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Chemistry Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

Chlorobenzene is prepared commercially by-

A
Grignard reaction

B
Raschig process

C
Wurtz-Fittig reaction

D
Friedel Crafts reaction

Solution
Question 2
1 / -0

Which one of the following cannot be prepared from B₂H₆ ?

A
NaBH₄

B
B₂(CH₃)₄H₂

C
B₂(CH₃)₆

D
H₃BO₃

Solution

$B_{2} H_{6}+3 O_{2} \rightarrow B_{2} O_{3}+3 H_{2} O+$ Heat
$B_{2} H_{6}+6 H_{2} O \rightarrow H_{3} B O_{3}+6 H_{2}$
$2 N a H+B_{2} H_{6} \rightarrow$ ether $N a B H_{4}$
From $B_{2} H_{6}$, all can be prepared except $B_{2}\left(C H_{3}\right)_{6}$
Hene option C is the correct answer.
Question 3
1 / -0

The electron energy in hydrogen atom is given by$\mathrm{E}_{\mathrm{n}}=-\frac{21 \cdot 7 \times 10^{-12}}{\mathrm{n}^{2}} \mathrm{erg}$. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength (in cm) of light that can be used to cause this transition?

A
$3.66 \times 1 0^{- 5}$

B
$3.00 \times 1 0^{- 5}$

C
$2.66 \times 1 0^{- 5}$

D
$2.00 \times 1 0^{- 5}$

Solution

As the electron is to be ejected from the atom, the final state will have 0 energy as $n_{2}$ tends to infinity. Hence, the amount of energy required for $t$
$\Delta E=21.7 \times 10^{-12}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \operatorname{erg}$
$=\frac{21.7}{4} \times 10^{-12} \mathrm{erg}$
$=5.425 \times 10^{-12} \mathrm{erg}$
1 erg $=10^{-7}$ joule
The longest wavelength that can cause above transition can be determined as :
$\begin{aligned} \lambda &=\frac{h c}{\Delta E}=\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{5.425 \times 10^{-12} \times 10^{-7}} \\ &=3.66 \times 10^{-7} \mathrm{m}=3.6 \times 10^{-5} \mathrm{cm} \end{aligned}$
Concepts :
Main Concept :
Bohr's ModelBohr Model:
Niels Bohr applied classical mechanics, electromagnetism and Planck’s quantum theory to modify the Rutherford’s model and proposed his atomic model in 1913. His model gives satisfactory explanation of stable atomic structure and emission of line spectra by hydrogen atom. He presented his theory in the form of 3 postulates.
Hence, the correct option is (A)
Question 4
1 / -0

Metal $(\mathrm{M})+\operatorname{air} \stackrel{\Delta}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow}(\mathrm{B}) \stackrel{\mathrm{HCl}}{\longrightarrow}$ White fumes. Metal (M) can be

A
Li, Mg or Al

B
Li, Al or K

C
Na, K or Mg

D
Li, Na or K

Solution

Reaction is:
$M g+\operatorname{air} \stackrel{\Delta}{\longrightarrow} M g_{3} N_{2} \stackrel{H_{2} O}{\longrightarrow} M g(O H)_{2}+N H_{3} \stackrel{H C l}{\longrightarrow} N H_{4} C l$
In the same way, aluminium and Lithium reacts to form nitrides
Sodium, Potassium do not react with nitrogen to form nitrides.
Hence, the correct option is (A)
Question 5
1 / -0

Titration of I₂ produced from 0.1045 g of primary standard KIO₃ required 30.72 mL of sodium thiosulphate as shown below :

$\mathrm{IO}_{3}^{-}+5 \mathrm{I}^{-}+6 \mathrm{H}^{+} \longrightarrow 3 \mathrm{I}_{2}+3 \mathrm{H}_{2} \mathrm{O}$

$\mathrm{I}_{2}+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$

The molarity of sodium thiosulphate ion is :

A
0.095

B
0.079

C
0.084

D
0.064

Solution

For every 1 mole of 13 you require 2 moles of $S_{2}0_{3}$
Thus, for every 1 mole of$\mathrm{KIO}_3$ that are present you require 6 moles of $\mathrm{Na} _2 \mathrm{S}_2 \mathrm{O}_3$
So, lets start by converting $0.1045 g$ of $\mathrm{KIO}_3$ to moles:
$0.1045 g \times \frac{1 \mathrm{mol}}{214 \mathrm{g}}=4.88 \times 10^{-4} \mathrm{moles}$
Next, we multiply this by 6 to determine the required number of moles of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ :
$4.88 \times 10^{-4} \cdot 6=0.0029 \mathrm{moles}$
Finally, we divide 0.0029 moles of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ by the volume (in Liters) to get the final
concentration:
$\frac{0.0029 \mathrm{moles}}{0.03072 L}=0.095 \mathrm{Molar}$
Hence, the correct option is (A)
Question 6
1 / -0

What is the pH of 0.01 M glycine solution?
[For glycine, $\boldsymbol{K}_{a_{1}}=4.5 \times 10^{-3}$ and $\boldsymbol{K}_{a_{2}}=1.72 \times 10^{-10}$ at $\left.298 \mathrm{k}\right]$

A
3.02

B
6.94

C
7.1

D
10.02

Solution

For glycine solution:
$K=K_{a_{1}} \times K_{a_{2}}=4.5 \times 10^{-3} \times 1.72 \times 10^{-10}=7.74 \times 10^{-13}$
$\left[H^{+}\right]=\sqrt{K \cdot C}$
$=\sqrt{7.74 \times 10^{-13} \times 0.01}=8.79 \times 10^{-8}$
Hence, $p H=-\log \left[H^{+}\right]$
$=-\log \left[8.79 \times 10^{-8}\right]$
$=-\left(\log 10^{-8}+\log 8.7\right)$
$=8-0.93=7.07 \approx 7.1$
Hence, the correct option is (C)
Question 7
1 / -0

An element has fcc structure with edge length 200 pm, calculate the density if 200g of this element contains 24 × 10²³ atoms. (N_A ⁼6 × 10²³)

A
4.16 g cm^–3

B
41.6 g cm^–3

C
4.16 kg cm^–3

D
41.6 kg cm^–3

Solution

$\because 24 \times 10^{23}$ atoms weigh $200 \mathrm{g}$
$\therefore \quad 6 \times 10^{23} \quad$ (Avogadro number) of atoms will weigh $=50 \mathrm{g} / \mathrm{mole}$
$\Rightarrow \quad d=\frac{N M}{N a}: N=$ Number of atoms per unit cell $=4$ for $\mathrm{FCC}$
$a=$ edge length $=200 \mathrm{pm}=2 \times 10^{-8} \mathrm{cm} \cdot a^{3}=8 \times 10^{-24}$
$\begin{aligned} d=\frac{4 \times 50}{6 \times 10^{23} \times 8 \times 10^{-24}} \mathrm{g} / \mathrm{cm}^{3} \\=& 41.67 \mathrm{g} / \mathrm{cm}^{3} \end{aligned}$
Hence, the correct option is (B)
Question 8
1 / -0

The treatment of $C H_{3} M g X$ with $C H_{3} C \equiv C-H$ produces:

A

$C H_{3} M g X=C-C H_{3}$

B

$C H_{3} C H=C H C H_{3}$

C

$C H_{4}$

D

$\left(C H_{3}\right)_{2} C=C H_{2}$

Solution

The treatment of $C H_{3} M g X$ (grignard reagent) with $C H_{3} C \equiv C-H$ (alkyne) produces $C H_{4}$(methane).
$
C H_{3} M g X+C H_{3} C \equiv C-H \rightarrow C H_{3} C \equiv C-M g X+C H_{4}
$
Note: This reaction is used for the alkylation of terminal alkynes.
Hence, the correct option is (C)
Question 9
1 / -0

A 50 litre vessel is equally divided into three parts with the help of two stationary semi permeable membrane (SPM). The vessel contains 60 g H₂gas in the left chamber, 160 g O₂ in the middle and 140 g N₂ in the right one. The left SPM allows transfer of only H₂ gas while the right one allows the transfer of both H₂ and N₂ . The final ratio of pressure in the three chambers.

A
4 : 7 : 5

B
5 : 4 : 7

C
7 : 4 : 5

D
5 : 7 : 4

Solution

$\begin{array}{lccc}\text { Intial } & \text { 30 } & \text { 5 } & \text { 5 } \\ \text { Moles } & \text { Moles } & \text { Moles } & \text { Moles }\end{array}$
$\begin{array}{lccc}\text { Final moles }\left(\mathrm{H}_{2}\right) & 10 & 10 & 10 \\ \text { Final moles }\left(\mathrm{O}_{2}\right) & 0 & 5 & 0 \\ \text { Final moles }\left(\mathrm{N}_{2}\right) & 0 & 2 \cdot 5 & 2 \cdot 5 \\ \text { Total moles } & 10 & 17 \cdot 5 & 12 \cdot 5\end{array}$
Ratio is 4:7:5
Hence, the correct option is (A)
Question 10
1 / -0

A compound that gives a positive iodoform test is:

A
1- pentanol

B
3-pentanone

C
2- pentanone

D
pentanal

Solution

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Chemistry Test - 41

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Chemistry Test - 41

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Question 1

Grignard reaction

Raschig process

Wurtz-Fittig reaction

Friedel Crafts reaction

Solution

Question 2

NaBH4

B2(CH3)4H2

B2(CH3)6

H3BO3

Solution

Question 3

3.66 × 1 0 - 5

3.00 × 1 0 - 5

2.66 × 1 0 - 5

2.00 × 1 0 - 5

Solution

Question 4

Li, Mg or Al

Li, Al or K

Na, K or Mg

Li, Na or K

Solution

Question 5

0.095

0.079

0.084

0.064

Solution

Question 6

3.02

6.94

7.1

10.02

Solution

Question 7

4.16 g cm–3

41.6 g cm–3

4.16 kg cm–3

41.6 kg cm–3

Solution

Question 8

\(C H_{3} M g X=C-C H_{3}\)

\(C H_{3} C H=C H C H_{3}\)

\(C H_{4}\)

\(\left(C H_{3}\right)_{2} C=C H_{2}\)

Solution

Question 9

4 : 7 : 5

5 : 4 : 7

7 : 4 : 5

5 : 7 : 4

Solution

Question 10

1- pentanol

3-pentanone

2- pentanone

pentanal

Solution

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NaBH₄

B₂(CH₃)₄H₂

B₂(CH₃)₆

H₃BO₃

$3.66 \times 1 0^{- 5}$

$3.00 \times 1 0^{- 5}$

$2.66 \times 1 0^{- 5}$

$2.00 \times 1 0^{- 5}$

4.16 g cm^–3

41.6 g cm^–3

4.16 kg cm^–3

41.6 kg cm^–3