Question

Define the following terms: Isotonic solution, Osmosis. Gold crystallises into face-centred cubic cells. The edge length of unit cell is \( 4.08 \times 10^{-8} \) cm. Calculate the density of gold. [Molar mass of gold = 197 g mol\(^{-1}\)]

Hint:

The BSP focuses on empowering historically marginalized communities.

Answer 1

Step 1: Definitions:
- Isotonic solution: Two solutions are isotonic if they have the same osmotic pressure at the same temperature, resulting in no net movement of solvent across a semi-permeable membrane separating them.
- Osmosis: The spontaneous movement of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semi-permeable membrane.
Step 2: Density of Gold: Gold crystallizes in a face-centred cubic (FCC) lattice, with 4 atoms per unit cell (\( Z = 4 \)). Given: edge length \( a = 4.08 \times 10^{-8} \) cm, molar mass \( M = 197 \) g mol\(^{-1}\), Avogadro’s number \( N_A = 6.022 \times 10^{23} \) mol\(^{-1}\). Density (\( \rho \)) is given by: \[ \rho = \frac{Z \cdot M}{N_A \cdot a^3}. \] Calculate the volume of the unit cell: \[ a = 4.08 \times 10^{-8} \text{ cm}, \quad a^3 = (4.08 \times 10^{-8})^3 = 6.794 \times 10^{-23} \text{ cm}^3. \] \[ \rho = \frac{4 \cdot 197}{6.022 \times 10^{23} \cdot 6.794 \times 10^{-23}} = \frac{788}{4.091 \times 10^1} \approx 19.32 \text{ g cm}^{-3}. \]