Chemistry Test

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Chemistry Test - 27

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Weekly Quiz Competition

Question 1
5 / -1

If the aluminum unit cell exhibits face-centered behavior then how many unit cells are present in 54g of aluminum?

A
1.2042 x 10²⁴

B
5.575 x 10²¹

C
3.011 x 10²³

D
2.4088 x 10²⁴

Solution

Atomic mass of Al = 27g/mole (contains 6.022 x 10²³ Al atoms)

Since it exhibits FCC, there are 4 Al atoms/unit cell.

If 27g Al contains 6.022 x 10²³ Al atoms then 54g Al contains 1.2044 x 10²⁴atoms.

Thus, if 1 unit cell contains 4 Al atoms then number of unit cells containing 1.2044 x 10²⁴ atoms=(1.2044 x 10²⁴ x 1)/4 = 3.011 × 10²³ unit cells.
Question 2
5 / -1

Which of the following is true regarding azeotropes?

A
An azeotrope does not exhibit the same concentration in the vapour phase and the liquid phase

B
In case of maximum boiling azeotropes, the boiling point of the azeotrope is lesser than the boiling point of either of the pure components

C
In case of maximum boiling azeotropes, the boiling point of the azeotrope is lesser than the boiling point of either of the pure components

D
Azeotropic mixtures cannot be separated into their constituents by fractional distillation

Solution

Azeotropes have the same concentration in the vapour phase and the liquid phase. In case of minimum boiling azeotropes, the boiling point of the azeotrope is lesser than the boiling point of either of the pure components. In case of maximum boiling azeotropes, the boiling point of the azeotrope is higher than the boiling point of either of the pure components.
Question 3
5 / -1

What is the radius of a metal atom if it crystallizes with body-centered lattice having a unit cell edge of 333 Pico meter?

A
1538.06 pm

B
769.03 pm

C
288.38 pm

D
144.19 pm

Solution

For body-centered unit cells, the relation between radius of a particle ‘r’ and edge length of unit cell ‘a’ is given as is the radius of the metal atom.
Question 4
5 / -1

If liquids A and B form an ideal solution, then what is the Gibbs free energy of mixing?

A
> 0

B
< 0

C
= 0

D
Not Defined

Solution

The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system. For ideal solutions, the value of the Gibbs Free energy is always negative as mixing of ideal solutions is a spontaneous process.
Question 5
5 / -1

How many parameters are used to characterize a unit cell?

A
Six

B
Three

C
Two

D
Nine

Solution

A unit cell is characterized by six parameters i.e. the three common edge lengths a, b, c and three angles between the edges that are α, β, γ. These are referred to as inter-axial lengths and angles, respectively. The position of a unit cell can be determined by fractional coordinates along the cell edges.
Question 6
5 / -1

5 moles of liquid X and 10 moles of liquid Y make a solution having a total vapour pressure 70 torr. The vapour pressures of pure X and pure Y are 64 torr and 76 torr respectively. Which of the following is true regarding the described solution?

A
The solution shows positive deviation

B
The solution shows negative deviation

C
The solution is ideal

D
The solution has volume greater than the sum of individual volumes

Solution

Observed pressure = 76 torr

According to Raoult’s law,

p_A = x_A x p_A⁰ = 5/15 x 64 = 21.33 torr

p_B = x_B x p_B⁰ = 10/15 x 76 = 50.67 torr

Therefore, pressure expected by Raoult’s law = 21.33 + 50.67 = 72 torr.

Thus, observed pressure (70 torr) is less than the expected value. Hence, the solution shows negative deviation.
Question 7
5 / -1

If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.

A
2923 kg/m³

B
5846 kg/m³

C
8768 kg/m³

D
1750 kg/m³

Solution

Edge length (a) = 500 pm = 500 x 10^-12 m

Atomic mass (M) = 110 g/mole = 110 x 10^-3 kg/mole

Avogadro’s number (NA) = 6.022 x 10²³/mole

z = 4 atoms/cell
Question 8
5 / -1

What is each point (position of particle) in a crystal lattice termed as?

A
Lattice index

B
Lattice point

C
Lattice lines

D
Lattice spot

Solution

Each point of the particle’s position is referred to as ‘lattice point’ or ‘lattice site’. Every lattice point represents one constituent particle which may be an atom, ion or molecule.
Question 9
5 / -1

Why is ‘raising of viscosity’ of a solution after addition of solute, not considered to be a colligative property?

A
The resultant viscosity depends on the nature of the solute

B
The resultant viscosity depends on the amount of solute

C
The resultant viscosity depends on the nature of solvent

D
The resultant viscosity depends on the amount of solvent

Solution

A colligative property is identified by the fact that it has no dependence on the nature of the particles of solute. However, in the case of viscosity it really depends on the solute that is added to the solvent. Thus, the change in viscosity after addition of a non-volatile solute cannot be considered to be a colligative property.
Question 10
5 / -1

At 70°C the vapor pressure of pure water is 31 kPa. Which of the following is most likely the vapor pressure of a 2.0 molal aq. glucose solution at 70°C?

A
30.001 kPa

B
29.915 kPa

C
28.226 kPa

D
32.392 kPa

Solution

Given, P⁰water = 31 kPa

Concentration of solution, c = 2 molal = 2 moles of glucose/kg of water

From law of relative lowering of vapor pressure, ΔP/P⁰ = X₂, where X₂ is the mole fraction of glucose in the solution.

Mass of water = 1 kg = 1000 g

Molecular weight of water = 18 g/mole

Moles of water = 1000/18 = 55.556 moles

X₂ = 2/(2 + 55.556) = 0.035

ΔP = 31 kPa x 0.035 = 1.085 kPa

Final pressure = 31 kPa – 1.085 kPa = 29.915 kPa.

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Chemistry Test - 27

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Chemistry Test - 27

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Question 1

1.2042 x 1024

5.575 x 1021

3.011 x 1023

2.4088 x 1024

Solution

Question 2

An azeotrope does not exhibit the same concentration in the vapour phase and the liquid phase

In case of maximum boiling azeotropes, the boiling point of the azeotrope is lesser than the boiling point of either of the pure components

In case of maximum boiling azeotropes, the boiling point of the azeotrope is lesser than the boiling point of either of the pure components

Azeotropic mixtures cannot be separated into their constituents by fractional distillation

Solution

Question 3

1538.06 pm

769.03 pm

288.38 pm

144.19 pm

Solution

Question 4

> 0

< 0

= 0

Not Defined

Solution

Question 5

Six

Three

Two

Nine

Solution

Question 6

The solution shows positive deviation

The solution shows negative deviation

The solution is ideal

The solution has volume greater than the sum of individual volumes

Solution

Question 7

2923 kg/m3

5846 kg/m3

8768 kg/m3

1750 kg/m3

Solution

Question 8

Lattice index

Lattice point

Lattice lines

Lattice spot

Solution

Question 9

The resultant viscosity depends on the nature of the solute

The resultant viscosity depends on the amount of solute

The resultant viscosity depends on the nature of solvent

The resultant viscosity depends on the amount of solvent

Solution

Question 10

30.001 kPa

29.915 kPa

28.226 kPa

32.392 kPa

Solution

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1.2042 x 10²⁴

5.575 x 10²¹

3.011 x 10²³

2.4088 x 10²⁴

2923 kg/m³

5846 kg/m³

8768 kg/m³

1750 kg/m³