Question

4. (6 pts) (a) You make a solution of 0.08 moles of a single molecule in water. The resulting taon was found to be +13.3° Your lab partner makes a different solution of 0.08 moles of your molecule and 0.2 moles of its enantiomer to the same concentration as your solution. What is the observed rotation of your partner's mixture? Show all work. No work = no credit. (b) (3 pts) How would your answer to (a) change if, instead of making up his solution to the same concentration as your original solution, your partner used the same VOLUME of water as you did? Show all work.

Answer 1

Let Oiginal solution A (a) other solution = B Enautio mel - Mole ule = N Path Lengta Idm Given i-For A No of moles of mole cule =O08 mol Observed Totation = +13 3 Let volume of solution ImL Concentration of solution=0.08 mal mL x path length observed aotation = (Specitic ro lation) x cone On SubsHituting ralues Specific rotation of molewle N =+13.30 008 +166-25 For B no of moles af moleule = no of moles of enantiomer = Giren O08 mol 0.2 mol no g moles of mole cule total no of moles mole faction af mol ule (N) = 0.08 0.2857 mole fraction off enautiomer (M) = 1-0 2857 = 07143 (-specific rotation af its enautiome) Specific rolation of moleule Rotation af Mixture (Frachon of N) Specific rota tion +ctaction fnspeaihe ota tion - (o.2851)(4625)+ CoT43)-i46 25) 47.449761U8 724 725480 71.-26

Cb) Observed rotation is directly proportional to concentsation of Solution Solution B has more no' of mole In Comperision to Solution A J equal volume of solveut I sed to make solutions, then cencentration will obseuved rotation allo get cha nged chauge Hen