Question

how many grams of potassium chloride, kcl, are required to prepare 500.0ml of a 1.50 m kcl solution? (molar mass of kcl = 74.55 g/mol)
a. 10.1 g
b. 24.9 g
c. 99.4 g
d. 224 g
e. 55.9 g

Explanation:

Step1: Calculate the number of moles of KCl

The formula for molarity \(M=\frac{n}{V}\) (where \(n\) is the number of moles and \(V\) is the volume in liters).
Given \(M = 1.50\space M\) and \(V=500.0\space mL=0.5000\space L\)
Rearrange the formula to \(n = M\times V\)
\(n=1.50\space mol/L\times0.5000\space L = 0.750\space mol\)

Step2: Calculate the mass of KCl

The formula for mass \(m=n\times M_m\) (where \(M_m\) is the molar mass).
Given \(n = 0.750\space mol\) and \(M_m=74.55\space g/mol\)
\(m=0.750\space mol\times74.55\space g/mol=55.9125\space g\approx55.9\space g\)

Answer:

E. 55.9 g