Percentage labelling Of oleum

The conversion formula for % acid to % oleum is as follows: $\text{ \% acid} = 100 + \frac{18}{80} \times \text{ \% oleum}$

So the first thing we know is that there are $200 g$ of $120\%$ oleum.

Because we know the % of oleum, that's all we need to calculate the % of the acid ( $\ce{H2O}$ is the acid here )

Using the conversion formula, we find that for any amount of $120\%$ oleum, there is $127\% \ce{ H2O}$

The next step is to compare the mass of oleum to its percentage with the mass of $\ce{H2O}$ to its percentage. We don't know the mass of $\ce{H2O}$ here, so we can solve for that value using the following equation: $LaTeX$

$\frac{mass \ce{H2O}}{mass \text{ oleum (g)}}$ = $\frac{\% \ce{ H2O}}{\% \text{ oleum}}$

Plugging in our numbers and solving for the mass of $\ce{H2O}$ , we get:

$mass$ $\ce{H2O}$ = $\frac{127\% \times 200 g}{120\%}$ $\approx 211.7 g$ $\ce{H2O}$

So now that we know that about there are about $212$ grams of \ce{H2O} in $200$ grams of $120\%$ oleum, we can figure out the amount of grams of \ce{H2O} in $200$ grams of $105\%$ oleum.

Converting $105\%$ of oleum to % acid, using the conversion formula, gives that there is $\approx 123.6\%$ $\ce{H2O}$ .

Using same formula as before, but replacing $127\%$ with $123.6\%$ , and replacing $120\%$ with $105\%$ , (and keeping $200$ g), we get…

$mass$ $\ce{H2O}$ = $\frac{123.6\% \times 200 g}{105\%}$ $\approx 241.9 g$ $\ce{H2O}$

So the question asks us to find how many more grams of $\ce{H2O}$ need to be added to the oleum. The final answer, then, is not the value we just calculated; it's the difference in the masses of $\ce{H2O}$ for $120\%$ oleum and $105\%$ oleum.

$241.9 \text{ g}$ - $211.7 \text{ g}$ $\approx 0.30 \text{ g}$