Q157 / 180 · KCET 2007 · Physics

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

  • A
    left handed rotation of $14^{\circ}$
  • B
    right handed rotation of $14^{\circ}$
  • C
    left handed rotation of $3^{\circ}$
  • Correct answer
    D
    right handed rotation of $3^{\circ}$

Solution

For liquid $\mathrm{A}$ $L_{1}=20 \mathrm{~cm}, \theta_{1}=38^{\circ}$; concentration $=C_{1}$ $\begin{aligned} \text { Specific rotation } \alpha_{1} &=\frac{\theta_{1}}{L_{1} C_{1}} \\ &=\frac{38^{\circ}}{20 \times C_{1}} \end{aligned}$ Similarly, for liquid $B$ $L_{2}=30 \mathrm{~cm} \theta_{2}=-24^{\circ}$, concentration $=C_{2}$ Specific rotation $\alpha_{2}=\frac{\theta_{2}}{L_{2} C_{2}}$ $$ =\frac{\left(-24^{\circ}\right)}{30 \times C_{2}} $$ The mixture has 1 part of liquid $A$ and 2 parts of liquid $B$. $$ \begin{aligned} \therefore \quad C_{1}^{\prime} &: C_{2}^{\prime}=1: 2 \\ \theta &=\left[\alpha_{1} C_{1}^{\prime}+\alpha_{2} C_{2}^{\prime}\right] l \\ &=\left\{\frac{38^{\circ}}{20 \times C_{1}} \times \frac{C_{1}}{3}+\frac{\left(-24^{\circ}\right)}{30 \times C_{2}} \times \frac{2 C_{2}}{3}\right\} \times 30 \\ &=19^{\circ}-16^{\circ}=3^{\circ} \end{aligned} $$ Thus, the optical rotation of mixture is $+3^{\circ}$ in right hand direction

Hence, the correct answer is option: 'D'.

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