When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = 
=74.8 %.
Answer:
1.43 M
Explanation:
We'll begin by calculating the number of mole of the solid. This can be obtained as follow:
Mass of solid = 8.60 g
Molar mass of solid = 21.50 g/mol
Mole of solid =?
Mole = mass / molar mass
Mole of solid = 8.60 / 21.50
Mole of solid = 0.4 mole
Next, we shall convert 280 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
280 mL = 280 mL × 1 L / 1000 mL
280 mL = 0.28 L
Thus, 280 mL is equivalent to 0.28 L.
Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:
Mole of solid = 0.4 mole
Volume = 0.28 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.4 / 0.28
Molarity = 1.43 M
Thus, the molarity of the solution is 1.43 M.
molar mass of the molecule is 60. Moles=mass/molar mass
92.02g/60=1.53366667mol
1.53mol(3sf)
