Ans. #4. Beer-Lambert’s Law, A = e C L - equation 1,
where,
A = Absorbance
e = molar extinction coefficient (M-1cm-1)
L = path length (in cm)
C = Molar concentration
Putting the values in equation 1-
A = (1.30 x 105 M-1cm-1) (5.50 x 10-6 M-1) x 1.50 cm
Hence, A = 1.07250
#5. Putting the values in equation 1-
1.300 = e x (2.50 x 10-6 M-1) x 2.0 cm
Or, e = 1.300 / (2.50 x 10-6 M-1 cm-1)
Hence, e = 2.600 x 105 M-1cm-1
#1. The frequency of an electromagnetic radiation or photon is given by the equation-
c = v l - equation 2
Where, c = speed of light = 299792458 m/s
v = frequency ; l = wavelength
Putting the value of wavelength in equation 2-
v = c / l
Or, v = (299792458 m/s) / (4.20 x 10-7 m) ; [1 m = 109 nm]
Hence, v = 7.1379 x 1014 s-1
# Energy of the photon is given by-
E = hv - equation 1
; where, h = Plank’s constant = 6.626 x 10-34 Js ; v = frequency of photon
Putting the value of v in above equation-
E = (6.626 x 10-34 Js) x (7.1379 x 10-14 s-1)
Hence, E = 7.7296 x 10-19 J
#2. % Transmittance = 100 % - % Absorbance = 100 % - 68.5 % = 31.5 %
Now,
Absorbance = 2 – log (%T) = 2 – log 31.5 = 0.5017
#3. Molar mass of the dye, C37H34N2Na2O9S3 = 792.862576 g/ mol
Mass of dye taken = 55.0 mg = 0.055 g
Volume of stock solution made upto = 250.0 mL = 0.250 L
Moles of dye taken = Mass / molar mass
= 6.9369 x 10-5 mol
Molarity of stock solution = Moles of dye / Volume of solution in liters
= 6.9369 x 10-5 mol / 0.250 L
= 2.7748 x 10-4 mol/ L
# Working solution: Using -
C1V1 (stock solution) = C2V2
Or, 2.7748 x 10-4 mol L-1 x 10.0 mL = C2 x 500.0 mL
Or, C2 = (2.7748 x 10-4 mol L-1 x 10.0 mL) / 500.0 mL
Hence, C2 = 5.5495 x 10-6 mol/L
Hence, [dye] in working solution = 5.5495 x 10-6 mol/L
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