Question

1.Using the absorbance of the spinach extract and the equation of the trendline, determine the concentration of the extract solution. Report the concentration in moles/L (M).
2. Calculate the number of rams of chlorophyll-a in the 25ml, spinach solution.
3. Calculate the concentration of the chloropyhll-a soultion in spinach (mg chlorophyll-a/g spinach)
Trendline: y=1609x + .0055
Absorbance spinach extract lamda max: .329
Absorbacne spinach extract, 750 nm: .023
Corrected absorbance: .306
Mass of Spinach: .1876g Total Volume of spinach: 25mL

Answer 1

Answer:

Explanation:

From the given information:

We are to make use of the spinach absorbance extract which is the corrected absorbance  (y) = 0.306

And also the trendline equation:

y = 1609x + 0.0055

where,

x = absorbance of the spinach extract.

∴

0.306 = 1609x + 0.0055

collecting the like terms

0.306 - 0.0055 = 1609x

0.3005  = 1609x

x = 0.3005/1609

x = 1.8676 × 10⁻⁴

x ≅ 0.0002 M

No. of grams for the chlorophyll can be computed as follows:

recall that:

molar mass of chlorophyll = 893.5 g/mol

the volume = 25ml = (25/1000) L = 0.025 L

∴

In spinach solution, the no. of grams for the chlorophyll:

= (0.0002) mol/L × (893.5 g/mol) × (0.025) L

= 0.0044675 g

≅ 0.0045 g

In the spinach, the concentration of chlorophyll = no of grams of chlorophyll/ mass of the spinach

= 4.5 mg/0.1876 g

= 23.987 mg/g

≅ 24 mg/g

Answer 2

From the given information:

• We are to make use of the spinach absorbance extract which is the corrected absorbance  (y) = 0.306

Chlorophyll

Chlorophyll is any member of the class of the green pigments involved in the photosynthesis process.

And also the trendline equation:

y = 1609x + 0.0055

where,

x = absorbance of the spinach extract.

so 0.306 = 1609x + 0.0055

collecting the like terms

0.306 - 0.0055 = 1609x

0.3005  = 1609x

x = 0.3005/1609

x = 1.8676 × 10⁻⁴

x ≅ 0.0002 M

2. No.of grams for the chlorophyll can be computed as follows:

recall that:

molar mass of chlorophyll = 893.5 g/mol

The volume = 25ml = (25/1000) L = 0.025 L

Therefore:

In spinach solution, the no. of grams for the chlorophyll:

= (0.0002) mol/L × (893.5 g/mol) × (0.025) L

= 0.0044675 g

≅ 0.0045 g

3. In the spinach, the concentration of chlorophyll = no of grams of chlorophyll/ mass of the spinach

= 4.5 mg/0.1876 g

= 23.987 mg/g

≅ 24 mg/g

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