Question

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

Answer 1

Answer:

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 =