Given:
Step 1: Calculate the volume of one unit cell (\(V_{\text{cell}}\)): \[ V_{\text{cell}} = a^3 = (4 \times 10^{-10} \, \text{m})^3 \] \[ = 4^3 \times (10^{-10})^3 \, \text{m}^3 \] \[ = 64 \times 10^{-30} \, \text{m}^3 \] Step 2: Calculate the mass of one unit cell (\(m_{\text{cell}}\)) using the density formula: \( \rho = \frac{m_{\text{cell}}}{V_{\text{cell}}} \). \[ m_{\text{cell}} = \rho \times V_{\text{cell}} \] \[ = 7.5 \times 10^{3} \, \text{kg/m}^3 \times 64 \times 10^{-30} \, \text{m}^3 \] \[ = (7.5 \times 64) \times (10^{3} \times 10^{-30}) \, \text{kg} \] \[ = 480 \times 10^{-27} \, \text{kg} \] \[ = 4.8 \times 10^{-25} \, \text{kg} \] Step 3: Calculate the number of unit cells present in 0.015 kg of metal: \[ \text{Number of unit cells} = \frac{\text{Total mass of metal}}{\text{Mass of one unit cell}} \] \[ = \frac{m}{m_{\text{cell}}} = \frac{0.015 \, \text{kg}}{4.8 \times 10^{-25} \, \text{kg}} \] \[ = \frac{0.015}{4.8} \times 10^{25} \] \[ = \frac{15 \times 10^{-3}}{48} \times 10^{25} \] \[ = \frac{15}{48} \times 10^{22} \] \[ = \frac{5}{16} \times 10^{23} \] \[ = 0.3125 \times 10^{23} \] \[ = 3.125 \times 10^{22} \] The number of unit cells present in 0.015 kg of the metal is \(3.125 \times 10^{22}\).
Correct Answer: (3) \(3.125 \times 10^{22}\)
Two statements are given below
Statement I: Benzanamine can be prepared from phthalimide.
Statement II: Benzanamine is less basic than phenyl methanamine.
What are X and Z in the following reaction sequence?
What is Y in the following reaction sequence?
Observe the following set of reactions:
Correct statement regarding Y and B is:
What is X in the following reaction?