Chemistry - Concentration Terms Question with Solution | TestHub

% by weight of${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{=}\frac{\text{wt}\text{.}\text{of}{\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{wt}\text{.}\text{of}\text{solution}}\text{×}\text{100}$
$=\frac{474}{1250}\times 100=37.92$