Question

A chemist prepares a solution of aluminum sulfate Al2SO43 by weighing out 101.g of aluminum sulfate into a 200.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /gdL of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits. gdL

Answer 1

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