Question

What is the angle of rotation when $5M\alpha -D$ mannopyranose is taken in of length 200mm?

• A. ${52.2}^{\circ }$
• B. ${29}^{\circ }$
• C. ${26.1}^{\circ }$
• D. ${14.5}^{\circ }$

Answer 1

The correct option is A ${52.2}^{\circ }$

As specific rotation is given by, $[\alpha {]}_{\lambda }^T=\frac{\alpha \left(observed\right)}{c\left(g/ml\right)\times d\left(dm\right)}$

Given here, $C=5M=\frac{5\times 180}{1000}g/ml$ $[concentrationing/L=\frac{molarity\times molarmass}{1000}]$

$d=200nm=2dm$ [Since $1mm=0.01dm]$

$[\alpha {]}_{\lambda }^T={29}^o$ (for $\alpha$ anomer)

$\therefore {29}^o=\frac{\alpha \left(observed\right)}{\frac{5\times 180}{1000}g/ml\times 2dm}$

$\Rightarrow \alpha$ (Observed for $\alpha$ anomer) $={52.2}^o$

Hence, the correct option is A.