91Ó°ÊÓ

On converting the given volume, we get:

$$\begin{gathered} V_{dil} of so\ln =3 gal so\ln ×\frac{4 qt so\ln }{1 gal so\ln }×\frac{0.946 L so\ln }{1 qt so\ln } \\                           =11.352 L s\ln \\                V_{conc}=\frac{M_{dil}×V_{dil}}{M_{conc}} \\                            =\frac{3.5 M×11.352 L}{11.7 m} \\                            =3.396 L solution. \end{gathered}$$

The first step is to prepare a 3.396 L of 11.7 M HCl, and water can be added till it reaches a volume of 11.352 L to get a diluted acid concentration of 3.5 M.

On converting the given mass of HCl to moles, we get:

$\begin{array}{rcl} Moles of HCl& =& 9.66 g HCl×\frac{1 mol HCl}{36.46 g HCl}\\                                   & =& 0.265 mol HCl\\ Volume of HCl& =& \frac{0.265 mol HCl}{11.7\frac{mol HCl}{L so\ln }}\\                                  & =& 0.0226 L solution\\                                  & =& 22.6 mL solution.\end{array}$