Question

(a) 15.8% (d) 23.8% (b) 0.492 (e) 0.085 (c) 39.4% (f) 5.38% *13-3 Calculate the percent transmittance of solutions having half the abse solutions in Problem 13-1. f the absorbance of the cent transmittance of *13-4 Calculate the absorbance of solutions having twice the percent transmiti those in Problem 13-2. *13-5 A solution containing 6.23 ppm KMnO, had a transmittance of 0.195 in a 10 cell at 520 nm. Calculate the molar absorptivity of KMnO4 at 520 nm. 13-6 A solution containing 5.24 mg/100 mL of A (335 g/molhas a transmittance of 55.2% in a 1.50-cm cell at 425 nm. Calculate the molar absorptivity of A at this! wavelength. *137 A solution containing the complex formed between Bi(III) and thiourea has a molar absorptivity of 9.32 x 103 L mol-1 cm-1 at 470 nm.

I need help with 5 & 7 please

Answer 1

(5) Given: Transmittance (T) = 0.195

Length of cell (l) = 1 cm

ppm = weight/ volume = 6.23 ppm

i.e Weight (w) = 6.23 g

Volume (V) = 106 ml

As we know,

A = € c l ---------(1)

Where A is absorbance , € is molar absorptivity, c is concentration and l is length.

To find concentration:

[concentration] = moles / volume

Moles = given weight / molar mass

Molar mass of KMnO4 = 158 g/mol

Moles = 6.23/158

= 0.0394 moles of KMnO4

​​​​​[concentration] = 0.0394 moles /106 ml

=0.0394 moles / 103 L

= 0.0394 x 10-3 M

To find absorbance:

A = - log T

= -log 0.195

= 0.7099

Putting values in (1)

€ = A/c l

= 0.7099/ (0.0394 x 10-3) x (1)

= 18017.76 L mol-1 cm-1

Molar absorptivity of KMnO4 is 18017.76 L mol-1 cm-1

(7) Given : Molar absorptivity (€) = 9.32 x 103 L mol-1cm-1

(a) : Concentration (c) = 3.79 x 10-5 M

Length (l) = 1cm

As, A = € c l

= 3.79 x 10-5 x 9.32 x 103 x 1

= 0.3532

So absorbance is 0.3532

(b) A = -log T

0.3532 = -log T

T = 10-0.3532

=0.4434

So, % T = 0.4434 x 100

= 44.34%

(c) Given : A = 0.3532

l = 2.50 cm

C = A / € l

= 0.3532 / ( 9.32 x 103) ( 2.50)

So, molar concentration of complex is 1.51 x 10-5 M.