NSEC CHEMISTRY

DAILY CHEMISTRY TEST QUESTIONS

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1 Electrochemistry

A metal rod is dipped in a solution of its ions. Its electrode potential is independent of

1) temperature of solutions.
2) concentration of solution
3) area of metal exposed
4) nature of metal

Option 1

Option 2

Option 3 Solution:

Option 4

2 Hydrocarbons

Arrange the following in decreasing order of reactivity towards EAS: (electrophilic aromatic substitution)

1) a > b > c
2) a > c > b
3) c > a > b
4) c > b > a

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

Order of +I effect: CT₃> CD₃ > CH₃

The compound containing group which has greater +I effect will be more reactive towards electrophilic aromatic substitution reaction.

3 Alcohols Phenols and Ethers

When sodium benzenesulphonate is fused with NaOH, followed by cidification with
HCl, the product formed is :

1) benzensulphonic acid
2) sodium
3) phenol
4) chlorobenzene

Option 1

Option 2

Option 3 Solution:

Phenol, also known as carbolic acid, was first isolated in the early nineteenth century from coal tar. Nowadays, phenol is commercially produced synthetically. In the laboratory, phenols are prepared from benzene derivatives by any of the following methods:

From haloarenes

Chlorobenzene is fused with NaOH at 623K and 320 atmospheric pressure. Phenol is obtained by acidification of sodium phenoxide so produced.

Option 4

4 Redox Reactions

How much amount of CuSO₄.5H₂O required for liberation of 2.54g I₂ when titrated with KI?

1) 2.5 gm
2) 4.99 gm
3) 2.4 gm
4) 1.2 gm

Option 1

Option 2 Solution:

SOLUTION

Molecular weight of CuSO₄.5H₂O = 249.5gmol⁻¹

The reaction for liberation of I₂;

2CuSO₄.5H₂O + KI = I₂+ Cu₂I₂+ 2K₂SO₄ + 5H₂O

2.54g of I₂= 2.54 /254 = 0.01 moles of I₂

Now, 1 mole of I₂ is liberated from 2 moles of CuSO₄.5H₂O

∴0.01 mole of I₂ will be liberated from 2×0.01 mole i.e. 0.02 mole of CuSO₄.5H₂O.

∴ Amount of CuSO₄.5H₂O=0.02 mole × 249.5gmol⁻¹=4.99gm

Option 3

Option 4

5 Hydrogen

5.0 cm³ of H₂O₂ liberates 0.508 g of iodine from an acidified KI solution. The strength of H₂ O₂ solution in terms of volume strength at STP is:

1) 6.48 volumes
2) 4.48 volumes
3) 7.68 volumes
4) none

Option 1

Option 2 Solution:

Solution

Moles of iodine= $$\cfrac {0.508}{254}$$ moles

Now, $$M_1V_1=M_2V_2$$

$$M_1\longrightarrow$$ Molarity of $$H_2O_2$$ solution

$$\therefore$$ $$V_1\longrightarrow$$ Volume of $$H_2O_2$$

$$M_2\longrightarrow$$ Molarity of iodine

$$V_2\longrightarrow$$ $$1000$$ $$ml$$

$$\therefore$$ $$M_1\times 5=\cfrac {0.508}{254}\times 1000$$

$$\therefore$$ $$M_1=0.4$$ $$M$$

The reaction can be given as,

$$H_2O_2+2I^{-}+2H^{+}\longrightarrow2H_2O+I_2$$

$$\therefore$$ n-factor for $$H_2O_2=2$$

$$\therefore$$ Normality of $$H_2O_2=2\times 0.4=0.8N$$

$$\therefore$$ Volume strength of $$H_2O_2=0.8\times 5.6$$

=$$4.48$$ volumes

Option 3

Option 4

6 Ionic Equilibrium

Which of the following salts on dissolution of water gives pH less than 7 at 25⁰C?

1) KCN.
2) CH₃COONa
3) NaBr
4) NH₄Cl

Option 1

Option 2

Option 3

Option 4 Solution:

Solution:

As NH₄Cl is a salt of strong acid and weak base, it's pH will be less than 7 since the salt is acidic in nature.

7 Alcohols Phenols and Ethers

When 10mL of ethyl alcohol (density =0.7893gmL⁻¹) is mixed with 20 mL of water (density 0.9971 gmL⁻¹) at 25^oC, the final solution has a density of 0.9571gmL⁻¹. The percentage change in total volume on mixing is :

1) 1%
2) 2.4%
3) 3.1%
4) none of these

Option 1

Option 2

Option 3 Solution:

Solution

Mass $$\displaystyle =$$ density $$\displaystyle \times $$ volume

Mass of ethyl aclohol $$\displaystyle = 0.7893 \times 10 = 7.893$$g

Mass of water $$\displaystyle = 0.9971 \times 20 = 19.942$$g

Total mass of solution $$\displaystyle = 7.893 + 19.942 = 27.835$$ g

Volume of solution after mixing $$\displaystyle = \dfrac {Mass}{Density} = \dfrac {27.835}{0.9571}=29.083mL $$

Total volume of ethanol and water before mixing $$\displaystyle = 10 + 20 = 30 mL$$

The percentage change in total volume on mixing is $$\displaystyle = \dfrac {30 - 29.083}{30} \times 100 = 3.1 $$ %.

Option 4

8 Aldehydes Ketones and Carboxylic Acids

Treatment of benzoic acid with Cl₂/FeCl₃will give________ acid.
1) o-chlorobenzoic.
2) p-chlorobenzoic.
3) m-chlorobenzoic.
4) 2,4-dichlorobenzoic.

Option 1

Option 2

Option 3 Solution:

Option 4

9 General Principles & Process of Isolation of Metals

Which of the following ore is best concentrated by froth flotation method ?

1) Galena
2) Cassiterite
3) Magnetite
4) none

Option 1 Solution:

Solution

Galena (PbS) are sulphide ores. Cassiterite SnO₂ (oxide ore), and Magnetite Fe₃O₄ (oxide ore) are not sulphide ore. The froth flotation process is used to concentrate sulphide ores, based on preferential wetting properties with frosting agent and water.

Option 2

Option 3

Option 4

10 Ionic Equilibrium

What volume of decinormal K₂Cr₂O₇ solution would be needed to convert 1.19 g of tin to stannic chloride in an acidic medium? (ASn = 119ASn = 119)

1) 400 ml
2) 200 ml
3) 100 ml
4) 2800 ml

Option 1 Solution:

SOLUTION

No.of moles of tin in 1.19g of tin=1.19=1.19

n=0.01

as Sn is converting to SnCl₄ i.e. Sn⁰→Sn⁺⁴

Thus n_f= 4

Normality of Kr₂Cr₂O₇ = O.1N

∴0.1N×V=0.01×4

V=0.42=400ml

400ml of volume of decinormal Kr₂Cr₂O₇solution would be needed

Option 2

Option 3

Option 4

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