How to write the amount of sodium-iodine?

Question

A chemist working on a flu vaccine needs to mix a 10% sodium-iodine solution with a 60% sodium-iodine solution to obtain a 50-milliliter mixture. Write the amount of sodium iodine in the mixture, S, in milliliters, as a function of the number of milliliters of the 10% solution used, x. Then find and interpret S(30)

Why we need to multiply x by 0.1 and 50-x by .6??

Answer 1

Since we want a $50\text{mL}$ mixture, we can:
Let $x$ represent the amount of $10\%$ sodium-iodine solution (of which $10\%$ is sodium-iodine); and
Let $50-x$ represent the amount of $60\%$ sodium-iodine solution (of which $60\%$ is sodium-iodine).

Answer 2

$\require{mhchem}$From a chemists point of view the question is ill-defined, I would even go as far as calling it absolute garbage. There are many ways to denote Concentration, the one used in the question is the most ambiguous (see for example Percent (%) Solutions Calculator).

Write the amount of sodium iodine in the mixture, S, in milliliters, as a function of the number of milliliters of the 10% solution used, x.

The term amount typically refers to amount of substance, but given they want it milliliters, they want a volume. Hence they must imply that they are using a volume concentration. Chemically that doesn't make a lot of sense, sodium iodide is a white solid at room temperature, and there will be volume contraction (e.g. Explain volume contraction in mixtures of alcohol and water).

The definition of the volume concentration is $$\sigma(i) = \frac{V(i)}{V_\mathrm{tot}}.$$

Given that both batches are the same chemicals at different dilutions, there should not be any reaction, therefore they should mix linearly. You need to find the volume of the solved $\ce{NaI}$ in both solutions, add them, and divide it by the total volume to find the new concentration \eqref{conc}. If you leave the last step, you get the volume \eqref{vol}.

\begin{align} \sigma_\mathrm{tot}(\ce{NaI}) &= \frac{\sigma_{10}(\ce{NaI}) \cdot V_{10} + \sigma_{60}(\ce{NaI}) \cdot V_{60}}{V_\mathrm{tot}}\tag1\label{conc}\\ V_\mathrm{tot}(\ce{NaI}) &= \sigma_{10}(\ce{NaI}) \cdot V_{10} + \sigma_{60}(\ce{NaI}) \cdot V_{60}\tag2\label{vol} \end{align}

We also know \begin{align} V_\mathrm{tot} &= V_{10} + V_{60}\\ \end{align}

Or let's rewrite it without units and the wanted labels: \begin{align} V_\mathrm{tot}(\ce{NaI}) &\rightarrow S(x)\\ V_\mathrm{tot} &\rightarrow 50\\ V_{10} &\rightarrow x\\ V_{60} &\rightarrow (50 - x)\\ \sigma_{10}(\ce{NaI}) &\rightarrow 0.1\\ \sigma_{60}(\ce{NaI}) &\rightarrow 0.6\\ S(x) &= 0.1x + 0.6(50 - x) \end{align}

For $x=30$ you get $1.5$, which refers to the volume of the dissolved sodium iodine. (Scientifically doubtful as it is.)

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A bit more science

Let's just assume that the question uses mass concentrations, like our friends from the biology department do.

In biology, the "%" symbol is sometimes incorrectly used to denote mass concentration, also called "mass/volume percentage." A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). The notation is mathematically flawed because the unit "%" can only be used for dimensionless quantities.

In other words: $$1\% = 0.01 \rightarrow \frac{\mathrm{1~g}}{\mathrm{100~mL}} = \mathrm{10~g/L} = \mathrm{10~kg/m^3}$$

More on this: May I treat units (e.g. joules, grams, etc.) in equations as variables?

Let's rewrite our quantities in proper units:

\begin{align} \rho_{10}(\ce{NaI}) &= \mathrm{100~kg/m^3} & \rho_{60}(\ce{NaI}) &= \mathrm{600~kg/m^3} \end{align}

The general formula for mass concentration is $$\rho(i) = \frac{m(i)}{V}.$$

Again, given that both batches are the same chemicals at different dilutions, there should not be any reaction, therefore they should mix linearly. Therefore you need to find the mass of the solved $\ce{NaI}$ in both solutions, add them, and divide it by the total volume to find the new concentration:

\begin{align} \rho_\mathrm{tot}(\ce{NaI}) &= \frac{\rho_{10}(\ce{NaI}) \cdot V_{10} + \rho_{60}(\ce{NaI}) \cdot V_{60}}{V_\mathrm{tot}}\\ m_\mathrm{tot} &= \rho_{10}(\ce{NaI}) \cdot V_{10} + \rho_{60}(\ce{NaI}) \cdot V_{60} \end{align}

This would make more sense, as we do not have to consider volume contraction.