A) left handed rotation of \[{{14}^{o}}\]
B) right handed rotation of \[{{14}^{o}}\]
C) left handed rotation of \[{{3}^{o}}\]
D) right handed rotation of \[{{3}^{o}}\]
Correct Answer: D
Solution :
For liquid A \[{{L}_{1}}=20\,cm,\,\,{{\theta }_{1}}={{38}^{o}}\]; concentration \[={{C}_{1}}\] Specific rotation \[{{\alpha }_{1}}=\frac{{{\theta }_{1}}}{{{L}_{1}}{{C}_{1}}}\] \[=\frac{{{38}^{o}}}{20\times {{C}_{1}}}\] Similarly, for liquid B \[{{L}_{2}}=30\,cm,\,\,{{\theta }_{2}}=-{{24}^{o}}\], concentration \[={{C}_{2}}\] Specific rotation \[{{\alpha }_{2}}=\frac{{{\theta }_{2}}}{{{L}_{2}}{{C}_{2}}}\] \[=\frac{(-{{24}^{o}})}{30\times {{C}_{2}}}\] The mixture has 1 part of liquid A and 2 parts of liquid B. \[\therefore \] \[{{C}_{1}}':{{C}_{2}}'=1:2\] \[\theta =[{{\alpha }_{1}}{{C}_{1}}'+{{\alpha }_{2}}{{C}_{2}}']\,l\] \[=\left\{ \frac{{{38}^{o}}}{20\times {{C}_{1}}}\times \frac{{{C}_{1}}}{3}+\frac{(-{{24}^{o}})}{30\times {{C}_{2}}}\times \frac{2\,{{C}_{2}}}{3} \right\}\times 30\] \[={{19}^{o}}-{{16}^{o}}={{3}^{o}}\] Thus, the optical rotation of mixture is \[+{{3}^{o}}\] in right hand direction.
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