The graph below shows the HPLC calibration curve for folic acid - VCE - SSCE Chemistry - Question 18 - 2023 - Paper 1

To find the concentration of folic acid in the original 1.00 mL sample, we first need to use the calibration curve. The peak area of the diluted sample is given as 0.35. By locating this value on the y-axis of the calibration curve and drawing a horizontal line until it intersects the curve, we can find the corresponding concentration on the x-axis.

Assuming the graph indicates that a peak area of 0.35 corresponds to a concentration of approximately 18 ng mL⁻¹, we can convert this concentration back to the original sample concentration.

Given that the sample was diluted to 100 L (or 100 mL), we apply the dilution factor:

$C_{1}V_{1} = C_{2}V_{2}$

Where:

• $C_{1}$ = concentration in the 1 mL sample
• $C_{2}$ = concentration in the diluted sample (18 ng mL⁻¹)
• $V_{1}$ = volume of the original sample (1 mL)
• $V_{2}$ = volume of the diluted sample (100 mL)

Rearranging the equation gives us:

$C_{1} = C_{2} \times \frac{V_{2}}{V_{1}}$

Substituting the values:

$C_{1} = 18 \text{ ng mL}^{-1} \times \frac{100 \text{ mL}}{1 \text{ mL}} = 1800 \text{ ng mL}^{-1}$

Converting ng mL⁻¹ to µg mL⁻¹ (since 1000 ng = 1 µg):

$C_{1} = \frac{1800 \text{ ng mL}^{-1}}{1000} = 1.8 \text{ µg mL}^{-1}$

Thus, the closest answer for the original concentration of folic acid in the 1 mL sample is 18 µg mL⁻¹, which corresponds to answer choice C.