A volume of 20 ml of 8.5% (w/v) H 2 O 2 solution is diluted to 50 ml. A volume of 10 ml of the diluted solution is react with excess of an oxidant. It will cause libration of _____ ml of ____ gas at 0 o C and 1 atm.
Answers
Answer:
8.5% (w/v) H2O2
∴ In 100 ml solution →8.5 gm H2O2
Now In 20 ml solution →58.5=1.7 gm H2O2
Now volume is made 50 ml from 20 ml
∴ Now 10 ml of this is taken.
∴ Cut of H2O2 in 10 ml=51.7 gm.
∴ nH2O2=5×341.7=0.01 mole
At 0oC and 1 atm i.e., STP
1 mole →22.4 let gas liberated 21×0.01 mole
∴ volume of gas at STP
=20.01×22.4 let =0.112 lit =112 ml
By R×4H
Answer:
8.5% (w/v) H
2
O
2
∴ In 100 ml solution →8.5 gm H
2
O
2
Now In 20 ml solution →
5
8.5
=1.7 gm H
2
O
2
Now volume is made 50 ml from 20 ml
∴ Now 10 ml of this is taken.
∴ Cut of H
2
O
2
in 10 ml=
5
1.7
gm.
∴ n
H
2
O
2
=
5×34
1.7
=0.01 mole
At 0
o
C and 1 atm i.e., STP
1 mole →22.4 let gas liberated
2
1
×0.01 mole
∴ volume of gas at STP
=
2
0.01
×22.4 let =0.112 lit =112 ml
By R×4H
2
O
2
→H
2
OH
2
O
2
so O
2
(oxygen is Release)
React with oxidant (oxidiasion aquent ) Self Respected
And other is oxidised
M+O
2
→MO
2
(Reduction oxidation)
So O
2
is Release
112 M of O
2