Question

a chemist prepares a solution of aluminum sulfate $(al_{2}(so_{4})_{3})$ by measuring out 36. g of aluminum sulfate into a 300. ml volumetric flask and filling the flask to the mark with water. calculate the concentration in mol/l of the chemists aluminum sulfate solution. round your answer to 2 significant digits?

Explanation:

Step1: Calculate the molar mass of \(Al_2(SO_4)_3\)

The molar mass of \(Al\) is \(27\space g/mol\), \(S\) is \(32\space g/mol\), and \(O\) is \(16\space g/mol\).

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Step2: Calculate the number of moles of \(Al_2(SO_4)_3\)

Use the formula \(n=\frac{m}{M}\), where \(m = 36\space g\) and \(M = 342\space g/mol\)

$$n=\frac{36}{342}\space mol$$

Step3: Convert the volume of the solution to liters

The volume \(V = 300\space mL=0.3\space L\)

Step4: Calculate the molarity of the solution

Use the formula \(c=\frac{n}{V}\), substitute \(n=\frac{36}{342}\space mol\) and \(V = 0.3\space L\)

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Answer:

\(0.35\space mol/L\)