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".