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

Since the original sample was diluted from 1.00 mL to 100 L, we need to calculate the concentration in the original sample:

Using the dilution formula: $C_1 V_1 = C_2 V_2$, where:

• $C_1$ = original concentration (unknown)
• $V_1$ = original volume (1.00 mL)
• $C_2$ = concentrated (diluted) concentration (18 ng mL⁻¹)
• $V_2$ = diluted volume (100 L = 100,000 mL)

Rearranging gives: $C_1 = \frac{C_2 V_2}{V_1}$. Substituting values: $C_1 = \frac{(18 \text{ ng mL}^{-1})(100,000 \text{ mL})}{1 \text{ mL}} = 1,800,000 \text{ ng mL}^{-1} = 1,800 \text{ µg mL}^{-1}$.

Thus, the approximate concentration of folic acid in the 1 mL sample is found to be 18 µg mL⁻¹.