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 Multiple Choice QuestionsShort Answer Type

271.

An element with density 2.8 g cm−3 forms of the f.c.c. unit cell with edge length 4 X10−8 cm. Calculate the molar mass of the element.
(Given: NA = 6.022 X 1023 mol −1)

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272.

(i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
(ii) What type of stoichiometric defect is shown by NaCl?

OR
How will you distinguish between the following pairs of terms:
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell.


i) The pink colour of LiCl is because of the metal excess defect caused by the anionic vacancies.
ii) NaCl shows Schottky defect. In the Schottky defect, an equal number of cations and anion are missing from their regular sites.

Or
(i) Tetrahedral void surrounded by four spheres is called a tetrahedral void.

Octahedral void surrounded by six spheres is called an octahedral void.

(ii) Crystal lattice: A crystal lattice is a diagrammatic representation of the constituent particles such as atoms, ions and molecules of a crystal in a regular three-dimensional arrangement.
Unit cell: A unit cell is the smallest three-dimensional portion of a crystal lattice. When it is repeated in different directions, it generates the entire crystal lattice.

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273.

An element has atomic mass 93 g mol–1 and density 11.5 g cm–3. If the edge
the length of its unit cell is 300 pm, identify the type of unit cell.

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274.

Write any two differences between amorphous solids and crystalline solids.

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275.

Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a
f.c.c. structure. (Atomic mass of Al = 27 g mol–1)

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276.

Give reasons :
In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel
defect.

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277.

Give reasons :
Silicon on doping with Phosphorus forms the n-type semiconductor.

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278.

Give reason:
Ferrimagnetic substances show better magnetism than antiferromagnetic
substances.

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279.

Analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. Give reason.


280.

An element ‘X’ (At. mass = 40 g mol–1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 1023 mol–1 )


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