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(D) For solution $A$: $L_{1} = 20 \ cm$,$	heta_{1} = +38^{\circ}$. Let concentration be $C_{1}$. The specific rotation is $\alpha_{1} = \frac{	heta_

Vedclass · Accepted Answer

right-handed rotation of $3^{\circ}$

Vedclass · Answer

(D) For solution $A$: $L_{1} = 20 \ cm$,$	heta_{1} = +38^{\circ}$. Let concentration be $C_{1}$. The specific rotation is $\alpha_{1} = \frac{	heta_{1}}{L_{1} C_{1}} = \frac{38^{\circ}}{20 C_{1}}$.For solution $B$: $L_{2} = 30 \ cm$,$	heta_{2} = -24^{\circ}$ (left-handed). Let concentration be $C_{2}$. The specific rotation is $\alpha_{2} = \frac{	heta_{2}}{L_{2} C_{2}} = \frac{-24^{\circ}}{30 C_{2}}$.In the mixture,the volume ratio is $1:2$. Thus,the new concentrations are $C_{1}' = \frac{C_{1}}{3}$ and $C_{2}' = \frac{2C_{2}}{3}$.The total optical rotation $	heta$ for a path length $l = 30 \ cm$ is given by $	heta = (\alpha_{1} C_{1}' + \alpha_{2} C_{2}') l$.Substituting the values: $	heta = \left( \frac{38^{\circ}}{20 C_{1}} \cdot \frac{C_{1}}{3} + \frac{-24^{\circ}}{30 C_{2}} \cdot \frac{2 C_{2}}{3} ight) 	imes 30$.$	heta = \left( \frac{38^{\circ}}{60} - \frac{48^{\circ}}{90} ight) 	imes 30 = \left( \frac{19^{\circ}}{30} - \frac{16^{\circ}}{30} ight) 	imes 30 = 19^{\circ} - 16^{\circ} = +3^{\circ}$.Since the result is positive,it is a right-handed rotation of $3^{\circ}$.

$A$ $20 \ cm$ length of a certain solution causes right-handed rotation of $38^{\circ}$. $A$ $30 \ cm$ length of another solution causes left-handed rotation of $24^{\circ}$. The optical rotation caused by a $30 \ cm$ length of a mixture of the above solutions in the volume ratio $1:2$ is

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