The specific rotatory power for sucrose dissolved in water at 20C ( 0 589 3 nm) is 66 45 per 10 cm of path traversed through a solution containing 1 g of active substance (sugar) per cm 3 of solution A vertical P state (sodium light) enters at one end of a 1 m tube containing 1000 cm 3 of solution, of which 10 g is sucrose At what orientation will the P state emerge

Question: The specific rotatory power for sucrose dissolved in water at 20C ( 0 = 589.3 nm) is +66.45 per 10 cm of path traversed through

The specific rotatory power for sucrose dissolved in water at 20°C (λ₀ = 589.3 nm) is +66.45° per 10 cm of path traversed through a solution containing 1 g of active substance (sugar) per cm³ of solution. A vertical P-state (sodium light) enters at one end of a 1-m tube containing 1000 cm³ of solution, of which 10 g is sucrose. At what orientation will the P-state emerge?

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