Chemistry - Concentration Terms Question with Solution | TestHub

The 118% label means 100 g oleum + 18 g water $\to$ 118 g ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$. The 18 g (1 mol) of water converts free $\mathrm{S}{\mathrm{O}}_3$, so free $\mathrm{S}{\mathrm{O}}_3$ = 1 mol = 80 g per 100 g sample: (a) true. Total acidity = 118/98 = 1.2041 mol ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ per 100 g. (b) 1 g needs 2 $x$ 0.012041 = 0.024081 mol $\mathrm{NaOH}$ $\to$ 24.08 ml of 1 M $\mathrm{NaOH}$ - true. (c) 5 g needs 0.1204 g-eq $\to$ 0.1204/0.1 = 1204 ml of 0.1 N $\mathrm{Ba}(\mathrm{OH}{)}_2$, not 12.04 ml - false. (d) 100 g sample gives 1.2041 mol ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ in 500 g (0.5 kg) of water $\to$ 2.408 m - true. Hence (a), (b), (d).