Why will the conjugate base of a weak acid affect pH? Select the correct answer below: it will react with hydroxide
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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.
<h3>How can the amount of solute deposited be found?</h3>
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.
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
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 =<u> 1,927.413 g </u>
Learn more about saturated solutions here:
Answer : The mass of of water present in the jar is, 298.79 g
Solution : Given,
Mass of barium nitrate = 27 g
The solubility of barium nitrate at is 9.02 gram per 100 ml of water.
As, 9.02 gram of barium nitrate present in 100 ml of water
So, 27 gram of barium nitrate present in of water
The volume of water is 299.33 ml.
As we know that the density of water at is 0.9982 g/ml
Now we have to calculate the mass of water.
Therefore, the mass of of water present in the jar is, 298.79 g