Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the
amount amount of solute that will be deposited is 1,927.413 grams.
How can the amount of solute deposited be found?
The volume of water 1.33 dm³ of water 70 °C.
The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C = 2.25 moles
At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles
Therefore;
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;
1.33 dm³ contains 2.25 moles.
[tex]Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}[/tex]
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;
1.33 dm³ contains 0.53 moles
[tex]Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}[/tex]
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles
The number of moles that precipitate out = The amount of solute deposited
Which gives;
Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles
The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g
The molar mass of Pb(NO₃)₂ = 331 g/mol
The amount of solute deposited = Number of moles × Molar mass
Which gives;
The amount of solute deposited = 5.823 moles × 331 g/mol = 1,927.413 g
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