Question

Given : 1 L, 2M solution of `C_(12)H_(22)O_(11)` under go first order reaction.
`C_(12)H_(22)O_(11)+H_(2)Orarr C_(6)H_(12)O_(6)+C_(6)H_(12)O_(6)`
`{:(,"Angle of rotation ",theta,20^(@),-10^(@),-40^(@)),(,"Time",t("min"),0,10,oo):}`
If osmotic pressure of solution is `pi` atm at `(300)/(0.821)K` temperature after 10 min then calculate the value of `(pi)/(10)` ?

Answer 1

Correct Answer - 9
`C_(12)H_(22)O_(11)+H_(2)OrarrC_(6)H_(12)O_(6)+C_(6)H_(12)O_(6)`
`{:(2,,,0,,0),(2-x,,,x,,x):}`
`K=(1)/(t)ln.(2)/(2-x)`
`K=(1)/(10).ln.(2)/(2-x)" "......(1)`
`K=(1)/(10).ln.(theta_(oo)-theta_(0))/(theta_(oo)-theta_(f))" "," "K=(1)/(10).ln.(-40-20)/(-40+10)`
`K=(1)/(10).ln2" "," "(1)/(10)ln2=(1)/(10).ln.(2)/(2-x)`
`2=(2)/(2-x)" "," "2-x=1impliesx=1`
`pi=RT(M_(S)+M_(G)+M_(F))" "," "pi=0.0821xx(300)/(0.821)xx(1+1+1)=90" ", " "(pi)/(10)=(90)/(10)=9`