# Numerical Problems Based on Chemistry Chapter 1 Solid State

Numerical Problems Based on Class 12 Chemistry Chapter 1 Solid State

Q.1. A cubic solid is made up of two clements X and Y. Atoms Y are present at the corners of the cube and atoms X at the body centre. What is the formula of the compound ?

Q.2. When atoms are placed at the corners of all 12 edges of a cube, how many atoms are present per unit cell?

Q.3. A unit cell consists of a cube in which there are A atoms at the corners and Batoms at the face centres and A atoms are missing from 2 corners in each unit cell. What is the simplest formula of the compound?

Q.4. A compound of X and Y crystallizes in the cubic structure in which Y atoms are at the corners and X atoms are at the alternate faces of the cube. Find the formula of the compound.

Q.5. Gold crystallizes in the face centred cubic lattice. Calculate the approximate number of unit cells in 2 mg of gold. (atomic mass of gold = 197 u).

Q.6. Xenon crystallizes in the face centred cubic lattice and the edge of the unit cell is 620 pm. What is the nearest neighbour distance and radius of xenon atom ?

Q.7. The radius of an atom is 220 pm. If it crystallizes as a face centred cubic lattice, what is the length of the side of the unit cell ?

Q.8. Silver crystallizes with face centred cubic unit cell. Each side of the unit cell has a length of 400 pm. Calculate the radius of the silver atom.

Q.9. A solid has a structure in which W atoms are present at the corners of the cubic unit cell, O atoms are located at the cube edges and Na atoms are present at cube centres. What is the formula of the compound?

Q.10. CsCl crystallizes in cubic structure in which Cl- ions are at the corners and Cs+ ions occupy centre of the unit cell. If radii of Cs+ and Cl-ions are 1.698 and 1.818 respectively, calculate the edge length of the unit cell.

Q.11. Tungsten crystallizes in body centred cubic lattice. Calculate the number of unit cells in 1.5 g of tungsten (Atomic mass of tungsten = 184 u).

Q.12. Copper crystallizes in a face centred cubic lattice. Calculate the number of unit cells in 12 g of copper (atomic mass of copper = 63,5 u).

Q.13. Gold (atomic radius 0.144 nm) crystallizes in face centred unit cell. What is the length of the side of the cell?

Q.14. In a face centred cubic arrangement of A and B atoms, A atoms occupy the corners and B atoms occupy the face centres of the unit cell. If one of the atoms is missing from the corner in each unit cell, what is the simplest formula of the compound

Q.15. The two ions A- and B-have radii 88 and 200 pm respectively. In the close packed crystal of compound AB, predict the coordination number of A+.

Q.16. A solid AB has NaCl structure. If the radius of cation A is 100 pm, what is the radius of anion B?

Q.17. If the radius of Brion is 0.182 nm, how large a cation can fit in each of the tetrahedral hole ?

Q.18. A solid is made of two elements X and Y. Atoms X are in fcc arrangement and Y atoms occupy all the octahedral sites and alternate tetrahedral sites. What is the formula of the compound ?

Q.19. A compound is formed by two elements X and Y. Atoms of element Y (as anions) make ccp and those of the element X (as cations) occupy all the octahedral voids. What is the formula of the compound.

Q.20. A compound is formed by two elements M and N. The element N forms ccp and M atom occupy 1/3 of the tetrahedral voids. What is the formula of the compound?

Q.21. In a crystalline solid, anions Y are arranged in cop arrangement. Cations X are equally distributed between tetrahedral and octahedral voids. If all the octahedral voids are occupied, what is the formula of the solid ?

Q.22. In a solid, oxide ions are arranged in ccp. One sixth of the tetrahedral voids are occupied by the cations A and one third of the octahedral voids are occupied by the cations B. What is the formula of the compound ?

Q.23. In corrundum, oxide ions are arranged in hcp arrangement and the aluminium ions occupy 2/3 of the octahedral voids. What is the formula of corrudum

Q.24. A unit cell of sodium chloride has four formula units per unit cell. The edge length of unit cell is 0.564 nm. Find out the density of sodium chloride.

Q.25. Tungsten has body centred cubic lattice. Each edge of the unit cell is 316 pm and density of the metal is 19.35 g cm-3. How many atoms are present in 50 g of the element?

Q.26. An element occurs in bcc structure with cell edge of 288 pm. Its density is 7.2 g cm-3. Calculate the atomic mass of the element.

Q.27. CsCl has body centred cubic lattice with the length of a side of a unit cell 412.1 pm and aluminium is face centred cubic lattice with length of the side of unit cell 405 pm. Which of the two has larger density ? (Atomic mass of Cs = 132.9, Cl = 29.9)

Q.28. An element crystallizes in a structure having fcc unit cell of an edge of 300 pm. Calculate its density if 200 g of this element contains 4.12 x 1024 atoms.

Q.29. Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro number. (At. mass of Fe = 55.845 u)

Q.30. Gold (atomic mass = 197 u, atomic radius = 0.144 nm) crystallizes in a face centred unit cell. Determine the density of gold, (NA = 6.022 x 1023 mol-1).

Q.31. An element (atomic mass = 60) having face centred cubic structure has a density of 6.23 g cm-3. What is the edge length of the unit cell?

Q.32. An element (density 6.8 g cm-) occurs in bcc structure with cell edge of 290 pm. Calculate the number of atoms present in 200 g of the element. (P.S.B. 2011)

Q.33. Lead (II) sulphide crystal has NaCl structure. What is its density ? The edge length of the unit cell of PbS crystal is 500 pm. (atomic masses : Pb = 207, S = 32)

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