1.33dm3 of water at 70degree C are saturated by 2.25mole of lead (ll) trioxonitrate, and 1.33dm3 of water at 18degree C are saturated by 0.53mole of the same salt. If 4.50dm3 of the saturated solution are cooled from 70degreeC to 18degreeC, calculate the amount of solute that will be deposited in

Question

A) mol
B) grams

Answer 1

To calculate the amount of solute that will be deposited when cooling the saturated solution from 70 degrees C to 18 degrees C, we first need to find the solubility of lead (II) trioxonitrate at both temperatures.

First, let's calculate the solubility of lead (II) trioxonitrate at 70 degrees C:
S1 = moles of lead (II) trioxonitrate / volume of water
S1 = 2.25 moles / 1.33 dm^3 = 1.69 mol/dm^3

Next, let's calculate the solubility of lead (II) trioxonitrate at 18 degrees C:
S2 = 0.53 moles / 1.33 dm^3 = 0.40 mol/dm^3

When the solution is cooled from 70 degrees C to 18 degrees C, the solubility of lead (II) trioxonitrate decreases. The difference in solubility between the two temperatures is:
ΔS = S1 - S2 = 1.69 mol/dm^3 - 0.40 mol/dm^3 = 1.29 mol/dm^3

Now, we need to calculate the amount of solute deposited when cooling from 70 degrees C to 18 degrees C using the volume of the saturated solution:
Volume = 4.50 dm^3

Amount of solute deposited = ΔS * Volume
Amount of solute deposited = 1.29 mol/dm^3 * 4.50 dm^3 = 5.80 mol

A) The amount of solute deposited when cooling the saturated solution from 70 degrees C to 18 degrees C is 5.80 mol.

To calculate the mass of the solute deposited, we need to determine the molar mass of lead (II) trioxonitrate (Pb(NO3)2):
Pb: 207.2 g/mol
N: 14.0 g/mol
O: 16.0 g/mol

Molar mass of Pb(NO3)2 = 207.2 + 2(14.0 + 3(16.0)) = 207.2 + 2(14.0 + 48.0) = 207.2 + 2(62.0) = 207.2 + 124.0 = 331.2 g/mol

B) Mass of solute deposited = moles of solute deposited * molar mass
Mass of solute deposited = 5.80 mol * 331.2 g/mol = 1921.76 g

Therefore, the amount of solute that will be deposited when cooling the saturated solution from 70 degrees C to 18 degrees C is:
A) 5.80 mol
B) 1921.76 grams