Solution Found!
A solution of 0.50 g of (see Figure 5-15) dissolved in
Chapter 6, Problem 9P(choose chapter or problem)
A solution of \(0.50 \mathrm{~g} \text { of }(-)\) -epinephrine (see Figure 5-15) dissolved in
\(10.0 \mathrm{~mL}\) of dilute aqueous \(H C l\) was placed in a \(20-\mathrm{cm}\) polarimeter tube. Using the sodium D line, the rotation was found to be at \(25^{\circ} C\). Determine the specific rotation of \((+)-\text { glyceraldehyde }\)
Equation Transcription:
Text Transcription:
0.50 g of (-)
10.0 mL
HCl
20-cm
25°C
(+)-glyceraldehyde
Questions & Answers
QUESTION:
A solution of \(0.50 \mathrm{~g} \text { of }(-)\) -epinephrine (see Figure 5-15) dissolved in
\(10.0 \mathrm{~mL}\) of dilute aqueous \(H C l\) was placed in a \(20-\mathrm{cm}\) polarimeter tube. Using the sodium D line, the rotation was found to be at \(25^{\circ} C\). Determine the specific rotation of \((+)-\text { glyceraldehyde }\)
Equation Transcription:
Text Transcription:
0.50 g of (-)
10.0 mL
HCl
20-cm
25°C
(+)-glyceraldehyde
ANSWER:
Answer :
Step 1
Given that
Weight of solution = 0.50 g .
Volume of aqueous dil.HCl dissolved = 10.0 mL
Length of polarimeter =20 cm
D line rotation found = -5.1° at 25° .
Specific rotation =?