Answer
In order to have a 50-milliliter solution that has 30mL of the NaI solution, we'd need to add $15mL$ of the 10% solution and $(50-15=35 mL$) of the 60% solution.
Work Step by Step
The exercise asks us to model the following: the sum of two solutions (10% and 60%) that gives a 50-milliliter mixture. Since the end result is discussed in terms of volume ($mL$), we can assume that the percentage concentration is in terms of ($\frac{mL_{NaI}}{mL_{solution}}$). We can, therefore, express the problem as follows: $$S(x) = 0.10\frac{mL}{mL}x + 0.60\frac{mL}{mL}(50-x)$$ $$S(x) mL= 0.1x + 30 - 0.6x$$$$S(x) = 30 - 0.5x$$ where $x$ represents the volume of 10% solution in milliliters, and $S(x)$ represents the final amount of Sodium Iodine in the $50mL$, expressed in $mL$. If we were to find $S(30)$: $$S(30) = 30 - 0.5(30)$$ $$S(30) = 15 mL$$ This means that, in order to have a 50-milliliter solution that has 30mL of the NaI solution, we'd need to add $15mL$ of the 10% solution and $(50-15=35 mL$) of the 60% solution.
