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Chemistry Test - 28

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Chemistry Test - 28
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Weekly Quiz Competition
  • Question 1
    5 / -1

    Atoms of element X form a BCC and atoms of element Y occupy 3/4th of the tetrahedral voids. What is the formula of the compound?

    Solution

    The number of tetrahedral voids form is equal to twice the number of atoms of element X. Number of atoms of Y is 3/4th the number of tetrahedral voids i.e. 3/2 times the number of atoms of X. Therefore, the ratio of numbers of atoms of X and Y are 2:3, hence X2Y3.

  • Question 2
    5 / -1

    Which of the following is true regarding non-ideal solutions with negative deviation?

    Solution

    The interactions between the components of a non-ideal solution showing negative deviation are greater than the pure components. The change in volume and enthalpy after mixing is negative, i.e., ΔVmixing = -ve, ΔHmixing = -ve.

  • Question 3
    5 / -1

    What is the total volume of the particles present in a body centered unit cell?

    Solution

    Since particles are assumed to be spheres and volume of one sphere is 4/3 πr3, total volume of all particles in 

  • Question 4
    5 / -1

    Which of the following cannot form an azeotrope?

    Solution

    H2O + C2H5OH forms an azeotrope with a boiling point of 351.15 K. CHCl3 + C2H5 OH forms an azeotrope with a boiling point of 332.3 K. HCl + H2O forms an azeotrope with a boiling point of 383 K. Benzene + Toluene is an ideal solution and hence does not form an azeotrope.

  • Question 5
    5 / -1

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

    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 6
    5 / -1

    Which of the following is true regarding azeotropes?

    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 7
    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?

    Solution

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

  • Question 8
    5 / -1

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

    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 9
    5 / -1

    How many parameters are used to characterize a unit cell?

    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 10
    5 / -1

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

    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.

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