Chemistry Test 213

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Chemistry Test 213

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Weekly Quiz Competition

Question 1
4 / -1

XeF₆ on partial hydrolysis with water, produces a compound X. The same compound X is formed when XeF₆ reacts with silica. The compound X is:

A
XeF₄

B
XeF₂

C
XeO₃

D
XeOF₄

Solution

During the hydrolysis the oxidation of xenon not changes in xenon hexafluoride.
Question 2
4 / -1

How many EDTA (ethylenediaminetetraacetate) unit(s) are required to make an octahedral complex with a Ca²⁺ ion?

A
Six

B
Three

C
One

D
Two

Solution

As EDTA is hexadentate ligand. Hence, only one EDT (ethylenediamine tetraacetate) molecule is required to make an octahedral complex with Ca²⁺.
Question 3
4 / -1

Electrometallurgical process is used to extract

A
Fe

B
Pb

C
Na

D
Ni

Solution

Because Na is very reactive and can not be extracted by means of the reduction by C, CO etc. So, it is extracted by electrolysis of molten NaCl Solution.
Question 4
4 / -1

For one mole of a Van der Waals gas, when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below. The value of the Van der Waals constant a (atm litre² mol^-2) is:

A
1.0

B
4.5

C
1.5

D
3.0

Solution
Question 5
4 / -1

Cr²⁺ and Mn³⁺ both have d⁴ configuration. Thus

A
both are reducing agents.

B
both are oxidizing agents.

C
Cr²⁺ is an oxidizing agent while Mn³⁺ is a reducing agent.

D
Mn³⁺ is an oxidizing agent while Cr²⁺ is a reducing agent.

Solution
Question 6
4 / -1

Three elements X, Y, and Z have atomic numbers 19, 37, and 55 respectively. Then the correct statements(s) is/are

A
Their ionization potential would increase with increasing atomic numbers

B
Y would have an ionization potential between those of X and Z

C
Y would have the highest ionization potential

D
Z would have the highest ionization potential

Solution

The elements X (19), Y (37) and Z (55) are the elements of same group (IA). As atomic number, increases, the ionization potential decrease down the group. Since, the position of Y is in between X and Z. Thus, the ionization potential of Y will also be in between X and Z.
Question 7
4 / -1

An electron in an atom jumps in such a way that its kinetic energy changes from x to x/9. The change in its potential energy (magnitude) will be-

A
x/9

B
16x/9

C
9x

D
9x/16

Solution

Change in kinetic energy is

For Bohr model of atom

The kinetic energy in a state is equal to half of the potential energy in magnitude.

∴ Potential energy

= 2 Kinetic energy
Question 8
4 / -1

The amount (in grams) of sucrose (mol wt. = 342 g) that should be dissolved in 100 g water in order to produce a solution with a 105.0 ^oC difference between the freezing point and boiling point is

(Given that K_f = 1.86 K kg mol⁻¹ and Kb = 0.51 K kg mol⁻¹for water)

A
34.2 g

B
72.2 g

C
342 g

D
460 g

Solution

Boiling point (T_b) = 100 + ΔT_b= 100 + k_bm

Freezing point (T_f) = 0 − ΔT_f= −k_fm

T_b − T_f= (100 + k_bm) − (−kfm)

105 = 100 + 0.51 m + 1.86 m
Question 9
4 / -1

The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant k₁ and k₂ respectively. The energy of activation for the two reactions are 152.30 kJ mol⁻¹and 157.7 kJ mol⁻¹ as well as frequency factors are 10¹³ and 10¹⁴ respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be the same for the two reactions.

A
256 K

B
354 K

C
282 K

D
674 K

Solution
Question 10
4 / -1

Choose from the indicated protons, the one that is most acidic.

A
1

B
2

C
3

D
4

Solution

Deprotonation of hydrogen labelled 44 produces a conjugate base which has more stable resonating structure and a resonance structure in which negative charge is present on the electronegative oxygen atom. Therefore, hydrogen labelled 44 is most acidic.