The observed rotation of 2 g of a compound in 10 mL of solution in polarimeter tube of 25 cm long is +134°. What is the specific rotation in degree?

Question

The observed rotation of 2 g of a compound in 10 mL of solution in polarimeter tube of 25 cm long is +134°. What is the specific rotation in degree? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

[α]Tλ=(q/d × l)=(134/(2)10 × (25)10=26.8/°)

Q. The observed rotation of $2 g$ of a compound in $10 m L$ of solution in polarimeter tube of $25 c m$ long is $+ 13 4^{\circ}$ . What is the specific rotation in degree?

$[α]_{T}^{λ} = \frac{q}{d \times l} = \frac{134}{\frac{2}{10} \times \frac{25}{10}} = 26. 8^{\circ}$

Q. The observed rotation of 2g of a compound in 10mL of solution in polarimeter tube of 25cm long is +134∘. What is the specific rotation in degree?

[α]Tλ​=d×lq​=102​×1025​134​=26.8∘

Q. The observed rotation of $2 g$ of a compound in $10 m L$ of solution in polarimeter tube of $25 c m$ long is $+ 13 4^{\circ}$ . What is the specific rotation in degree?

$[α]_{T}^{λ} = \frac{q}{d \times l} = \frac{134}{\frac{2}{10} \times \frac{25}{10}} = 26. 8^{\circ}$