Answer : The molar absorptivity coefficient is,
![1.505* 10^(2)M^(-1)cm^(-1)](https://img.qammunity.org/2020/formulas/physics/high-school/mnjt23frsrttqjd6g0oysq6tku7i5x0ug0.png)
Explanation :
Using Beer-Lambert's law :
Formula used :
![A=\epsilon * C* l](https://img.qammunity.org/2020/formulas/physics/high-school/ahbtaienyxsv9uuwfun4glfrzwijnuzh2f.png)
![A=\log (I_o)/(I)](https://img.qammunity.org/2020/formulas/physics/high-school/xzd457e3wqo2pav4qedqcbc6cs5icexmxs.png)
![\log (I_o)/(I)=\epsilon * C* l](https://img.qammunity.org/2020/formulas/physics/high-school/ccr1pkgf2q2re59y8bratq1ufxv37h65q6.png)
where,
A = absorbance of solution
C = concentration of solution =
![2.00* 10^(-3)M](https://img.qammunity.org/2020/formulas/physics/high-school/wg3lbiyl2ojt0s6nwxn1a7lu4k1ugpvmv0.png)
l = path length = 1.00 cm
= incident light
= transmitted light
= molar absorptivity coefficient = ?
A compound absorb 50 % of the light that means,
Incident light =
![I_o](https://img.qammunity.org/2020/formulas/physics/high-school/79qsex4bveyr85l1lagg3mszvuozyzlqgh.png)
Transmitted light =
![0.5* I_o](https://img.qammunity.org/2020/formulas/physics/high-school/17b4l9cjg70ka92h4wjytnl154fic56ujj.png)
Now put all the given values in the above formula, we get the molar absorptivity coefficient.
![(I_o)/(0.5* I_o)=\epsilon * (2.00* 10^(-3)M)* (1.00cm)](https://img.qammunity.org/2020/formulas/physics/high-school/xt0aqmztrjod0k2pzd2ok6y5r56j72bgrp.png)
![\epsilon=1.505* 10^(2)M^(-1)cm^(-1)](https://img.qammunity.org/2020/formulas/physics/high-school/furpft90m2nhsi0bkvzix0f5lhosfwjxza.png)
Therefore, the molar absorptivity coefficient is,
![1.505* 10^(2)M^(-1)cm^(-1)](https://img.qammunity.org/2020/formulas/physics/high-school/mnjt23frsrttqjd6g0oysq6tku7i5x0ug0.png)