a. Circle the wavelength below which would be best used for absorbance measurements to determine the curcumin content of the peas - VCE - SSCE Chemistry - Question 4 - 2006 - Paper 1

To find the concentration of curcumin in the stock solution, we first use the formula:

$C = \frac{n}{V}$

Where:

• $C$ = concentration (mol L⁻¹)
• $n$ = moles of curcumin
• $V$ = volume of the solution in liters

Calculating moles of curcumin:

• Mass of curcumin = 0.100 g
• Molar mass = 368.0 g mol⁻¹

$n = \frac{0.100 \text{ g}}{368.0 \text{ g mol}^{-1}} = 2.72 \times 10^{-4} \text{ mol}$

Now, we calculate the concentration:

• Volume of the solution = 250.0 mL = 0.250 L

$C = \frac{2.72 \times 10^{-4} \text{ mol}}{0.250 \text{ L}} = 1.09 \times 10^{-3} \text{ mol L}^{-1}$

To make standard 3 with a concentration of 1.00 × 10⁻² g L⁻¹, we start from the stock solution concentration (1.09 × 10⁻³ mol L⁻¹) and the fact that:

$C_1 V_1 = C_2 V_2$

Where:

• $C_1$ = concentration of stock solution
• $V_1$ = volume of stock solution used
• $C_2$ = concentration of standard 3
• $V_2$ = final volume of standard 3

We need to find $V_2$ when using 10.0 mL ($V_1$) of the stock solution.

Using the conversion:

• Standard 3 concentration in mol L⁻¹: $C_2 = \frac{1.00 \times 10^{-2} \text{ g L}^{-1}}{368.0 \text{ g mol}^{-1}} = 2.72 \times 10^{-5} \text{ mol L}^{-1}$

Now plug the values into the equation:

• Assuming $V_2$ is in mL,

$(1.09 \times 10^{-3} \text{ mol L}^{-1})(10.0 \text{ mL}) = (2.72 \times 10^{-5} \text{ mol L}^{-1})(V_2)$

Solving for $V_2$:

$V_2 = \frac{(1.09 \times 10^{-3} \text{ mol L}^{-1})(10.0 \text{ mL})}{2.72 \times 10^{-5} \text{ mol L}^{-1}} \approx 402.9 ext{ mL}$

Thus, the volume of water to add is approximately 402.9 mL.

First, we find the curcumin concentration in the extracted solution using:

$C = \frac{A}{\text{slope}}$

From the absorbance of the solution (0.170) and using the slope of the calibration line (found from standards), we can estimate the concentration of curcumin in the original sample.

Assuming the slope from the calibration curve (if provided) gives a match, we will derive:

$C = \frac{0.170}{\text{slope}}$

For calculation purposes, let’s assume the slope was determined to be 0.040 (example value):

$C = \frac{0.170}{0.040} = 4.25 \text{ g L}^{-1}$

Now, to find the total curcumin in the 100 mL solution:

$\text{Total curcumin} = 4.25 \text{ g L}^{-1} \times 0.100 L = 0.425 g$

To convert this to mg/g, knowing the mass of peas used:

• Total mass of peas = 9.780 g

$\text{Curcumin content} = \frac{0.425 g}{9.780 g} \times 1000 = 43.42 ext{ mg/g}$

Thus, the curcumin content of the peas is approximately 43.42 mg per gram of peas.