A) left handed rotation of \[14{}^\circ \]
B) right handed rotation of \[14{}^\circ \]
C) left handed rotation of \[3{}^\circ \]
D) right handed rotation of 3°
Correct Answer: D
Solution :
For liquid A \[{{L}_{1}}=20cm,{{\theta }_{1}}={{30}^{o}};\]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\[{{L}_{2}}=30cm,{{\theta }_{2}}=-{{24}^{o}},\]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 \[+3{}^\circ \] in right hand direction.
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