## Chemistry: The Molecular Science (5th Edition)

$V_{conc}=100.\,mL$ $c_{conc}=1.44\,M$ (obtained in part (a)) $c_{dil}=1.10\,M$ $c_{conc}\times V_{conc}= c_{dil}\times V_{dil}$ $\implies V_{dil}=\frac{c_{conc}\times V_{conc}}{c_{dil}}=\frac{1.44\,M\times100\,mL}{1.10\,M}$ $=131\,mL$