Question

Could you please help me to solve these question with all parts, including graph? thanks so much!!

Answer 1

ANSWER - Part (1):

1) concentration of solution 1

• Drops of Stock solution = 1 drop
• Volume of solution 1 = 50.0 mL

Then, using equation 6:

Conc. Sol 1 (drops.mL-1number of drops Vsol (mL)

I drop Conc. Sol 1 (drops.mL 50.0 mL 0.02 drops.mL

2) Concentration of solution 2

• These solutions are prepared by dilution of solution 1
• Preparation: 5.0 mL of Sol 1 + 5.0 mL H₂O
• Total volume of solution = 10.0 mL

Then, using equation 7:

Vsol 1 X Conc. Sol1 Vsol 2 Conc. Sol 2 (drops.mL-

5.0mL x 0.02 drops.mL Conc. Sol 2 (drops.mL) 0.01 drops.mL 10.0 mL

3) Concentration of solution 3

• These solutions are prepared by dilution of solution 2
• Preparation: 5.0 mL of Sol 2 + 5.0 mL H₂O
• Total volume of solution = 10.0 mL

Then, using equation 7:

Vsol 2 X Conc. Sol2 Vsol 3 Conc. Sol 3 (drops.mL-

5.0mL x 0.01 drops.mL 5x10 drops.mL Conc. Sol 3 (drops.mL) 10.0 mL

4) Concentration of solution 4

• These solutions are prepared by dilution of solution 3
• Preparation: 5.0 mL of Sol 3 + 5.0 mL H₂O
• Total volume of solution = 10.0 mL

Then, using equation 7:

Conc. Sol 4 (drops.mL-)sol 3 X Conc. Sol 3

5.0mL x 5x10 drops.mL 10.0 mL 3 Conc. Sol 4 (drops.mL) 2.5x10 drops.mL

5) Concentration of solution 5

• These solutions are prepared by dilution of solution 4
• Preparation: 5.0 mL of Sol 4 + 5.0 mL H₂O
• Total volume of solution = 10.0 mL

Then, using equation 7:

Conc. Sol 5 (drops.mL-1)Vsol 4 X Conc. Sol 4 Vsol 5

5.0 mL x 2.5x10 drops.mL 3 Conc. Sol 5 (drops.mL) 1 1.25x10 drops.mL 10.0 mL

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ANSWER - Part (2):

• The absorbance is defined as:

$absorbance=-log \left ( \frac{\%T}{100} \right )$

then, we can complete the table

Solution	concentration (drops.mL-1)	%T	A
1	0.02	5.5	$A=-log \left ( \frac{5.5}{100} \right )=1.260$
2	0.01	22.6	22.6 A=log100 = 0.646
3	5x10-3	46.2	$A=-log \left ( \frac{46.2}{100} \right )=0.335$
4	2.5x10-3	68.4	$A=-log \left ( \frac{68.4}{100} \right )=0.165$
5	1.25x10-3	82.8	$A=-log \left ( \frac{82.8}{100} \right )=0.0820$
Unknown	Unknown	51.3	$A=-log \left ( \frac{51.3}{100} \right )=0.290$

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ANSWER - Part (3):

• Using data from previous table, we plot Absorbance (y-axis) vs concentration (x-axis)
• This is done because the absorbance of any solution is proportionally to concentration of solution and it follows a straight-line equation:

A mx c A

where A = absorbance, c = concentration of solution, A₀ = y-intercept and m = slope

y 62.636x+ 0,0121 R2 0.9997 A vs Conc 1.400 1.200 1.000 0.800 0.600 0.400 0.200 0.000 0 0.005 0.01 0.015 0.02 0.025 concentration (drops.mL-1) Absorbance

Plotting the data, we get this equation:

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ANSWER - Part (4):

Using data from part (3) and the previous plot, we can calculate the concentration of the unknown stock solution of red food dye.

• Absorbance of unknown stock solution = 0.290

Using the equation obtained fro plot A vs concentration:

$c=\frac{A-0.0121}{62.636}$

$c=\frac{0.290-0.0121}{62.636}=\mathbf{4.45x10^{-3}\, drops.mL^{-1}}$

The concentration of the unknown stock solution of red food dye is 4.45x10^-3 drops.mL^-1