If 24.4 g of lead (ll) trioxonitrate (V) were dissolved in 42 g of distilled water at 20oC; calculate the solubility of the solute in gdm-3.

Answer Details

To calculate the solubility of lead (II) trioxonitrate (V) in water, we need to use the formula: Solubility (g/dm³) = (mass of solute ÷ volume of solvent) First, let's convert the temperature to Kelvin (K) using the formula K = °C + 273.15: 20°C + 273.15 = 293.15 K Next, we need to calculate the volume of water that the lead (II) trioxonitrate (V) was dissolved in. We can assume that the density of water is 1 g/cm³, so: Volume of water = mass of water ÷ density of water Volume of water = 42 g ÷ 1 g/cm³ Volume of water = 42 cm³ Now, we can calculate the solubility of lead (II) trioxonitrate (V) in water: Solubility (g/dm³) = (mass of solute ÷ volume of solvent) Solubility (g/dm³) = 24.4 g ÷ (42 cm³ ÷ 1000) Solubility (g/dm³) = 581 g/dm³ Therefore, the solubility of lead (II) trioxonitrate (V) in water is 581 g/dm³. The correct option is the one that reads "581.000".