Question

The absorbance of a 0.45 mol dm–3 solution of an aromatic amino acid, 3.0 cm thick is 0.22 at a wavelength of 295 nm:

(a) Calculate the molar absorption coefficient. [10 marks]

(b) What fraction of the incident light is transmitted by the solution if the absorbance was doubled to 0.44? [5 marks]

(c) What thickness of a 0.75 mol dm–3 solution would be required to transmit 65% of the incident radiation? [7 marks]

(d) Can you calculate the absorbance at 590 nm from the data provided? [3 marks]

Answer 1

Explanation:

a) Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution = 0.22

C = concentration of solution = $0.45 mol/dm^3=0.45 mol/L=0.45 M$

$1 dm^3 = 1 L$

l = length of the cell = 3.0 cm

$\epsilon$ = molar absorptivity of this solution = ?

Now put all the given values in the above formula, we get the molar absorptivity of this solution.

$0.22=\epsilon \times (0.45 M)\times (3.0 cm)$

$\epsilon=0.163 M^{-1}cm^{-1}$

Therefore, the molar absorptivity of this solution is, $1.93\times 10^{4}M^{-1}cm^{-1}$

b) $A=\log \frac{I_o}{I_t}$

$T=\frac{I_t}{I_o}$

$A=\log \frac{1}{T}$

A = 2 × 0.22 =0.44

$I_o,I_t$ = Intensities of Incident light and transmitted light respectively

T = Transmittance

$0.44=\log \frac{1}{T}$

T = 0.3630

c) $I_o=x$

$I_t=65\% of x=0.65 x$

Thickness of cell = l' =?

$c = 0.75 mol/ dm^3=0.75 mol/L=0.75 M$

$A=\log \frac{I_o}{I_t}=\epsilon \times C\times l$

$\log \frac{x}{0.65x}=0.163 M^{-1}cm^{-1}\times 0.45 M\times l'$

l' = 1.53 cm

d) No, we cannot calculate the absorbance at 590 nm from the given data. This is because absorbance at this wavelength  can be observe experimentally.