Question

Silver crystallizes in fcc structure. If edge length of unit cell is 400 pm, calculate density of silver (Atomic mass of Ag = 108). Write a note on Halform reaction.

Hint:

For fcc crystals, the density depends on the number of atoms per unit cell and the edge length. In the Haloform reaction, halogenation occurs at the methyl group adjacent to a carbonyl group.

Answer 1

Step 1: Formula for Density of a Cubic Unit Cell.
The density (\(\rho\)) of a crystal is given by the formula: \[ \rho = \frac{Z \cdot M}{N_A \cdot V} \] Where: - \(Z\) is the number of atoms per unit cell (for fcc structure, \(Z = 4\)), - \(M\) is the molar mass of the substance (108 g/mol for silver), - \(N_A\) is Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol), - \(V\) is the volume of the unit cell. Step 2: Volume of the Unit Cell.
For fcc structure, the edge length \(a\) is related to the volume \(V\) of the unit cell: \[ V = a^3 \] Where \(a = 400 \, \text{pm} = 400 \times 10^{-12} \, \text{m}\). The volume of the unit cell is: \[ V = (400 \times 10^{-12})^3 = 6.4 \times 10^{-29} \, \text{m}^3 \] Step 3: Calculate the Density.
Substitute the values into the density formula: \[ \rho = \frac{4 \cdot 108 \, \text{g/mol}}{6.022 \times 10^{23} \cdot 6.4 \times 10^{-29} \, \text{m}^3} \] \[ \rho = 10.5 \, \text{g/cm}^3 \] Thus, the density of silver is \(10.5 \, \text{g/cm}^3\). Step 4: Halform Reaction.
The Haloform reaction is a chemical reaction where a haloform (CHCl3, CHBr3, or CHI3) is produced when a methyl group is attached to a carbonyl group, typically in the presence of halogens. For example, when chloroform is reacted with a base, it forms chloroform and a carboxylate ion: \[ CH_3CH(OH)COOH + 3Cl_2 \rightarrow CHCl_3 + \text{carboxylate ion} \]