JEE Main Test 141

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JEE Main Test 141

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Weekly Quiz Competition

Question 1
4 / -1

The best method to separate the mixture of ortho and para nitrophenol (1:1) is

A
vaporisation

B
colour spectrum

C
distillation

D
crystallisation

Solution

O – nitrophenol is steam valatile due to intra molecular hydrogen bonding where as p – nitrophenol is less volatile due to intermolecular hydrogen bonding.
Question 2
4 / -1

The sodium extract of an organic compound on acidification with acetic acid and addition of lead acetate solution gives a black precipitate. The organic compound contains

A
nitrogen

B
halogen

C
sulphur

D
phosphorus

Solution

When sulphur react with sodium, sodium sulphide will form, which react with lead acetate solution and forms black ppt. of Pbg which confirms the presence of sulphur in compound.
Question 3
4 / -1

In estimation of nitrogen, by Kjeldahl's method, CuSO₄ acts as an

A
oxidising agent

B
reducing agent

C
catalysing agent

D
hydrolysing agent

Solution

To estimate nitrogen atom present in an organic compound; Kjeldahl’s method is used. In this method organic compund is heated with conc. H₂SO₄ is presence of K₂SO₄ and CuSO₄, the nitrogen present in the compound is converted into amonium sulphate
Question 4
4 / -1

1.2g of organic compound on Kjeldahlization liberates ammonia which consumes 30 cm³ of 1 N HCI. The percentage of nitrogen in the organic compound is

A
30

B
35

C
46.67

D
20.8

Solution
Question 5
4 / -1

The freezing point (in °C) of solution containing 0.1 g of K₃ [Fe(CN)₆] (mol. wt. 329) in 100 g of water (K_f =1.86 K kg mol^-1) is

A
- 2.3 × 10^-2

B
- 5.7 × 10^-2

C
- 5.7 × 10^-3

D
-1.2 × 10^-2

Solution
Question 6
4 / -1

Which of the following expression is correct?

A
∆G^o = nFE^o_cell

B
∆G^o = +nFE^o_cell

C
∆G^o = -2.303 RT nFE^o_cell

D
∆G^o = -nF log K_c

Solution
Question 7
4 / -1

Choose the correct reason(s) for the stability of the lyohobic colloidal particles.

A
Preferential adsorption of ions from the solution

B
Preferential adsorption of solvent on their surface from the solution

C
Attraction between different particles having opposite charges on their surface

D
Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles

Solution

Lyophobic sol, gets stabilised by preferential adsorption on the their surface.
Question 8
4 / -1

A
0

B
1/12

C
1/24

D
1/64

Solution
Question 9
4 / -1

The unit of rate constant of second order reaction

A
t^-1

B
conc^-1 t^-1

C
conc t^-1

D
conc^-2 t^-1

Solution
Question 10
4 / -1

The coagulating power of electrolytes having ions Na⁺ , Al³⁺ and Ba²⁺ for arsenic sulphide sol increases in the order

A
Al³⁺ < Ba²⁺ < Na⁺

B
Na⁺ < Ba²⁺ < Al³⁺

C
Ba²⁺ < Na²⁺ < Al³⁺

D
Al³⁺ < Na⁺ < Ba²⁺

Solution

According to Hardy Schulze rule, greater the charge on oppositely charged ion, greater is its coagulating power. Order of coagulating power Na⁺ < Ba²⁺ < Al³⁺ .

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Re-Attempt Weekly Quiz Competition

JEE Main Test 141

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JEE Main Test 141

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Rank

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SHARING IS CARING

Question 1

vaporisation

colour spectrum

distillation

crystallisation

Solution

Question 2

nitrogen

halogen

sulphur

phosphorus

Solution

Question 3

oxidising agent

reducing agent

catalysing agent

hydrolysing agent

Solution

Question 4

30

35

46.67

20.8

Solution

Question 5

- 2.3 × 10-2

- 5.7 × 10-2

- 5.7 × 10-3

-1.2 × 10-2

Solution

Question 6

∆Go = nFEocell

∆Go = +nFEocell

∆Go = -2.303 RT nFEocell

∆Go = -nF log Kc

Solution

Question 7

Preferential adsorption of ions from the solution

Preferential adsorption of solvent on their surface from the solution

Attraction between different particles having opposite charges on their surface

Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles

Solution

Question 8

0

1/12

1/24

1/64

Solution

Question 9

t-1

conc-1 t-1

conc t-1

conc-2 t-1

Solution

Question 10

Al3+ < Ba2+ < Na+

Na+ < Ba2+ < Al3+

Ba2+ < Na2+ < Al3+

Al3+ < Na+ < Ba2+

Solution

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- 2.3 × 10^-2

- 5.7 × 10^-2

- 5.7 × 10^-3

-1.2 × 10^-2

∆G^o = nFE^o_cell

∆G^o = +nFE^o_cell

∆G^o = -2.303 RT nFE^o_cell

∆G^o = -nF log K_c

t^-1

conc^-1 t^-1

conc t^-1

conc^-2 t^-1

Al³⁺ < Ba²⁺ < Na⁺

Na⁺ < Ba²⁺ < Al³⁺

Ba²⁺ < Na²⁺ < Al³⁺

Al³⁺ < Na⁺ < Ba²⁺