NSEC CHEMISTRY

DAILY CHEMISTRY TEST QUESTIONS

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1 Electrochemistry

The standard reduction potential for two reaction are given below
AgCl(s) + e^- → Ag(s) + Cl^-(aq), E°=0.22V
Ag+(aq) + e^- → Ag(s) , E°=0.80V
the solubility product of AgCl under standard condition of temperature (298K) is given by

1) 1.6 ×10^-10
2) 1.5 ×10^-10
3) 3.2 ×10^-8
4) 3.2 ×10^-10

Option 1

Option 2 Solution:

SOLUTION
Subtracting second equation from first we get
AgCl(s) → Ag+(aq) + Cl-(aq) , E°=-0.50V
From Nernst equation

As AgCl is solid its concentration is 1 and equilibrium E=0 Thus E°=0.0591 log[Ag+][Cl-]
E°=0.0591 log Ksp
-0.5=0.0591 log Ksp
log Ksp=-9.8139 Or Ksp1.535 ×10^-10

Option 3

Option 4

2 Aldehydes Ketones and Carboxylic Acids

The acid which reduces Fehling's solution is :
1) Ethanoic acid.
2) Methanoic acid.
3) Propanoic acid.
4) Butanoic acid

Option 1

Option 2 Solution:

Option 3

Option 4

3 Ionic Equilibrium

At a particular temperature, the ratio of molar conductance to specific conductance of 0.01 M NaCl solution is:

1) 10⁵ cm³mol⁻¹
2) 10³ cm³mol⁻¹
3) 10 cm³mol⁻¹
4) 10⁵ cm²mol⁻¹

Option 1 Solution:

SOLUTION

∧_m= (K×1000 )/ M

where,

∧_m= Molar conductance

K= Specific conductance

⇒∧_{m /}K

Option 2

Option 3

Option 4

4 Basic Principles of Organic Chemistry

Petroleum consists mainly of

1) aromatic hydrocarbons

2) aliphatic alcohols

3) aliphatic hydrocarbons

4) none of these

Option 1

Option 2

Option 3 Solution:

Option 4

5 Mole Concept

A person adds 1.71 gram of sugar (C₁₂H₂₂O₁₁) in order to sweeten his tea. The number of carbon atoms added are (mol. mass of sugar = 342)

1) 7.2 x 10²¹

2) 3.6 x 10²²

3) 6.6 x 10²²

4) 0.05

Option 1

Option 2 Solution:

SOLUTION

Sugar is nothing but sucrose, which has a formula of C??H??O?? and a molecular weight of 342.

number of moles of sucrose in 1.71 grams = 0.005

1 mole of sucrose contains Avogadro number(6 x 10²³) of sucrose molecules.
As each molecule of sucrose has 12 carbon atoms,
1 mole of sucrose contains 12( 6 x 10²³) of carbon atoms.

Therefore 0.005 moles of sucrose contains 0.005 x 12( 6 x 10²³)
= 3.6 x 10²² atoms of carbon are added

Option 3

Option 4

6 Ionic Equilibrium

The compound whose 0.1 M solution is basic is

1) Sodium acetate

2) Ammonium sulphate

3) Ammonium chloride

4) Ammonium acetate

Option 1 Solution:

Solution

CH3COONa is a salt of weak acid, (CH3COOH) and strong base (NaOH).

Option 2

Option 3

Option 4

7 General Principles & Process of Isolation of Metals

When ZnS and PbS minerals are present together, then NaCN is added to separate them in the froth flotation process as a depressant, because:

1) $$Pb(CN)_2 \,S$$ precipitated while no effect on $$ZnS$$
2) $$ZnS$$ forms soluble complex $$Na_2 [Zn(CN)_4]$$
3) $$PbS$$ forms soluble complex $$Na_2[Pb(CN)_4]$$
4) they cannot be separated by adding $$NaCN$$

Option 1

Option 2 Solution:

SOLUTION

$$NaCN$$ prevents $$ZnS$$ from forming the froth by reacting with it to from complex $$Na_2[Zn(CN)_4]$$ and acts as a depressant. While it does not prevent $$PbS$$ from forming froth and both the ores can easily be separated.

Option 3

Option 4

8 Surface Chemistry

Which property of colloidal solution is independent of charge on the colloidal particles?

1) Coagulation
2) Electrophoresis
3) Electroosmosis
4) Tyndall effect

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

The Tyndall effect is light scattering by particles in a colloid or in a very fine suspension.
Coagulation is a chemical process that involves neutralization of charge whereas flocculation is a physical process and does not involve neutralization of charge.
Electrophoresisstrong - > is the motion of dispersed particles relative to a fluid under the influence of a spatially uniform electric field. Electrophoresis of positively charged particles is sometimes called cataphoresis, while electrophoresis of negatively charged particles is sometimes called anaphoresis.
Electro-osmosisstrong - > refers to the movement of liquid in a porous material due to an applied electric field

9 Chemical Kinetics

A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial velocity is ten times the permissible value, after how many days will it be safe to enter the room?

1) 100 days
2) 1000 days
3) .300 days
4) 10 days

Option 1

Option 2 Solution:

Solution

Say Permissible rate is $$\vartheta$$

then $$\vartheta _{2}=100$$

$$\vartheta =kA$$

$$=kA_{0}e^{-kt}$$

$$kA_{0}e^{\frac{-0.693t}{t_{\frac{1}{2}}}}

$$\vartheta =\vartheta _{0}e^{\frac{-0.693t}{t_{\frac{1}{2}}}}$$

$$\vartheta =10\vartheta e^{\frac{-0.693t}{t_{\frac{1}{2}}}}$$

$$l_{n}(10)=\frac{-0.693}{t_{\frac{1}{2}}}t$$

$$t=\frac{t_{\frac{1}{2}}*2.303}{0.693}*10$$

$$=1000$$days

Option 3

Option 4

10 Mole Concept

The mass of CO₂obtained by the complete thermal decomposition of 80% of 40 pure sample of CaCO₃ is nearly equal to?

1) 4.4g
2) 44g
3) 32g
4) 14g

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Weight of pure $$CaCO_3=\cfrac {40}{100}\times 80=32$$

Number of moles of $$CaCO_3=\cfrac {32}{100}=0.32$$

$$CaCO_3 \longrightarrow CaO+CO_2$$

$$1$$ mole of $$CaCO_3$$ gives $$1$$ mole of $$CO_2$$

$$\therefore$$ Moles of $$CO_2=\cfrac {32}{100}$$

$$\therefore$$ Weight of carbon dioxide= $$\cfrac {32}{100}\times 44= 14.08 gm$$

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