Answer:
Solution's mass = 200.055 g
[PbSO₄] = 275 ppm
Explanation:
Solute mass = 0.055 g of lead(II) sulfate
Solvent mass = 200 g of water
Solution mass = Solvent mass + Solution mass
0.055 g + 200 g = 200.055 g
ppm = μg of solute / g of solution
We convert the mass of solute from g to μg
0.055 g . 1×10⁶ μg/ 1g = 5.5×10⁴μg
5.5×10⁴μg / 200.055 g = 275 ppm
ppm can also be determined as mg of solute / kg of solution
It is important that the relation is 1×10⁻⁶
Let's verify: 0.055 g = 55 mg
200.055 g = 0.200055 kg
55 mg / 0.200055 kg = 275 ppm
Mass of ammonium sulfate = 660.7 g
<h3>Further explanation</h3>
Given
3.01 x 10²⁴ molecules of ammonium sulfate
Required
mass
Solution
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
mol ammonium sulfate (NH₄)₂SO₄ :
n = N : No
n = 3.01 x 10²⁴ : 6.02 x 10²³
n = 5
mass ammonium sulfate :
= mol x MW
= 5 x 132,14 g/mol
= 660.7 g
Answer: 72 g of 
is formed.
Explanation:
To calculate the moles :
As 
is the excess reagent, 
is the limiting reagent as it limits the formation of product.
According to stoichiometry :
1 mole of produce= 2 moles of
Thus 2 moles of will require=
of
Mass of 
Thus 72 g of is formed.
