Question

A solution of three non-interacting compounds P, Q, and R is taken in a cuvette of 1 cm path length. Their concentrations are [P] = \(1\times10^{-6}\) M, [Q] = \(2\times10^{-6}\) M, [R] = \(3\times10^{-6}\) M and the molar extinction coefficients at 300 nm are \(\varepsilon_P = 1\times10^{5}\ \mathrm{M^{-1}\,cm^{-1}}\), \(\varepsilon_Q = 2\times10^{5}\ \mathrm{M^{-1}\,cm^{-1}}\) and \(\varepsilon_R = 3\times10^{5}\ \mathrm{M^{-1}\,cm^{-1}}\). The % transmittance at 300 nm is ....................... (rounded off to two decimal places)

Hint:

- For mixtures, total absorbance is additive: \(A_\text{total} = \sum \varepsilon c l\).
- Convert using \(T = 10^{-A}\) and \(%T = 100\times T\).

Answer 1

Step 1: Use Beer–Lambert law and add absorbances.
For non-interacting species, \( A = \sum \varepsilon_i\,c_i\,l \). With \( l = 1 \, \text{cm} \):
\[ A = (1 \times 10^{5})(1 \times 10^{-6}) + (2 \times 10^{5})(2 \times 10^{-6}) + (3 \times 10^{5})(3 \times 10^{-6}) \] \[ A = 0.10 + 0.40 + 0.90 = 1.40 \] Step 2: Convert absorbance to transmittance.
\( T = 10^{-A} = 10^{-1.40} \approx 0.03981 \). Hence, \( \%T = 100\,T \approx 3.981\% \).
\[ \boxed{\%T \approx 3.98\%} \]