## Chemistry and Chemical Reactivity (9th Edition)

$5.25\ M$
Working from the titration backwards: Number of moles of $Ag^+$ required: $0.100\ mol/L\times26.25\ mL\div 1000mL/L=0.002625\ mol$ Which is the same number of moles of $Cl^-$ in the 50.0 mL sample: $0.002625\ mol\div50.0\ mL\times1000mL/L=0.0525\ M$ Step c is a 10 times dilution, so the concentration from b is: $10\times0.0525\ M=0.525\ M$ Similarly from a to b: $10\times0.525\ M=5.25\ M$