily than in an uncoordinated water molecule. Thus, a hydrated metal cation functions as a Bronsted acid or proton donor.
\(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})\right]^{3+}+\mathrm{H}_{3} \mathrm{O}^{+}\)
Hence, the correct option is (C)
Question 9
1 / -0

ZnCO₃ is thermally more stable than MgCO₃ because:

A

Mg(OH)2exhibits only basic properties while Zn(OH)2 is amphoteric

B
polarising action of Zn2+ with 18-electron configuration on the anion is larger than that of Mg2+ with a noble-gas electron configuration and of same size and charge.

C
both (a) and (b) are correct

D
none of the above is correct

Solution

ZnCO₃ is thermally more stable than MgCO₃because polarising action of Zn²⁺ with 18-electron configuration on the anion is larger than that of Mg²⁺ with a noble-gas electron configuration and of same size and charge. Higher is the polarizing power, higher will be the lattice enthalpy and higher will be the thermal stability.
Hence, the correct option is (B)
Question 10
1 / -0

The IUPAC name of the given compound is:

A
Bicyclo [2.2.2] heptane

B
Bicyclo [2.0.2] octane

C
Bicyclo [2.2.0] nonane

D
Bicyclo [2.2.2] octane

Solution

1. Start with one of the bridgehead carbons and number it 1 .
2. Proceed round the longest chain of carbons to the second bridgehead.
3. Number the second bridgehead carbon and continue on round the next longest chain of carbons back towards the first bridgehead carbon.
4. Pass over the first bridgehead carbon (it already has the number 1 ) and along the shortest chain of carbons to the second bridgehead carbon again.
So here total carbon atoms are 8 , hence octane and the IUPac name will be Bicyclo \([2.2 .2]\) octane.
Option D is correct.
Hence, the correct option is (D)

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Chemistry Test - 36

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Chemistry Test - 36

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Question 1

3.62 x 102

1.00

9.94 x 10-1

2.80 x 10-3

Solution

Question 2

Both flasks contain the same number of molecules

The pressure in the nitrogen flask is greater than in the oxygen flask

More molecules are present in the oxygen flask

Molecules in the oxygen flask are moving faster on the average than the ones in the nitrogen flask

Solution

Question 3

1.05×10−13m

1.13×10−13m

1.15×10−13m

1.25×10−13m

Solution

Question 4

0.08 M

0.04 M

0.02 M

0.02 M

Solution

Question 5

FFTT

FTTT

TFTF

TFTT

Solution

Question 6

only (P)and(Q) are correct

only (P)and(R)(P)and(R) are correct

only (R)and(S) are correct

(P),(Q)and(S) are correct

Solution

Question 7

20 mL

30 mL

40 mL

50 mL

Solution

Question 8

[Cu(NH3)4]2+

[FeCl4]−

[Fe(H2O)6]3+

[Zn(OH)4]2−

Solution

Question 9

Mg(OH)2exhibits only basic properties while Zn(OH)2 is amphoteric

polarising action of Zn2+ with 18-electron configuration on the anion is larger than that of Mg2+ with a noble-gas electron configuration and of same size and charge.

both (a) and (b) are correct

none of the above is correct

Solution

Question 10

Bicyclo [2.2.2] heptane

Bicyclo [2.0.2] octane

Bicyclo [2.2.0] nonane

Bicyclo [2.2.2] octane

Solution

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3.62 x 10²

9.94 x 10^-1

2.80 x 10^-3

1.05×10⁻¹³m

1.13×10⁻¹³m

1.15×10⁻¹³m

1.25×10⁻¹³m

[Cu(NH₃)₄]²⁺

[FeCl₄]⁻

[Fe(H2O)₆]³⁺

[Zn(OH)₄]²⁻