Chemistry Test

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Chemistry Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Calculate the ratio of $\mathrm{HCOO}^{-}$ and $\mathrm{F}^{-}$ in a mixture of $0.2 \mathrm{M} \mathrm{HCOOH}\left(\mathrm{K}_{\mathrm{g}}=2 \times 10^{-4}\right)$ and $0.1 \mathrm{M} \mathrm{HF}\left(\mathrm{K}_{\mathrm{a}}=6.6 \times 10^{-4}\right):$

A
1 : 6.6

B
1 : 3.3

C
2 : 3.3

D
3.3 : 2

Solution
Question 2
1 / -0

Which of the following is soluble in water?

A
CS₂

B
C₂H₅OH

C
CCl₄

D
CHCl₃

Solution

Ethanol is soluble in water due to its ability to form intermolecular H-bonds with water:

Hence, the correct option is (B)
Question 3
1 / -0

When gold is dissolved in aqua - regia which of the following is formed :

A
Aurous chloride

B
Chloroauric acid

C
Auric chloride

D
None of these

Solution

When gold is dissolved in aqua regia, it forms chloroauric acid ( $H A u C l_{4}$ ). The chemical reaction

shows gold reacting with both components of aqua regia: nitric acid $\left(H N O_{3}\right)$ and hydrochloric

acid $(H C l)$

In this reaction gold is oxidised.

$A u+3 H N O_{3}+4 H C l \longrightarrow 2 H A u C l_{4}+3 N O_{2}+3 H_{2} O$
Hence, the correct option is (B)
Question 4
1 / -0

When the applied voltage is 10,000 V, the de-Broglie wavelength of an electron is (Planck’s constant: h = 6.626 × 10^–34Js , Mass of electron m_e : = 9. 1 × 10^–31Kg)

A
0.0123 Å

B
12.3 Å

C
0.123 Å

D
1.23 Å

Solution

$\lambda=\sqrt{\frac{150}{V}}=\sqrt{\frac{150}{10,000}}=\sqrt{0.015}$
$V$ is accelerating voltage $\&$
$\lambda \operatorname{in} A$
$\sqrt{\left(1.5 \times 10^{-2}\right)}=1.23 \times 10^{-1} \mathrm{A}$
$=0.123 $
Hence, the correct option is (C)

Question 5

1 / -0

Among the electrolytes Na₂SO₄ CaCl₂, Al₂ (SO₄)₃ and NH₄Cl, the most effective coagulating agent for Sb₂S₃ sol is

Na₂SO₄

CaCl₂

Al₂ (SO₄ )₃

NH₄Cl

Solution

Sb₂S₃ is a negative (anionic) sol. According to Hardy Schulze rule, greater the valency of cationic coagulating agent, higher its coagulating power. Therefore, Al₂ (SO₄ )₃ will be the most effective
coagulating agent in the present case.
Concepts :
Main Concept :
Types of Sols,- Lyophilic and Lyophobic Soli) Lyophilic sol Dispersion medium has high affinity for dispersed phase.

ii) Lyophobic sol Dispersion medium has very less affinity (hate) for dispersed phase.

Lyophilic Sols	Lyophobic Soles
Reversible in nature.	Irreversible in nature.
Quite stable, nor easily coagulated by addition of electrolytes.	Unstable, coagulated easily by addition of a small amount of electrolyte.
Obtained mainly from organic materials such as starch, gelatin etc.	Obtained mainly from inorganic materials such as metals, metal sulphides, metal hydroxides etc.
Self - stabilized.	Require stabilizer (protective agents).
Particles are heavily solvated, have high viscosity.	Particles are less solvated, have low viscosity.

Hence, the correct option is (C)

Question 6
1 / -0

Wave number of a spectral line for a given transition is x cm^–1 for He⁺, then its value for Be³⁺ (isoelectronic of He⁺) for same transition is :

A
x cm^–1

B
4x cm^–1

C
$\frac{x}{4} \mathrm{cm}^{-1}$

D
2x cm^–1

Solution

$\bar{v}($ wavenumber $)=\bar{R}_{H} Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \bar{v}_{1}(\mathrm{He}, Z=2)=\bar{R}_{H}(2)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
$\bar{\nu}_{2}\left(\mathrm{Be}^{3} Z=4\right)=\mathrm{R}_{\mathrm{H}}(4)^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right] \frac{\bar{v}_{2}}{v_{1}}=4 \therefore \bar{v}_{2}=4 \bar{v}_{1}=4 x$
hence $4 \mathrm{x} \mathrm{cm}^{-1}$
Hence, the correct option is (B)
Question 7
1 / -0

When salicylic acid is heated with acetic anhydride, we get :

A
Aspirin

B
Salol

C
Paracetamol

D
None of these

Solution

Acetyl salicylic acid is known as aspirin.

Concepts :
Main Concept :
Examples on Preparation of esters by Fischer esterification

The Lewis or Brønsted acid-catalyzed esterification of carboxylic acids with alcohols to give esters is a typical reaction in which the products and reactants are in equilibrium.

The equilibrium may be influenced by either removing one product from the reaction mixture (for example, removal of the water by azeotropic distillation or absorption by molecular sieves) or by employing an excess of one reactant.

Mechanism of the Fischer Esterification

Addition of a proton $(e . g . : p - T s O H, H_{2} S O_{4})$ or a Lewis acid leads to a more reactive electrophile. Nucleophilic attack of the alcohol gives a tetrahedral intermediate in which there are two equivalent hydroxyl groups. One of these hydroxyl groups is eliminated after a proton shift (tautomerism) to give water and the ester.

Alternative reactions employ coupling reagents such as DCC (Steglich Esterification), preformed esters (transesterification), carboxylic acid chlorides or anhydrides (see overview). These reactions avoid the production of water. Another pathway for the production of esters is the formation of a carboxylate anion, which then reacts as a nucleophile with an electrophile (similar reactions can be found here). Esters may also be produced by oxidations, namely by the Baeyer-Villiger oxidation andoxidative esterifications.

Alternative reactions employ coupling reagents such as DCC (Steglich Esterification), preformed esters (transesterification), carboxylic acid chlorides or anhydrides (see overview). These reactions avoid the production of water. Another pathway for the production of esters is the formation of a carboxylate anion, which then reacts as a nucleophile with an electrophile (similar reactions can be found here). Esters may also be produced by oxidations, namely by the Baeyer-Villiger oxidation and oxidative esterifications.

Hence, the correct option is (A)
Question 8
1 / -0

The percentage dissociation of a 0.011 m aqueous solution of K₃[Fe(CN)₆] which freezes at - 0.063^oC is (K_f for water = 1.86^oC m^-1)

A
75%

B
67%

C
33%

D
50%

Solution

$\Delta \mathrm{T}_{f}=\mathrm{m} \times \mathrm{K}_{f}=0.011 \times 1.86=0.021$
$\Delta \mathrm{T}_{f}($ calculated $)=0.021^{\circ} \mathrm{C}$
$\Delta \mathrm{T}_{f}($ observed $)=0.063^{\circ} \mathrm{C}$
$\mathrm{i}=\frac{\text { Observed } \Delta \mathrm{T}_{f}}{\text { Calculated } \Delta \mathrm{T}_{f}}=\frac{0.063}{0.021}=3$
Degree of dissociation, $\alpha=\frac{i-1}{n-1}$
$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] 3 \mathrm{K}^{+}+\left[\mathrm{Fe}\left(\mathrm{CN}_{6}\right)\right]^{3-}$
thus $n=4$
$\alpha=\frac{3-1}{4-1}$
Percent dissociation $=67 \%$
Hence, the correct option is (B)
Question 9
1 / -0

The compressibility of a gas is less than unity at STP.
Therefore,

A
V_m = 22.4 L

B
V_m = 44.8 L

C
V_m > 22.4 L

D
V_m < 22.4 L

Solution

$Z=\mathrm{d} \frac{\left(\mathrm{V}_{\text {oberved }}\right)_{\mathrm{STP}}}{\left(\mathrm{V}_{\mathrm{m}}\right)_{\mathrm{STP}}}$ or $\frac{\mathrm{V}_{\text {obverved }}}{\mathrm{V}_{\text {Idesl }}}$
$$
Z<1 \quad \therefore \frac{V_{\text {oberved }}}{V_{\text {Ideal }}}<1
$$
$\mathrm{V}_{\text {observed }}<22.4 \mathrm{L}$ at STP.
Real Gas Gases that do not obey pV = nRT equation of state are known as real gas
$$
\text { Compression factor }(Z)=\frac{\mathrm{v}}{\mathrm{v}(\text { ideal })}=\frac{\mathrm{PV}}{\mathrm{nRT}}
$$
For ideal gas, $Z=1$ and for non-ideal gas (real gas), $Z \neq 1$ If $Z<1,$ there exist net attraction between the gas molecules. If $Z>1,$ there exist net repulsion between the gas molecules.
Hence, the correct option is (D)
Question 10
1 / -0

An acid-base indicator which is a weak acid has a pK_a value=5.35. At what concentration ratio of sodium acetate to acetic acid would the indicator show a colour half-way between those of its acid and conjugate base forms? pK_a of acetic acid =4.75. [log2=0.3].

A
4:1

B
7:1

C
5:1

D
2:1

Solution

The indicator is a weak acid. its dissociation equilibrium is as given below. $H I n \rightleftharpoons H^{+}+I n^{-}$
The expression for the dissociation constant $K_{a}$ is $K_{a}=\frac{\left[H^{+}\right]\left[I n^{-}\right]}{[H I n]}$
Colour is observed at halfway of ionization. $\left[I n^{-}\right]=[H I n]$ Hence $K_{a}=\left[H^{+}\right]$ or $p H=p K_{a}=5.35$
The expression for the pH of buffer containing acetic acid and sodium acetate is $p H=p K_{a}+\log \frac{[S a l t]}{[A c i d]}$
Substitute values in this expression. $5.35=4.75+\log \frac{[\text {Salt}]}{[A c i d]} \Rightarrow \frac{[S a l t]}{[A c i d]}=\frac{4}{1}$
Hence, $5.35=4.75+\log \frac{[\text {Salt}]}{[A c i d]} \Rightarrow \frac{[\text {Salt}]}{[A c i d]}=\frac{4}{1}$
Hence, the correct option is (A)

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Chemistry Test - 39

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Chemistry Test - 39

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Question 1

1 : 6.6

1 : 3.3

2 : 3.3

3.3 : 2

Solution

Question 2

CS2

C2H5OH

CCl4

CHCl3

Solution

Question 3

Aurous chloride

Chloroauric acid

Auric chloride

None of these

Solution

Question 4

0.0123 Å

12.3 Å

0.123 Å

1.23 Å

Solution

Question 5

Na2SO4

CaCl2

Al2 (SO4 )3

NH4Cl

Solution

Question 6

x cm–1

4x cm–1

\(\frac{x}{4} \mathrm{cm}^{-1}\)

2x cm–1

Solution

Question 7

Aspirin

Salol

Paracetamol

None of these

Solution

Question 8

75%

67%

33%

50%

Solution

Question 9

Vm = 22.4 L

Vm = 44.8 L

Vm > 22.4 L

Vm < 22.4 L

Solution

Question 10

4:1

7:1

5:1

2:1

Solution

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CS₂

C₂H₅OH

CCl₄

CHCl₃

Na₂SO₄

CaCl₂

Al₂ (SO₄ )₃

NH₄Cl

x cm^–1

4x cm^–1

2x cm^–1

V_m = 22.4 L

V_m = 44.8 L

V_m > 22.4 L

V_m < 22.4 L