a. Circle the wavelength below which would be best used for absorbance measurements to determine the curcumin content of the peas. 400 nm 450 nm 550 nm 600 nm 650 nm b. In one analysis, a stock solution of curcumin was prepared by dissolving 0.100 g of curcumin (C21H20O6; molar mass = 368.0 g mol−1) in 250.0 mL of water. The stock solution was then diluted to make the following three standard solutions. | standard solution | curcumin concentration in g L−1 | |------------------|------------------------------| | standard 1 | 1.00 × 10−3 | | standard 2 | 5.00 × 10−3 | | standard 3 | 1.00 × 10−2 | The absorbances of the standard solutions at the chosen wavelength were measured and the following calibration line was obtained. The curcumin in a 9.780 g sample of peas was extracted into solution. The solution was filtered and made up to 100 mL in a volumetric flask. The absorbance of the solution at the wavelength used in the above calibration was found to be 0.170. Calculate the curcumin content of the peas in milligram per gram of peas.

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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

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a. Circle the wavelength below which would be best used for absorbance measurements to determine the curcumin content of the peas. 400 nm 450 nm 550 nm 600 nm 650 n... show full transcript

Worked Solution & Example Answer: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

Step 1

Circle the wavelength below which would be best used for absorbance measurements to determine the curcumin content of the peas.

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Answer

The best wavelength to determine the curcumin content is 450 nm, as it is where curcumin absorbs most strongly while minimizing interference from the bright blue agent.

Step 2

i. Calculate the concentration of curcumin in the stock solution in mol L−1.

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Answer

To find the concentration of curcumin in the stock solution, we start with the formula:

C = \frac{n}{V}

Where:

$C$ is the concentration in mol L−1
$n$ is the number of moles; $\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.100 \, \text{g}}{368.0 \, \text{g mol}^{-1}} = 2.72 \times 10^{-4} \, \text{mol}$ .
$V$ is the volume in liters; $250.0 \, \text{mL} = 0.250 \, \text{L}$ .

Thus,

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

Step 3

ii. What volume of water must be added to 10.0 mL of the stock solution to make standard 3?

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Answer

To make standard 3, we need a final concentration of $1.00 \times 10^{-2} \, \text{g L}^{-1}$ . First, convert this to mol L−1:

C = \frac{1.00 \times 10^{-2}}{368.0} = 2.72 \times 10^{-5} \, \text{mol L}^{-1}.

Given that we have $1.09 \times 10^{-3} \, \text{mol L}^{-1}$ in our stock solution, we can use dilution calculations:

$C_1V_1 = C_2V_2$

Where:

$C_1 = 1.09 \times 10^{-3} \, \text{mol L}^{-1}$
$V_1 = 10.0 \, \text{mL}$
$C_2 = 2.72 \times 10^{-5} \, \text{mol L}^{-1}$
$V_2 = ?

Rearranging:

$V_2 = \frac{C_1V_1}{C_2} = \frac{(1.09 \times 10^{-3})(10.0)}{2.72 \times 10^{-5}} = 402.9 \, \text{mL}.$

The volume of water required is:

$402.9 - 10.0 = 392.9 \, \text{mL}.$

Step 4

iii. Calculate the curcumin content of the peas in milligrams per gram of peas.

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Answer

Using the calibration curve, the absorbance is related to concentration. Given an absorbance of 0.170:

Using the slope of the calibration line (derived from standards) we can find the equivalent concentration. Assuming the slope is represented as $m$ (determined from prior standard solutions), the concentration can be calculated as $C = m \times 0.170.$

From earlier, if the concentration found =~ $5.00 \times 10^{-3} \, g \, L^{-1}$ , then We can compute the content for the total 100 ml of solution:

$C_{extract} = C \times V = 5.00 \times 10^{-3} \, g \, L^{-1} \times 0.100 \, L = 5.00 \times 10^{-4} \text{g} = 0.500 mg.$

Now divide by the weight of the peas: $\frac{0.500 \, mg}{9.780 \, g} \times 1000 = 0.051 \, mg/g.$

So the final answer is approximately 0.051 mg of curcumin per gram of peas.

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

Worked Solution & Example Answer: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

Circle the wavelength below which would be best used for absorbance measurements to determine the curcumin content of the peas.

i. Calculate the concentration of curcumin in the stock solution in mol L−1.

ii. What volume of water must be added to 10.0 mL of the stock solution to make standard 3?

iii. Calculate the curcumin content of the peas in milligrams per gram of peas.

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