Question

Find the concentration of [X] if the absorbance are 0.700 at 272 nm and

0.550 at 327 nm.

Short answer

The concentration of [X] if the absorbance are 0.700 at 272 nm and

0.550 at 327 nm is .

Step-by-step solution

Calculate the concentration of [X]:

The analysis of mixture is solved using equation,

$\left[\mathrm{X}\right]=\frac{\left|\begin{array}{cc}\mathrm{A}*& {\mathrm{\zeta }}_{-}-\mathrm{b}\\ \mathrm{A}*& {\mathrm{\zeta }}_{-}-\mathrm{b}\end{array}\right|}{\left|\begin{array}{cc}{\mathrm{\zeta }}_{\mathrm{x}}-\mathrm{b}& {\mathrm{\zeta }}_{-}-\mathrm{b}\\ {\mathrm{\zeta }}_{\mathrm{x}}-\mathrm{b}& {\mathrm{\zeta }}_{-}-\mathrm{b}\end{array}\right|}$

$\left[\mathrm{X}\right]=\frac{\left|\begin{array}{cc}0.700& 3870\\ 0.550& 6420\end{array}\right|}{\left|\begin{array}{cc}16400& 3870\\ 3900& 6420\end{array}\right|}$

$$\begin{gathered} \left[\mathrm{X}\right]=\frac{\left(0.700\right)\left(6420\right)-\left(3870\right)\left(0.500\right)}{\left(16400\right)\left(6420\right)-\left(3870\right)\left(3990\right)} \\ \left[\mathrm{X}\right]=2.63\times {10}^{-5}\mathrm{M} \end{gathered}$$

The solution for concentration of [X] is$2.63\times {10}^{-5}\mathrm{M}$.