Polarized light, stereoisomers?
Hello dear community,
Can someone help me with this problem? I don't know how to solve it. Is there a formula for it? I would appreciate a solution!
In a measuring setup, a 1 mol/l solution of the stereoisomer S1 rotates the plane of polarized light by +112°, a 1 mol/l solution of S2 by +37.
An optically active solution containing S1 and S2 with a total concentration of 1 mol/l rotates the plane of linearly polarized light by +52°.
What proportion of the S2 isomers do the solution have?
The correct answer is: 80%
Thank you in advance!!
(No Ratings Yet)
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The angle of rotation changes linearly with the concentration.
Angle=c1•112+c2•37.
In addition, c1+c2 is=1 mol/l.
Simply dissolve after c2 and insert it into the first equation.
When I convert the second equation according to c2 and put it into the first one I do not come to a result. Can you look where my fault is/rew the way of solving briefly with numbers used? I am grateful to you!!
My approach:
c1*112 + (1mol/l)* 37= 52
c1*112 + 37mol/l – c1*37 = 52
c*75 + 37mol/l = 52
From here I don’t know how to go on
c1=1-c2
(c2)•112+c2•37=52
112-112c2+37c2=52
60=75c2
However, it is rounded to 79%.
Sorry, of course, are exactly 80%. 4/5 you should be able to count in your head. I don’t know what I was typing in there.