Question

which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: 150.0 ml of 0.60 m alcl3, 75.0 ml of 1.2 m mgcl2, or 300.0 ml of 0.80 m nacl?

Explanation:

Step1: Calculate moles of chloride ions for \(AlCl_3\)

Use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume in liters). For \(AlCl_3\), each \(AlCl_3\) dissociates into \(Al^{3 +}+3Cl^{-}\).
\(V = 150.0\space mL=0.150\space L\), \(C = 0.60\space M\)
\(n_{AlCl_3}=0.60\times0.150 = 0.09\space mol\)
\(n_{Cl^{-}}=3\times n_{AlCl_3}=3\times0.09 = 0.27\space mol\)

Step2: Calculate moles of chloride ions for \(MgCl_2\)

For \(MgCl_2\), each \(MgCl_2\) dissociates into \(Mg^{2+}+2Cl^{-}\).
\(V = 75.0\space mL = 0.075\space L\), \(C = 1.2\space M\)
\(n_{MgCl_2}=1.2\times0.075=0.09\space mol\)
\(n_{Cl^{-}}=2\times n_{MgCl_2}=2\times0.09 = 0.18\space mol\)

Step3: Calculate moles of chloride ions for \(NaCl\)

For \(NaCl\), each \(NaCl\) dissociates into \(Na^{+}+Cl^{-}\).
\(V = 300.0\space mL=0.300\space L\), \(C = 0.80\space M\)
\(n_{NaCl}=0.80\times0.300 = 0.24\space mol\)
\(n_{Cl^{-}}=n_{NaCl}=0.24\space mol\)

Answer:

\(150.0\space mL\) of \(0.60\space M\space AlCl_3\)