Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

The Solid State

Long Answer Type

151. An element occurs in BCC structure with cell edge of 300 pm. The density of the element is 5.2 g cm^–3. How many atoms of the element does 200 g of the element contain?

1177 Views

Short Answer Type

152. Iron (II) oxide has a cubic structure and each unit cell has side 5A. If the density of the oxide is 4 g cm^–3, calculate the number of Fe²⁺ and O^2– ions present in each unit cell. (Molar mass of FeO = 72 g mol^–1, N_A = 6.02 x 10²³ mol^–1).

2407 Views

153. A metallic element exists as a cubic lattice. Each edge of the unit cell is 2.88 A°. The density of the metal is 7.20 g cm^–3. How many unit cells will be there in 100 g of the metal?

2490 Views

154. The unit cell of an element of atomic mass 96, and density 10.3 g cm^–3 is a cube with edge length of 314 pm. Find the structure of crystal lattice (simple cubic, F.C.C. or B.C.C.) Avogadro’s constant. N_A = 6.023 x 10²³ mol^–1?

2325 Views

Long Answer Type

155. An element a crystallises in fcc structure. 200 g of this element has 4.12 x 10²⁴atoms. The density of A is 7.2 g cm^-3. Calculate the edge length of the unit cell?

1979 Views

Short Answer Type

156. An element crystallises in BCC structure. The edge length of its unit cell is 288 pm. If the density of crystal is 7.2 g cm^–3, what is the atomic mass of the element?

1664 Views

Long Answer Type

157. Caesium chloride crystallises as a body centred cubic lattice and has a density of 4.0 g cm^–3. Calculate the length of the edge of the unit cell of caesium chloride crystal.
[Molar Mass of CsCl = 168.5 g mol^–1, N_A = 6.02 x 10²³ mol^–1]

564 Views

Short Answer Type

158. The density of chromium is 7.2 g cm^–3. If the unit cell is cubic with edge length of 289 pm, determine the type of the unit cell (Atomic mass of Cr = 52 amu).

1682 Views

Long Answer Type

159. Iron (II) oxide has a cubic structure and each of the unit cell is 5.0 A°. If density of the oxide is 4.0 g cm^-3, calculate the number of Fe²⁺and O^2– ions present in each unit cell.

Volume of the unit cell = (5 A°)
= (5 x 10^–8 cm)³= 125 x 10^–24 cm³= 1.25 x 10^–22 cm³Density of FeO = 4.0 g cm^–3
Therefore, mass of the unit cell
= 1.25 x 10^–22 cm³ x 4 g cm^–3= 5 x 10^–22g
Mass of all molecules of FeO
$= \frac{72}{6.022 \times 10^{23}} = 1.195 \times 10^{- 22} g$

Number of FeO molecules/unit cell
$= \frac{5 \times 10^{- 22} g}{1.195 \times 10^{- 22} g} = 4.19 \approx 4$

Thus, there are 4 Fe²⁺ ions and 4O^2– ions in each unit cell.