Correct Answer - 0.158M
Decomposition of sucrose is of first order reaction
∴k=
2.303
t
log
a
a-x
=
2.303
t
log
[A]0
[A]
Fraction of sucrose remaining after 8hr=
[A]
[A]
∴k=
2.303
t1/2
=
0.693
3hr
=0.231hr-1
Hence, 0.231hr-1=
2.303
8hr
lot
[A]0
[A]
orlog
[A]0
[A]
=0.8024
or
[A]0
[A]
=Antilog(0.8024)=6.345
or
[A]
[A]0
=
1
6.345
=0.158M
Second method
Let x%
of the sucrose decomposes. Thuen use the relation :
t1/2
tx%
=
0.3
log(
100
100-x
)
∴log(
100
100-x
)=
t(x%)×0.3
t1/2
=
8hr×0.3
3hr
=0.8
∴
100
100-x
=
a
a-x
=Antilog(0.8)=6.345
