$$PV = nRT$$

Where:

* P is the pressure in atmospheres (atm)

* V is the volume in liters (L)

* n is the number of moles of gas

* R is the ideal gas constant (0.08206 L atm/mol K)

* T is the temperature in Kelvin (K)

We need to convert the given mass of sulfur dioxide gas to moles using its molar mass (64.06 g/mol):

$$16.0 \text{ g} \ SO_2 \times \frac{1 \text{ mol} \ SO_2}{64.06 \text{ g} \ SO_2} = 0.250 \text{ mol} \ SO_2$$

Now we can calculate the volume of the gas by rearranging the ideal gas law equation and substituting the given values:

$$V = \frac{nRT}{P}$$

$$V = \frac{(0.250 \text{ mol})(0.08206 \text{ L atm/mol K})(308 \text{ K})}{97 \text{ atm}}$$

$$V = 0.657 \text{ L}$$

Therefore, the volume of 16.0 grams of sulfur dioxide gas at 35 C and 97 atm is 0.657 L.