Answer:
50.5 g/dL
Explanation:
From the question given above, the following data were obtained:
Mass of Al₂(SO₄)₃ = 101 g
Volume (V) = 200 mL
Concentration of solution (g/dL) =?
Next, we shall convert 200 mL to decilitre (dL).
This is illustrated below:
1 mL = 0.01 dL
Therefore,
200 mL = 200 mL / 1 mL × 0.01 dL
200 mL = 2 dL
Therefore, 200 mL is equivalent to 2 dL.
Finally, we shall determine the concentration of the solution in g/dL as follow:
Mass of Al₂(SO₄)₃ = 101 g
Volume (V) = 2 dL
Concentration of solution (g/dL) =?
Concentration of solution (g/dL) = mass /volume
Concentration of solution (g/dL) = 101/2
= 50.5 g/dL
Therefore, the concentration of the Al₂(SO₄)₃ solution is 50.5 g/dL
