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

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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.