JEE CHEMISTRY

DAILY CHEMISTRY TEST QUESTIONS

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1 P � Block Elements

Amphoteric oxide is:

1) Al₂O_3?
2) SiO₂?
3) P_4?O₁₀?
4) ClO₇

Option 1 Solution:

SOLUTION

An amphoteric compound is a molecule or ion that can react both as an acid as well as a base. Many metals (such as copper, zinc, tin, lead, aluminium, and beryllium) form amphoteric oxides or hydroxides.

In acid: Al₂O_3?+6HCl ? 2AlCl₃?+3H₂?O

In base: Al₂O_3?+2NaOH+3H₂O ? 2Na[Al(OH)_4?]

Hence, Al₂?O_3?? is an amphoteric oxide.

Option 2

Option 3

Option 4

2 Surface Chemistry

What is the correct sequence of the increasing effectiveness of the following electrolyte for the coagulation of ferric hydroxide sol?
K₄[Fe(CN)₆]
AlPO₄
CaSO₄
SnCl₄
Select the correct answer using the codes given below:

1) d,b,a,c
2) d,c,b,a
3) a,b,c,d
4) c,d,b,a

Option 1 Solution:

Solution

Ferric hydroxide sol is positively charged.

Hence, more the negative charge of electrolyte, more is its effectiveness in coagulation.

SnCl₄ has 4Cl⁻ ions, AlPO₃has PO₄³⁻ ion, K4[Fe(CN)₆has [Fe(CN)]⁴⁻ion, CaSO₃ has SO₄²⁻ ion.

∴ Order of effectiveness:-

SnCl₄> AlPO₄> K₄[Fe(CN)₆] > CaSO₄or d > b > a > c

Option 2

Option 3

Option 4

3 D and F Block Elements

Select the correct order with respect to the basic character of the transition metal monoxides.

(Atomic number in amu of Ti=22, V=23, Cr=24, Fe=26)

1) VO>CrO>TiO>FeO
2) CrO>VO>FeO>TiO
3) TiO>VO>CrO>FeO
4) TiO>FeO>VO>CrO

Option 1

Option 2

Option 3 Solution:

Solution

The correct order of basic character of the transition metal monoxides is TiO>VO>CrO>FeO, because basic character of oxides decreases with an increase in atomic number. Atomic number increases from TiTi to FeFe. Hence, oxides of transitional metals in low oxidation state, i.e., +2 and +3 are generaly basic except Cr₂O₃

Option 4

4 Hydrocarbons

Which of the following compound contains intermolecular H-bonds?

1) resorcinol

2) ethanoic acid

3) o-nitrophenol

4) phenol

Option 1

Option 2

Option 3 Solution:

Option 4

5 Alcohols Phenols and Ethers

When ethanol is subjected to dehydration in presence of H₂SO₄, the intermediate formed is a:

1) carbonium ion

2) carbanion

3) radical

4) carbene

Option 1 Solution:

SOLUTION

CH₃−CH₂−OH ? CH₃−⁺CH₂? CH₂=CH₂

The intermedate formed is a carbonium ion or carbocation

Option 2

Option 3

Option 4

6 Chemical Kinetics

The rate of a reaction quadruples when the temperature changes from 300 to 310K. What is the activation energy of this reaction?
(Assume activation energy and preexponential factor are independent of temperature ; In 2= 0.693; R= 8.314 Jmol⁻¹K⁻¹)

1) 0.60 ×10⁵Jmol⁻¹
2) 2.46×10⁵Jmol⁻¹
3) .246×10⁵Jmol⁻¹
4) 4.00 ×10⁵Jmol⁻¹

Option 1

Option 2 Solution:

Solution

$$\log \left(\cfrac {k_2}{k_1}\right)=\cfrac {E_A}{2.303R} \left[\cfrac {1}{T_1}-\cfrac {1}{T_2}\right]$$

$$\Rightarrow \log (4)=\cfrac {E_A}{2.303\times 8.314}\times \left[\cfrac {1}{300}-\cfrac {1}{310}\right]$$

$$\Rightarrow 2\times 0.693= \cfrac {E_A}{2.303}\times \cfrac {0.0001075}{8.314}$$

$$\Rightarrow E_A=\cfrac {26.53793}{0.0001075}=2.46\times 10^{5}Jmol^{-1}$$

Option 3

Option 4

7 Titrations

In order to prepare one litre 1N solution of KMnO4?, how many grams of KMnO4? are required, if the solution to be used in acid medium for oxidation?

1) 128 g

2) 41.75 g

3) 31.60 g

4) 62.34 g

Option 1

Option 2

Option 3 Solution:

Solution

One liter 1N solution of KMnO4? corresponds to 1 g equivalent. In acid medium for oxidation, the equivalent weight of KMnO₄? is one fifth of its molecular mass. It is 15 / 58?=31.6.
Hence, 1 g equivalent of KMnO₄? corresponds to 31.6 g.

Hence, in order to prepare one liter 1N solution of KMnO₄?, 31.6 grams of KMnO4? are required, if the solution to be used in acid medium for oxidation.

Option 4

8 Chemical Kinetics

A first order reaction was started with a decimolar solution of the reactant. 8 minutes and 20 seconds later its concentration was found to be M/100. So the rate constant of the reaction is

1) 2.303 × 10^-5/sec
2) 5 × 10^-3 /sec
3) 2.303 × 10^-4/sec
4) 4.606 × 10^-3/sec

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

9 Stoichiometry

An organic compound is found to contain C = 54.5% O = 36.4% and H = 9.1% by mass. Its empirical formula is :

1) C₂O
2) CHO₂
3) C₂H₄O
4) C₃H₄O

Option 1

Option 2

Option 3 Solution:

Option 4

10 P � Block Elements

Match metal ion Column I with colour Column II in the presence of alizarin.

1) A = 1, B = 4, C = 3, D = 2
2) A = 2, B = 1, C = 4, D = 3
3) A = 3, B = 2, C = 1, D = 4
4) A = 2, B = 3, C = 4, D = 1

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTIONS

Alizarin is a red pigment present in madder root, used in dying and gives different colors with different colors
i.e.
Sr²⁺→ Pink, Mg²⁺ → Violet
Al³⁺ → Red, Ba²⁺ → blue.

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