The mass of lead bromide that is dissolved in 250 mL of a saturated solution is____ grams.
Depends on the temperature of the solution. At zero C, .455g of lead bromide can be dissolved in 100g water. At 100C, about ten times that can be dissolved.
For salts in pure water & with low solubilities, the molar amount of compound delivered into solution can be calculated from simple formulas based on ionization ratios ...
Salt--(Ionz'n Ratio)--Solubility(S)
AB => 1:1 => S = (Ksp)ª; a = ½
AB₂ & A₂B => 1:2 or 2:1 => S = (Ksp/4)ª; a = ⅓
AB₄ or A₄B => 1:4 or 4:1 => S = (Ksp/256)ª; a =⅕
A₂B₃ or A₃B₂ => 2:3 or 3:1 => S = (Ksp/108)ª; a =⅕
To determine the mass of lead bromide dissolved in a 250 mL saturated solution, you need to know the solubility of lead bromide at a specific temperature. The solubility is typically given in grams per liter (g/L).
Once you have the solubility value, you can convert it to grams per milliliter (g/mL) since you are working with a 250 mL solution. To do this, divide the solubility in g/L by 1000 to convert from grams to milligrams (mg), and then divide the result by 1000 to convert from milligrams to grams.
Finally, multiply the solubility in g/mL by 250 mL to find the mass of lead bromide dissolved in the solution.
However, without the specific solubility value at a given temperature, it is not possible to determine the mass of lead bromide dissolved.