1.33dm^3 of water at 70°c are saturated by 2.25 mole of lead 2.2trioxocarbonate(v) pb(No3) and 1.33dm^3 of water at 18°c are saturated by 0.53 mole of the same salt.If 4.5dmdm^3 of the saturated solution are cool from 70°c to 18°c.Calculate the amount of solute that will be deposited in;
a) mols
b)gram .(pb=207,N=14)
To solve this problem, we need to calculate the solubility of lead 2.2trioxocarbonate(v) pb(No3) at both 70°C and 18°C, and then determine the amount of solute that will be deposited when the saturated solution is cooled from 70°C to 18°C.
1. Calculate the solubility at 70°C:
From the given data, we know that 1.33 dm^3 of water at 70°C is saturated by 2.25 moles of lead 2.2trioxocarbonate(v) pb(No3). We can calculate the molarity of the saturated solution at 70°C as follows:
Molarity = moles of solute / volume of solution (in L)
Molarity = 2.25 moles / 1.33 dm^3 = 2.25 moles / 1.33 L = 1.69 M
2. Calculate the solubility at 18°C:
Similarly, at 18°C, 1.33 dm^3 of water is saturated by 0.53 moles of lead 2.2trioxocarbonate(v) pb(No3). Calculate the molarity at 18°C:
Molarity = moles of solute / volume of solution (in L)
Molarity = 0.53 moles / 1.33 dm^3 = 0.53 moles / 1.33 L = 0.40 M
3. Calculate the amount of solute that will be deposited when the saturated solution cools from 70°C to 18°C:
First, calculate the amount of solute in 4.5 dm^3 of the saturated solution at 70°C:
Amount of solute = Molarity * Volume of solution
Amount of solute at 70°C = 1.69 M * 4.5 L = 7.60 moles
Next, calculate the amount of solute in 4.5 dm^3 of the saturated solution at 18°C:
Amount of solute at 18°C = 0.40 M * 4.5 L = 1.80 moles
Finally, calculate the amount of solute deposited when cooling from 70°C to 18°C:
Amount deposited = Amount at 70°C - Amount at 18°C
Amount deposited = 7.60 moles - 1.80 moles = 5.80 moles
4. Convert the amount of solute deposited to grams:
To convert moles of lead 2.2trioxocarbonate(v) pb(No3) to grams, use the molar mass of the compound:
Molar mass = 207 (Pb) + 2(14) (N) + 9(16) (O) + 3(16) (O) = 331 g/mol
Mass of solute deposited = Amount deposited * Molar mass
Mass of solute deposited = 5.80 moles * 331 g/mol = 1919.80 grams
Therefore, the amount of solute that will be deposited when cooling the saturated solution from 70°C to 18°C is:
a) 5.80 moles
b) 1919.80 grams