Beer-Lambert's Law is given by the equation:

A = εlc

where:
A is the absorbance,
ε is the molar absorptivity or extinction coefficient (in M^-1 cm^-1),
l is the path length through which the light passes (in cm), and
c is the concentration of the solution (in M).

Given in the problem, we have:

ε = 41500 M^-1 cm^-1,
l = 1 cm, and
c = 15 µM = 15 x 10^-6 M (since 1 µM = 10^-6 M).

Substituting these values into the equation gives:

A = (41500 M^-1 cm^-1) * (1 cm) * (15 x 10^-6 M) = 0.6225

So, the expected absorbance at 620 nm of a 15 µM solution of Bromocresol Green is 0.6225 (to four decimal places). However, the question asks for the answer to three decimal places, so we round 0.6225 to 0.623.

Therefore, the expected absorbance at 620 nm of a 15 µM solution of Bromocresol Green is 0.623 (to three decimal places).

Question

Beer-Lambert's Law is given by the equation:

A = εlc

where:
A is the absorbance,
ε is the molar absorptivity or extinction coefficient (in M^-1 cm^-1),
l is the path length through which the light passes (in cm), and
c is the concentration of the solution (in M).

Given in the problem, we have:

ε = 41500 M^-1 cm^-1,
l = 1 cm, and
c = 15 µM = 15 x 10^-6 M (since 1 µM = 10^-6 M).

Substituting these values into the equation gives:

A = (41500 M^-1 cm^-1) * (1 cm) * (15 x 10^-6 M) = 0.6225

So, the expected absorbance at 620 nm of a 15 µM solution of Bromocresol Green is 0.6225 (to four decimal places). However, the question asks for the answer to three decimal places, so we round 0.6225 to 0.623.

Therefore, the expected absorbance at 620 nm of a 15 µM solution of Bromocresol Green is 0.623 (to three decimal places).

Knowee AI · Accepted Answer

Beer-Lambert's Law is given by the equation:

A = εlc

where:
A is the absorbance,
ε is the molar absorptivity or extinction coefficient (in M^-1 cm^-1),
l is the path length through which the light passes (in cm), and
c is the concentration of the solution (in M).

Given in the problem, we have:

ε = 41500 M^-1 cm^-1,
l = 1 cm, and
c = 15 µM = 15 x 10^-6 M (since 1 µM = 10^-6 M).

Substituting these values into the equation gives:

A = (41500 M^-1 cm^-1) * (1 cm) * (15 x 10^-6 M) = 0.6225

So, the expected absorbance at 620 nm of a 15 µM solution of Bromocresol Green is 0.6225 (to four decimal places). However, the question asks for the answer to three decimal places, so we round 0.6225 to 0.623.

Therefore, the expected absorbance at 620 nm of a 15 µM solution of Bromocresol Green is 0.623 (to three decimal places).

Use Beer-Lamberts' Law to calculate (to three decimal places) the expected absorbance at 620nm of a 15uM solution of Bromocresol Green. The extinction co-efficient for BCG is 41500M-1cm-1 and the cell path length is 1cm.

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