An oleum sample is labelled $118%$ , meaning that $100 g$ of oleum yields $118 g$ of pure $H X_{2} SO X_{4}$ on adding the required amount of water. Calculate the volume of $1.00 N$ $NaOH$ needed to neutralise $1.00 g$ of this oleum completely, in $mL$ (two decimals; $H X_{2} SO X_{4} = 98 g mo l^{- 1}$ ).

Numerical answer type, answer in mL, tolerance 0.05.

Answer: 24.08 mL

1.00 g

oleum gives

1.18 g

H X_{2} SO X_{4}

after hydration. Equivalent mass of

H X_{2} SO X_{4} = 98/2 = 49

, so equivalents

= 1.18/49 = 0.02408

. For

1.00 N

NaOH

, volume

= 24.08 mL