Question

calculate the volume in milliliters of a 1.67m aluminum chloride solution that contains 150. g of aluminum chloride (alcl₃). be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate the molar mass of \(AlCl_3\)

The molar mass of \(Al\) is \(26.98\ g/mol\), and the molar mass of \(Cl\) is \(35.45\ g/mol\).
For \(AlCl_3\), \(M = 26.98+(3\times35.45)=26.98 + 106.35=133.33\ g/mol\)

Step2: Calculate the number of moles of \(AlCl_3\)

Use the formula \(n=\frac{m}{M}\), where \(m = 150\ g\) and \(M=133.33\ g/mol\)
\(n=\frac{150}{133.33}\approx1.125\ mol\)

Step3: Calculate the volume of the solution

Use the formula \(V=\frac{n}{C}\), where \(n = 1.125\ mol\) and \(C = 1.67\ mol/L\)
\(V=\frac{1.125}{1.67}\approx0.674\ L\)

Step4: Convert liters to milliliters

Since \(1\ L=1000\ mL\), \(V = 0.674\times1000 = 674\ mL\)

Answer:

\(674\)