NSEC CHEMISTRY

DAILY CHEMISTRY TEST QUESTIONS

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1 S � Block Elements

The metal carbonate that is thermally least stable is

1) BeCO₃

2) BaCO₃

3) CaCO₃

4) MgCO₃

Option 1 Solution:

Option 2

Option 3

Option 4

2 Solutions

An aqueous solution of non-electrolyte 'A' with molecular mass 60 contains 6 g in 500 ml and has a density equal to 1.05. The molality of solution is

1) 0.30

2) 0.25

3) 0.19

4) 1.25

Option 1

Option 2

Option 3 Solution:

Solution

Mass of solution =500×1.05=525g
Mass of solvent =525−6=519g

Molality = $$\displaystyle=\frac{{{6}\times{1000}}}{{{60}\times{519}}}={0.192}{M}$$

Option 4

3 Purification and Characterisation of Organic Compounds

The fragrance of the flower is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in the vapour phase. A suitable method for the extraction of these oil from the flower is:

1) distillation with reduced pressure

2) distillation

3) steam distillation

4) chromatography

Option 1

Option 2

Option 3 Solution:

SOLUTION

As the component is miscible with water vapour in vapour phase so steam distillation will be most suitable method for extraction.

Option 4

4 Surface Chemistry

A catalyst:
(i) increases the average kinetic energy of reacting molecules
(ii) decrease the activation energy
(iii) alters the reaction mechanism
(iv) increases the frequency of collisions of reacting species.
The true statements are:

1) i, ii, iii
2) ii, iv
3) ii, iii
4) all are correct

Option 1

Option 2

Option 3 Solution:

Solution

A catalyst is a substance which when added to a reaction, finds an alternative route (hence, altering the reaction mechanism) with lower activation energy as shown in the diagram. Remember that with a catalyst, the average kinetic energy of the molecules remains the same but the required energy decreases.

Option 4

5 Haloalkanes and Haloarenes

Acetyl bromide reacts with excess of CH₃MgI followed by treatment with a saturated solution of NH₄Cl gives:

1) acetone
2) acetamide
3) 2-Methyl-2-propanone
4) acetyl iodide

Option 1

Option 2

Option 3 Solution:

Solution

2-methyl-2-propanol
Acetyl chloride reacts with two equivalents of Grignard reagent to form tertiary alcohol, two of the groups bonded to C having OH group are from the Grignard reagent.

Option 4

6 Ionic Equilibrium

At 298K a 0.1 M CH3COOHsolution is 1.34% ionized. The ionization constant K_a for acetic acid will be

1) 0.182×10⁻⁵
2) 18.2×10⁻⁵
3) 1.82×10⁻⁵
4) none

Option 1

Option 2

Option 3 Solution:

Solution:

K = cα²= 0.1 × (1.34 / 100)² = 1.82×10⁻⁵

Option 4

7 Titrations

The one which decreases with dilution is

1) molar conductance

2) conductance

3) specific conductivity

4) None of these

Option 1

Option 2

Option 3 Solution:

Solution

Conductivity through a solution is generallly through the ions that are present in it. Pure water is not very conductive at all, but salt water is quite conductive . Therefore the more salt or ions dissolved in solution, higher is the conductivity (within limits).

As you dilute the solution , the concentration of ions goes down, and the ability to pass a current is diminished. Hence the specific conductance decreases upon dilution

Option 4

8 S � Block Elements

When carbon monoxide is passed over solid caustic soda to heated to 200⁰C, it form:

1) H-COONa
2) CH₃COONa
3) Na₂CO₃
4) NaHCO₃.

Option 1 Solution:

Option 2

Option 3

Option 4

9 Electrochemistry

The specific conductance of saturated solution of CaF₂ is 3.86 × 10⁻⁵ ohm cm⁻¹ and that of water used for solution is 0.15×10⁻⁵. The specific conductance of CaF₂ alone is:

1) 3.71 × 10⁻⁵
2) 4.01 × 10⁻⁵
3) 3.7 × 10⁻⁴
4) 3.86 × 10⁻⁴

Option 1 Solution:

SOLUTION

k (CaF₂ solution) = 3.86 × 10−5 Ω⁻¹cm⁻¹

k (water solution) = 0.15×10−5 Ω⁻¹cm⁻¹

k (CaF₂ alone) = ?

k (CaF₂ alone) = k (CaF₂ solution) −k (Water solution)

=(3.86 − 0.15) × 10⁻⁵Ω⁻¹cm⁻¹

=3.71 × 10⁻⁵ Ω⁻¹ cm⁻¹

The specific conductance of CaF₂ alone is 3.71×10⁻⁵ Ω⁻¹cm⁻¹

Option 2

Option 3

Option 4

10 Haloalkanes and Haloarenes

The reaction of 1-hexene with HBr in presence of peroxide yields 1-bromo cyclohexane. The mechanism of the reaction involves :

1 attack on alkene by a Br+ ion
2) attack on alkene by a bromide ion
3) attack on alkene by a bromine free radical
4) None of these

Option 1

Option 2

Option 3 Solution:

SOLUTION

The reaction of 1-Hexene with HBr in presence of peroxide yields 1-bromo cyclohexane.

The mechanism of the reaction involves an attack on alkene by a bromine atom. The reaction proceeds through a free radical mechanism.

In the presence of peroxide, bromine free radical is generated. This bromine free radical then attacks alkene.

Option 4

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