1.33dm^3 of water at 70°c are saturated by 2.25 mole of lead 2.2 trioxocarbonate(v)pb(no3)and 1.33dm^3 of water at 18°cqre saturated by 0.53 mole of the same salt.if 4.5dm^3 of the saturated solution are cool from 70°c to 18°c . calculate the amount of solute that will be deposited in:a)mole b)gram.(pb=207,n=14).
To calculate the amount of solute that will be deposited when the saturated solution is cooled from 70°C to 18°C, we need to determine the amount of solute that is no longer soluble in the cooler temperature.
1. Calculate the solubility at 70°C:
From the given information, 2.25 moles of Pb(NO3)2 saturate 1.33 dm^3 of water at 70°C.
Molar mass of Pb(NO3)2 = 207 + 14 + (3 x 16) x 2 = 331 g/mol
The solubility at 70°C is:
2.25 moles x 331 g/mol = 743.25 g
2. Calculate the solubility at 18°C:
0.53 moles of Pb(NO3)2 saturate 1.33 dm^3 of water at 18°C.
The amount of salt that would saturate 1 dm^3 of water at 18°C is:
0.53 moles / 1.33 dm^3 = 0.3985 moles per dm^3
The molar mass of the salt is 331 g/mol.
The solubility at 18°C is:
0.3985 moles x 331 g/mol = 131.6 g
3. Calculate the amount of solute to be deposited:
The solubility decreases from 743.25 g at 70°C to 131.6 g at 18°C when cooling the solution.
The amount of solute to be deposited when cooling from 70°C to 18°C is:
743.25 g - 131.6 g = 611.65 g
4. Convert the amount of solute to be deposited into moles:
611.65 g / 331 g/mol = 1.85 moles
Therefore, the amount of solute that will be deposited when cooling the saturated solution from 70°C to 18°C is:
a) 1.85 moles
b) 611.65 g