A chemist prepares a solution of aluminum sulfate (Al2 (SO4) by weighing out 116.0 g of aluminum sulfate into a 450. mL volumetric flask and filing the flask to the mark with water Calculate the concentration in g/dL of the chemist's aluminum sulfate solution. Round your answer to 3 significant digits

0.0733 mol / dL

Explanation:

Mole -

Moles is denoted by given mass divided by the molecular mass,

Hence,

n = w / m

n = moles,

w = given mass,

m = molecular mass.

From the information of the question,

w = 116.0 g

As we know, molecular mass of aluminum sulfate,

m = 342.15 g/mol

n = w / m

n = 116.0 g / 342.15 g/mol = 0.33 mol

Molarity -

Molarity of a substance, is the number of moles present in a liter of solution.

M = n / V

M = molarity

V = volume of solution in liter,

n = moles of solute,

The unit is mol / L,

Hence, from the question, the value of molarity is asked,

From the information of the question,

V = 450 mL,

Since,

1 mL = 0.01 dL

As, the unit asked in the question is in dL,

V = 4.5 dL

n = 0.33 mol (as calculated above)

Using the above formula,

M = n / V

M = 0.33 mol / 4.5 dL

M = 0.0733 mol / dL

M =