16.55g of lead (ll) trioxonitrate (V) was dissolved in 100g of distilled water at 20oC, calculate the solubility of the solute in moldm-3 [Pb = 207, N = 14,...

Answer Details

The problem provides the amount of lead (II) trioxonitrate (V) dissolved in 100g of water at a specific temperature. The question asks for the solubility of the solute in moldm-3. To calculate the solubility, we need to determine the amount of the solute (lead (II) trioxonitrate (V)) that is present in one liter of the solution. We can use the given amount of the solute and the molar mass of the compound to calculate the number of moles present in the solution. First, we calculate the molar mass of lead (II) trioxonitrate (V): 207 (mass of lead) + 3(16) (mass of oxygen) + 2(14) (mass of nitrogen) = 406 g/mol Next, we calculate the number of moles of the solute present in the solution: 16.55 g / 406 g/mol = 0.0407 mol Finally, we can calculate the solubility of the solute in moldm-3 by dividing the number of moles by the volume of the solution in liters: 0.0407 mol / 0.1 L = 0.407 moldm-3 Therefore, the solubility of the solute in moldm-3 is 0.407. The correct option is: -  0.05 g (none of the above) None of the options provided match the calculated solubility value.