## Chemistry Homework Help

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Solution

The molar mass of ZnI2 is 319 g/mol. (The formula weight is 319 amu, which is obtained by summing the atomic weight in the formula) Thus

319 g ZnI2

0.5 mol ZnI2 x __________

1 mol ZnI2

= 159.5 gm ZnI2

The molar mass of PbCrO4 is 323 g/mol (i.e) 1 mol PbCrO4 = 323 g PbCrO4

Therefore,

323 g PbCrO4 x 1 mol.PbCrO4

___________________________

323 g PbCrO4

= 1.001 mol PbCrO4

Solution:

(i) The primitive unit cell consists of one atom at each of the 8 corners; each atom is thus shared by 8 unit cells. Hence n = 8 x (1/8) = 1

(ii) The b.c.c unit cell consists of 8 atoms at the 8 corners and one atom at the centre. At each corner only 1/8th of the atom is within the unit cell. Thus the contribution of the 8 corners is 8 x (1/8) = 1 while the number of atoms in the body is 1. Hence, n = 1+1 =2

(iii) The 8 atoms at the corners contribute 8 x (1/8) = 1 atom. There is one atom each of the 6 faces, which is shared by 2 unit cells each. Therefore, the contribution face-centred atoms = 6x (1/2) = 3 Hence, n = 1+3 = 4.

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**Example of Homework problem and solution****Problem****1**. ZnI2, can be prepared by the direct combination of elements. A chemist determines from the amounts of elements that 0.5 mol ZnI2 can be formed.Solution

The molar mass of ZnI2 is 319 g/mol. (The formula weight is 319 amu, which is obtained by summing the atomic weight in the formula) Thus

319 g ZnI2

0.5 mol ZnI2 x __________

1 mol ZnI2

= 159.5 gm ZnI2

**Problem 2:**In the preparation of lead(II)chromate PbCrO4, 323 g of lead(II)chromate is obtained as a precipitate. How many moles of PbCrO4 is this?The molar mass of PbCrO4 is 323 g/mol (i.e) 1 mol PbCrO4 = 323 g PbCrO4

Therefore,

323 g PbCrO4 x 1 mol.PbCrO4

___________________________

323 g PbCrO4

= 1.001 mol PbCrO4

**Problem 3:**Calculate the number (n) of atoms contained within (i) a primitive cubic unit cell (ii) a body –centred cubic unit cell and (iii) a face-centred cubic (f.c.c) unit cellSolution:

(i) The primitive unit cell consists of one atom at each of the 8 corners; each atom is thus shared by 8 unit cells. Hence n = 8 x (1/8) = 1

(ii) The b.c.c unit cell consists of 8 atoms at the 8 corners and one atom at the centre. At each corner only 1/8th of the atom is within the unit cell. Thus the contribution of the 8 corners is 8 x (1/8) = 1 while the number of atoms in the body is 1. Hence, n = 1+1 =2

(iii) The 8 atoms at the corners contribute 8 x (1/8) = 1 atom. There is one atom each of the 6 faces, which is shared by 2 unit cells each. Therefore, the contribution face-centred atoms = 6x (1/2) = 3 Hence, n = 1+3 = 4.

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