An aqueous solution containing `10 gm` of optically pure fructose was diluted to `500 ml` with water and placed in a polarimeter tube `20 cm` long. The measured roatation was `-5.0^(@)`. To this solution, `500 ml` of a solution containing `10 gm` of racemic fructose is added. What will be the change in specific rotation?

An aqueous solution containing `10 gm` of optically pure fructose was

1.	An aqueous solution containing `10 gm` of optically pure fructose was diluted to `500 ml` with water and placed in a polarimeter tube `20 cm` long. The measured roatation was `-5.0^(@)`. To this solution, `500 ml` of a solution containing `10 gm` of racemic fructose is added. What will be the change in specific rotation?
Answer» As we know, `20 cm= 2dm`. `[alpha]_(D)=(theta)/(lxxC)=(-5.0^(@))/(2dmxx10gm//500ml)=-125^(@)` Total volume of solution `=500+500= 1000 ml` Mass per ml of pure fructose`=10//1000` `= 0.01 gm m1^(-1)` Mass per `m1` of mixtrue `= (10 + 10 = 20) = 20//1000` `= 0.02 gm m1^(-1)` `OP` (optical purity) = `([alpha]_(obs.)"of mixtrue")/([alpha]_(D) "of pure form") prop("Mass per ml of pure form")/("Mass per ml of mixtrue")` `=(0.01)/(0.02)= 0.5( because l "constant")` New `[alpha]_(D)= -125^(@)xx0.5= -62.5^(@)` Change in specific rotation `=-62.5- (-125)= +62.5^(@)`