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 moles of Ca(NQ3)2 = 7.0 g / 308.29 g/mol = 0.0227 mol. The molar mass of H2O is 18.015 g/mol. So, the moles of H2O = 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 = moles of Ca(NQ3)2 / (moles of Ca(NQ3)2 + moles of H2O) = 0.0227 mol / (0.0227 mol + 5.55 mol) = 0.0041. The mole fraction of H2O = moles of H2O / (moles of Ca(NQ3)2 + moles of H2O) = 5.55 mol / (0.0227 mol + 5.55 mol) = 0.9959.

Step 3: Calculate the vapor pressure of the solution According to Raoult's Law, the vapor pressure of the solution = mole fraction of H2O * vapor pressure of pure H2O = 0.9959 * 760 mm Hg = 757.47 mm Hg.

Therefore, the vapor pressure of the solution is approximately 757.47 mm Hg.