Question 36d85

Sep 1, 2017

$\text{0.14 g}$

Explanation:

The idea here is that a solution's parts per million concentration, ppm, tells you the number of grams of solute present in exactly

${10}^{6} = 1 , 000 , 000$

parts of solution. Since the mass of the solvent is much, much bigger than the mass of the solute, which I'll assume is the chromium(III) cation, ${\text{Cr}}^{3 +}$, you can assume that the mass of the solution is equal to the mass of the solvent, i.e. of water.

You know that $\text{1 L}$ of water has a mass of $\text{2.5 kg}$, so you can say that your solution will contain

2.5 color(red)(cancel(color(black)("L"))) * (1.00color(red)(cancel(color(black)("kg"))))/(1color(red)(cancel(color(black)("L")))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 2.5 * 10^3# $\text{g}$

of water, the solvent. This means that you assume that the mass of the solution is equal to $2.5 \cdot {10}^{3}$ $\text{g}$.

Now, your solution is $\text{57 ppm}$ chromium(III) cations, which means that you get $\text{57 g}$ of chromium(III) cations for every ${10}^{6}$ $\text{g}$ of solution.

You can thus say that your solution will contain

$2.5 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g solution"))) * overbrace("57 g Cr"^(3+)/(10^6color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 57 ppm Cr"^(3+))) = color(darkgreen)(ul(color(black)("0.14 g Cr}}^{3 +}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution and for its ppm concentration.