iv) The degree of dissociation of Ca(NQ3)2 in a dilute aqueous solution, containing 7.0 g of salt per 100 g of water at 100°Cis 70%. If the vapour pressure of water at 100°C is 760 mm of Hg, calculate the vapour pressure of the solution.

To solve this problem, we need to use Raoult's Law, which states that the partial pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

Step 1: Calculate the moles of Ca(NQ3)2 and H2O The molar mass of Ca(NQ3)2 is approximately 308.29 g/mol. So, the number of moles of Ca(NQ3)2 is 7.0 g / 308.29 g/mol = 0.0227 mol. The molar mass of H2O is 18.015 g/mol. Since the density of water is approximately 1 g/mL, 100 g of water is about 100 mL of water. So, the number of moles of H2O is 100 g / 18.015 g/mol = 5.55 mol.

Step 2: Calculate the mole fraction of Ca(NQ3)2 and H2O The mole fraction of Ca(NQ3)2 is 0.0227 mol / (0.0227 mol + 5.55 mol) = 0.0041. The mole fraction of H2O is 5.55 mol / (0.0227 mol + 5.55 mol) = 0.9959.

Step 3: Apply Raoult's Law The vapor pressure of the solution is equal to the sum of the vapor pressures of the components, each multiplied by its mole fraction. Since the degree of dissociation of Ca(NQ3)2 is 70%, we consider it as a non-volatile solute, so it does not contribute to the vapor pressure. Therefore, the vapor pressure of the solution is just the vapor pressure of the water component, which is 760 mm Hg * 0.9959 = 757.1 mm Hg.