# What is the concentration in "%w/v" of sodium sulfate when "9.74 g" of it is placed into a 165 volume?

Well... $165$ what? I assume $\text{mL}$ (because who would ever drop $\text{9.74 g}$ of something in a huge $\text{165 L}$ vat?)...
${c}_{N {a}_{2} S {O}_{4}} = \left(\text{9.74 g Na"_2"SO"_4)/("165 mL soln}\right)$
%w//v is defined for a solute mass in $\text{g}$ and a solution volume in $\text{mL}$, so we are basically done...
color(blue)(c_(Na_2SO_4)) = 0.0590 xx 100% = color(blue)(5.90% w//v)