Question

yellow 5 dye
determining concentration
the calibration curve for a set of yellow-5 food dye samples is shown below, including the linear best fit line.

an unknown sample from a sports drink is tested and has an absorbance of 0.313. note the concentration values carefully.

absorbance vs concentration at 425 nm
$y = 0.9645x$

what is the concentration of the yellow dye in the sample if the absorbance is 0.313?
$3.25 \times 10^{-4}$ m
0.325 m
$3.01 \times 10^{-4}$ m
0.301 m

Explanation:

Step1: Identify calibration equation

The linear fit is $y = 0.9645x$, where $y$ = absorbance, $x$ = concentration ($\text{M} \times 10^{-3}$).

Step2: Substitute absorbance value

Set $y=0.313$, solve for $x$:
$0.313 = 0.9645x$

Step3: Calculate raw concentration value

$x = \frac{0.313}{0.9645} \approx 0.3245$

Step4: Convert to actual concentration

$x$ is in $\text{M} \times 10^{-3}$, so actual concentration is $0.3245 \times 10^{-3}\ \text{M} = 3.25 \times 10^{-4}\ \text{M}$

Answer:

3.25 x 10⁻⁴ M