3 Marks Question

Question 13 Marks

Silver crystallises in fcc lattice. If edge length of the cell is $ 4.077\times10^{-8} $ cm and density is 10.5 g $ cm^{-3} $, calculate the atomic mass of silver.

Answer

For a $f c c$ lattice,
$Z =4$
Edge length, $a=4.077 \times 10^{-8} cm$.
Density, $\rho=10.5 g cm ^{-3}$
Using the relationship, density,
$\rho=\frac{ ZM }{ N _{ A } \times a^3}$ or $M =\frac{\rho \times N _{ A } \times a^3}{ Z }$
This gives,
$M =\frac{10.5 g cm ^{-3} \times 6.02 \times 10^{23} mol^{-1} \times\left(4.077 \times 10^{-8} cm\right)^3}{4}$
$=107.14 g mol ^{-1}$
So, the atomic mass of silver is 107.14

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Question 23 Marks

An element 'X' (At mass = 40 g $ mol^{-1} $) having fcc structure has unit cell edge length of 400 pm. Calculate the density of 'X' and the number of unit cell in 4 g of 'X'. $ (N_{A}=6.022\times10^{23}mol^{-1}) $.

Answer

We know $ d = \frac{Z \times M}{N_{A} \times a^{3}} $
For fcc, $ Z = 4 $.
$ d = \frac{4 \times 40}{6.022 \times 10^{23} \times (400 \times 10^{-10})^{3}} $
$= 4.1514 \text{ g cm}^{-3} $
Volume of 4 g 'X' = $ \frac{4}{4.1514} = 0.96 \text{ cm}^{3} $
Volume of 1 unit cell = $ (400 \times 10^{-10})^{3} $$cm ^3$
$= 64 \times 10^{-24} \text{ cm}^{3} $
Number of unit cells = $ \frac{0.96}{64 \times 10^{-24}} = 1.5 \times 10^{22} $.

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Question 33 Marks

Iron has a body centred unit cell with a cell edge of 286.6 pm. The density of iron is 7.87 g $ cm^{-3} $ Use these information to calculate Avogadro's number. (Atomic weight of Fe, 56 g $ mol^{-1} $)

Answer

We know that $\rho=\frac{ ZM }{ N _{ A } a^3}, Z =2$
Given, $d=7.87 g cm ^{-1}, M =56 g mol ^{-1}$,
$a=286.65 \times 10^{-10} cm$
$\therefore \quad N _{ A }=\frac{2 \times 56}{7.87 \times(286.65)^3 \times 10^{-30}}=6.04 \times 10^{23} mol^{-1}$

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Question 43 Marks

Write down three differences between crystalline and amorphous solids.

Answer

Distinction between Crystalline and Amorphous Solids

Property	Crystalline Solids	Amorphous Solids
1. Shape	Definite characteri-stics and geometrical shape.	Irregular shape.
2. M.Pt.	Melt at a sharp and characteristics temp.	Gradually soften over a range of temperature.
3. Cleavage property	When cut with a sharp edged tool, they split into the pieces and the newly generated surfaces are plain and smooth.	When cut with sharp a edged tool, they cut into two pieces with irregular surfaces.

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Question 53 Marks

Calculate the void space in a primitive unit cell and also the fraction of the total volume occupied.

Answer

In a primitive unit cell suppose the edge length of the unit cell $=a$ and the radius of sphere $=r$. As spheres are touching each other so $a=2 r$.
No. of spheres per unit cell $=\frac{1}{8} \times 8=1$.
Volume of the sphere $=\frac{4}{3} \pi r^3$
Volume of the cube $=a^3=(2 r)^3=8 r^3$
$\therefore \quad$ Fraction occupied $=\frac{4}{3} \pi r^3 / 8 r^3$
$=0.524$ or $52.4 \%$.

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