Question

a chemist adds 305.0 ml of a 3.79 mol/l potassium iodide (ki) solution to a reaction flask. calculate the mass in kilograms of potassium iodide the chemist has added to the flask. be sure your answer has the correct number of significant digits. kg

Explanation:

Step1: Calculate the number of moles of KI

Use the formula \(n = c\times V\), where \(c = 3.79\space mol/L\) and \(V=305.0\space mL=305.0\times10^{- 3}\space L\).
\(n = 3.79\space mol/L\times305.0\times10^{-3}\space L\)
\(n=3.79\times0.3050\space mol\)
\(n = 1.15595\space mol\)

Step2: Calculate the molar mass of KI

The molar mass of \(K\) is \(M_{K}=39.10\space g/mol\), and the molar mass of \(I\) is \(M_{I}=126.90\space g/mol\).
The molar mass of \(KI\), \(M = M_{K}+M_{I}\)
\(M=(39.10 + 126.90)\space g/mol=166.0\space g/mol\)

Step3: Calculate the mass of KI

Use the formula \(m=n\times M\), where \(n = 1.15595\space mol\) and \(M = 166.0\space g/mol\)
\(m=1.15595\space mol\times166.0\space g/mol\)
\(m = 192.9877\space g\)

Step4: Convert grams to kilograms

Since \(1\space kg = 1000\space g\), then \(m=\frac{192.9877\space g}{1000}\)
\(m=0.1929877\space kg\approx0.193\space kg\) (three significant digits)

Answer:

\(0.193\space kg\)