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°c are 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).
First, we need to calculate the solubility of Pb(NO3)2 at both 70°C and 18°C using the given information:
At 70°C:
2.25 moles of Pb(NO3)2 are dissolved in 1.33 dm^3 of water, so the molarity of the solution is:
Molarity = moles of solute / volume of solution (in liters)
Molarity = 2.25 moles / 1.33 dm^3 = 1.69 mol/dm^3
At 18°C:
0.53 moles of Pb(NO3)2 are dissolved in 1.33 dm^3 of water, so the molarity of the solution is:
Molarity = moles of solute / volume of solution (in liters)
Molarity = 0.53 moles / 1.33 dm^3 = 0.40 mol/dm^3
Next, we need to calculate how much Pb(NO3)2 will precipitate when the solution is cooled from 70°C to 18°C. The solubility of Pb(NO3)2 at 18°C is 0.40 mol/dm^3, which means that when the solution is cooled to 18°C, the excess Pb(NO3)2 (1.69 - 0.40 = 1.29 mol/dm^3) will precipitate.
a) Amount of solute deposited in moles:
1.29 mol/dm^3 x 4.5 dm^3 = 5.80 moles of Pb(NO3)2 will precipitate
b) Amount of solute deposited in grams:
The molar mass of Pb(NO3)2 is:
(1*207) + (2*14) + (6*16) = 331 g/mol
So, the mass of Pb(NO3)2 that will precipitate is:
5.80 moles x 331 g/mol = 1919.8 grams or approximately 1920 grams.