Chemistry Test

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Chemistry Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product are respectively

A
Redox reaction; -3and -5

B
Redox reaction; 3 and 5

C
Disproportionation reaction; -3 and 1

D
Disproportionation reaction; -3 and 3

Solution

The balanced disproportionationreaction involving white phosphorus with aqueousNaOH is
at low concentration of NaOH

Hence, the correct option is (C)
Question 2
1 / -0

Propyne and propene can be distinguished by

A
Concentrated H₂SO₄

B
Br₂ in CCl₄

C
Dilute KMnO₄

D
ammonical AgNO₃

Solution

The terminal hydrogen is acidic inH₃C-C ≡ CH(propyne) and it reacts with ammoniacal AgNO₃. In propene, CH₃CH = CH₂ there is no acidic hydrogen.

Concepts :
Main Concept :
Acetylides formation of external alkynes with Tollen's reagentSilver acetylide is obtained as a white precipitate when acetylene is passed through ammoniacal silver nitrate (Tollens reagent)
Hence, the correct option is (D)
Question 3
1 / -0

A

B

C

D

Solution

In any reaction mechanism the no. of activated complexes showing maxima of potential energies is the no. of steps in that mechanism and in between these maxima is a valley where lies a more stable intermediate. There is one reaction intermediate lying at valley of two maxima representing two steps. E_a for first step should be higher than second step as first is slow step. These requirements fulfilled by choice (ii) & not choice (i) the choice (iv) is not possible because that is for endothermic reaction $(∆ H v e)$ & (ii) is single step equation so not possible.
Concepts :
Main Concept :
Reaction mechanism potential energy v/s time diagramA potential-energy profile is a diagram used to describe the mechanism of a reaction. This diagram is used to better illustrate the concepts of activation energy and the arrhenius equation, as well as to show the changing potential energy between the reactant and product that occur during a chemical reaction.

1. The potential energy of the reactant.
2. The difference in potential energy between the reactant and the activated complex (also known as the activation energy of the forward reaction).
3. The potential energy of the activated complex (the transition state).
4. The energy difference between the product and the activated complex.
5. The difference in potential energy between the reactant and the product.
6. The potential energy of the product.
Hence, the correct option is (B)
Question 4
1 / -0

For a monatomic gas kinetic energy = E, the relation with rms velocity is

A
$u=\left(\frac{3 E}{2 m}\right)^{1 / 2}$

B
$u=\left(\frac{E}{2 m}\right)^{1 / 2}$

C
$u=\left(\frac{E}{3 m}\right)^{1 / 2}$

D
$u=\left(\frac{2 E}{m}\right)^{1 / 2}$

Solution

The expression for the kinetic energy of a monoatomic gas is $(E)=\frac{3}{2} k T$
The expression for the kinetic energy is $=\sqrt{\frac{3 k T}{m}}$
Thus, the rms velocity is given as:
$u=\sqrt{\frac{2 E}{m}}$
Hence, the correct option is (D)
Question 5
1 / -0

In Langmuir's model of adsorption of a gas on a solid surface

A
The rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered

B
The adsorption at a single site on the surface may involve multiple molecules at the same time

C
The mass of gas striking a given area of surface is proportional to the pressure of the gas

D
The mass of gas striking a given area of surface is independent of the pressure of the gas

Solution

Assuming the formation of a monolayer of the adsorbate on the surface of the adsorbant, it was derived by Langmuir that the mass of the gas adsorbed per gram of the adsorbant is related to the equilibrium pressure according to the equation. $\frac{x}{m}=\frac{a p}{(1+b p)}$
$\mathrm{x}=$ mass of the gas adsorbed on $\mathrm{m}$ gram of the adsorbent. $\mathrm{p}=$ Pressure $, \mathrm{a}$ and $\mathrm{b}$ are constant.
Hence, the correct option is (C)
Question 6
1 / -0

An electron moving with velocity 'v' is found to have a certain value of de-Broglie wavelength. The velocity to be possessed by the neutron to have the same de-Broglie wavelength is-

A
$\frac{1840}{v}$

B
1840v

C
$\frac{v}{1840}$

D
v

Solution

$\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}$ and $\lambda_{\mathrm{m}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{a}} \mathrm{v}_{\mathrm{n}}}$
$\therefore \quad \lambda_{\mathrm{e}}=\lambda_{\mathrm{n}}$
$\therefore \quad \frac{1}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{e}}=\frac{1}{\mathrm{m}_{\mathrm{a}} \mathrm{v}_{\mathrm{n}}}$
or $\frac{\mathrm{u}_{\mathrm{u}}}{\mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{n}}}=\frac{1}{1840}$
or $\mathrm{v}_{\mathrm{n}}=\frac{1}{1840} \times \mathrm{v}_{\mathrm{e}}=\frac{\mathrm{v}}{1840}$
Hence, the correct option is (C)
Question 7
1 / -0

$x \mathrm{CIO}_{4}^{-}+y \mathrm{MnO}_{2}+z \mathrm{OH}^{-} \longrightarrow x \mathrm{CIO}^{-}+y \mathrm{MnO}_{4}^{2-}+\frac{z}{2} \mathrm{H}_{2} \mathrm{O}$

In this balanced equation x, y, z are-

A
2,3,6

B
1,3,4

C
1,3,6

D
2,6,6

Solution

$6 H^{\oplus}+C l O_{4}^{-} \rightarrow C l O^{-}+3 H_{2} O \ldots \ldots$
$2 H_{2} O+M n O_{4}^{-} \rightarrow M n O_{4}^{-2}+4 H^{\oplus}+2 e^{-} \ldots \ldots(2)$
$e q^{n}(2)$ multiply with 3 and adding with eq $^{n}(1)$ we get
$3 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{MnO}_{2}+\mathrm{ClO}_{4}^{-} \rightarrow \mathrm{ClO}^{-}+3 \mathrm{MnO}_{4}^{-2}+6 \mathrm{H}^{\oplus}$
$a d d 6 O H^{-}$ in both side $e$
$3 \mathrm{MnO}_{2}+\mathrm{ClO}_{4}^{-}+6 \mathrm{OH}^{-} \rightarrow \mathrm{ClO}^{-}+3 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{MnO}_{2}^{-2}$
$\operatorname{So} x=1$
$y=3$
$z=6$
Hence, the correct option is (C)
Question 8
1 / -0

In Cannizzaro’s reaction, the intermediate which is the best hydride donor is-

A

B

C

D

Solution

Dioxoanion is better hydride donor. Electron donating group at ortho/para position further promote H^- transfer.

Concepts :
Main Concept :
Preparation of Carboxylic Acids from Cannizzaro's ReactionCannizzaro reaction
The Cannizzaro reaction, named after its discoverer Stanislao Cannizzaro, is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the alpha position.

The oxidation product is a salt of a carboxylic acid and the reduction product is an alcohol.
Mechanism
The reaction involves a nucleophilic acyl substitution on an aldehyde, with the leaving group concurrently attacking another aldehyde in the second step. First, hydroxide attacks a carbonyl. The resulting tetrahedral intermediate then collapses, re-forming the carbonyl and transferring hydride to attack another carbonyl. In the final step of the reaction, the acid and alkoxide ions formed exchange a proton. In the presence of a very high concentration of base, the aldehyde first forms a doubly charged anion from which a hydride ion is transferred to the second molecule of aldehyde to form carboxylate and alkoxide ions. Subsequently, the alkoxide ion acquires a proton from the solvent.

Overall, the reaction follows third-order kinetics. It is second order in aldehyde and first order in base:
Rate $=\mathrm{k}[\mathrm{RCHO}]^{2}\left[\mathrm{OH}^{-}\right]$
At very high base a second path (k') becomes important that is second order in base:
Rate=k[RCHO]2[OH−]+k'[RCHO]2[OH−]²

The k' pathway implicates a reaction between the doubly charged anion $\left(\mathrm{RCHO}_{2}^{2-}\right)$and the aldehyde. The direct transfer of hydride ion is evident from the observation that the recovered alcohol does not contain any deuterium attached to the -carbon when the reaction is performed in the presence of D₂O.
Hence, the correct option is (D)
Question 9
1 / -0

A piston is cleverly designed so that it extracts the maximum amount of work out of a chemical reaction, by matching P_external to the P_internal at all times. This 8cm diameter piston initially holds back 1 mol of gas occupying 1 L, and comes to rest after being pushed out a further 2 L at 25^oC.After exactly half of the work has been done, the piston has travelled out a total of-

A
10.0 cm

B
11.2 cm

C
14.5 cm

D
20.0 cm

Solution

$v_{i}=1$ lit
$v_{f}=1+2=3$ lit
$\mathbf{W}=-\mathbf{n} \mathbf{R} \mathbf{T} \ln \frac{3}{1}=-2 \cdot 303 \mathbf{R} \mathbf{T} \log 3$
$-\frac{w}{2}=-\frac{2 \cdot 303 R T}{2} \log 3=-2 \cdot 303 \mathrm{RT} \log \frac{\mathrm{V}}{1}$
$\frac{1}{2} \log 3=\log \mathrm{V}(\mathrm{V}$ is new final volume $)$
$\log \sqrt{3}=\log \mathrm{V}$
$\therefore \mathrm{V}=1 \cdot 732 \mathrm{L}$
Piston change in volume $=1 \cdot 732-1 \cdot 000$
$=0 \cdot 732 \mathrm{L}=732 \mathrm{cm}^{3}$
V of cylinder $=\pi r^{2} h$
$732 \mathrm{cm}^{3}=3 \cdot 142 \times(4)^{2} \times \mathrm{h}$
$\frac{732}{50-24} \approx 14 \cdot 5 \mathrm{cm}$
Hence, the correct option is (C)
Question 10
1 / -0

Which of the following substance can be used as a raw material for obtaining alcohol?

A
Potatoes

B
Molasses

C
Maize

D
All of the above

Solution

All of them contain starch in certain quantities which can be fermented to give alcohol.
Hence, the correct option is (D)