Question

A solution of $I_2$in ethanol had a density of $0.794g/c{m}^3$. A $1.50cm$layer was found to transmit 33.2% of the radiation from a Mo $K_{\alpha }$ source. Mass absorption coefficients for I, C, H, and O are 39.2, 0.70, 0.00, and 1.50, respectively. (a) Calculate the percentage of $I_2$ present, neglecting absorption by the alcohol. (b) Correct the results in part (a) for the presence of alcohol

Short answer

Part (a) Percent of $I_2$is localid="1646233641637" $1.18\%$

Part (b) Percent of$I_2$in presence of alcohol islocalid="1646233656094" $0.0503\%$

Step-by-step solution

Step 1. Given information 

Density is given as $0.794g/c{m}^3$

percent transmittance is given as $33.2\%$

And mass absorption coefficients of different elements are given.

Part (a) Step 1. Substitute the given values in the formula

$$\begin{gathered} \ln \frac{P_0}{P}={\mu }_M\rho x \\ \ln \left(\frac{1}{0.332}\right)={\mu }_M\times 0.794\times 1.50 \\ {\mu }_M=0.925c{m}^2/g \end{gathered}$$

Therefore, mass absorption coefficient of the solution in ethanol is$1.09c{m}^2/g$

Part (a) Step 2. Find the percentage 

$$\begin{gathered} {\mu }_M=W_{I_2}{\mu }_{I_2} \\ 0.925=W_{I_2}\times 2\times 39.2 \\ {W}_{I_2}=1.18 \end{gathered}$$

Hence, Percentage of $I_2$is localid="1646233681698" $1.18\%$

Part (b) Step 1. Calculate the percentage of iodine in the presence of alcohol

$$\begin{gathered} {\mu }_M=W_{I_2}{\mu }_{I_2}+W_{ETOH}+{\mu }_{ETOH} \\ 0.925=78.4W_{I_2}+0.887W_{EtOH} \end{gathered}$$

Also, we know that $W_{I_2}+W_{ETOH}=1$

Therefore, solving above two equations, we get

localid="1646233570133" $W_{I_2}=0.503\times {10}^{-3}$

localid="1646233601094" $W_{I_2}=0.0503\%$

Therefore, in the presence of alcohol percentage of iodine present is 0.0503%