20 mL of `H_(2)O_(2)` is reacted completely with acidified `K_(2)Cr_(2)O_(7)` solution 40 mL of `K_(2)Cr_(2)O_(7)` solution is required to oxidise the `H_(2)O_(2)` completely. Also, 2.0 mL of the same `K_(2)Cr_(2)O_(7)` solution is required to oxidise 5.0 mL of a 1.0 M `H_(2)C_(2)O_(4)` solution to reach equivalent point. Which of the following staement is`//`are correct ?A. The `H_(2)O_(2)` solution is 5MB. The volume strength of `H_(2)O_(2)` is 56 VC. The volume strength of `H_(2)O_(2)` is 112 VD. If 40 mL more `H_(2)O_(2)` solution, the volume strength of the resulting solution is changed to 16.8 V

20 mL of `H_(2)O_(2)` is reacted completely with acidified `K_(2)Cr_(2

1.	20 mL of `H_(2)O_(2)` is reacted completely with acidified `K_(2)Cr_(2)O_(7)` solution 40 mL of `K_(2)Cr_(2)O_(7)` solution is required to oxidise the `H_(2)O_(2)` completely. Also, 2.0 mL of the same `K_(2)Cr_(2)O_(7)` solution is required to oxidise 5.0 mL of a 1.0 M `H_(2)C_(2)O_(4)` solution to reach equivalent point. Which of the following staement is`//`are correct ?A. The `H_(2)O_(2)` solution is 5MB. The volume strength of `H_(2)O_(2)` is 56 VC. The volume strength of `H_(2)O_(2)` is 112 VD. If 40 mL more `H_(2)O_(2)` solution, the volume strength of the resulting solution is changed to 16.8 V
Answer» Correct Answer - A::B::D

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Strength of the sample of `H_(2)O_(2)` is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of `H_(2)O_(2)` . The concentration of `H_(2)O_(2)` in a solution can also expressed as percentage of `H_(2)O_(2)` in solution . Normality of this solution can be calculated if the equivalent mass of `H_(2) O_(2)` is known . 25 mL of `H_(2)O_(2)` solution was added to the excess of acidified KI solution . The iodine so liberated required 40 mL of `0.1 N Na_(2)S_(2)O_(3)` solution . what is normality of `H_(2)O_(2)` solution ?A. `0.08`B. `0.02`C. `0.16`D. `0.20`
Strength of the sample of `H_(2)O_(2)` is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of `H_(2)O_(2)` . The concentration of `H_(2)O_(2)` in a solution can also expressed as percentage of `H_(2)O_(2)` in solution . Normality of this solution can be calculated if the equivalent mass of `H_(2) O_(2)` is known . Percentage strength of the above `H_(2)O_(2)` solution isA. `0.126`B. `0.272`C. `0.544`D. `0.136`
Strength of the sample of `H_(2)O_(2)` is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of `H_(2)O_(2)` . The concentration of `H_(2)O_(2)` in a solution can also expressed as percentage of `H_(2)O_(2)` in solution . Normality of this solution can be calculated if the equivalent mass of `H_(2) O_(2)` is known . Volume strength of above `H_(2)O_(2)` solution is ?A. `0.448`B. `0.632`C. `0.896`D. `0.556`
What is the molarity of `H_2O_2` of the `11.2V` (volume strength)?
What is the molarity of `H_(2)O_(2)` of the 11.2 V (volume strength) ?
The volume of perhydrol which on decomposition gives 1.5 Iit of `O_(2)` gas at STP isA. 25mlB. 15mlC. 10mlD. 0ml
Perhydrol isA. `10%(w//v)H_(2)O_(2)`B. `30%(w//v)` of `H_(2)O_(2)`C. `3%(w//v)H_(2)O_(2)`D. `100%(w//v)H_(2)O_(2)`
The volume of perhydrol which on decompostion gives 1L of `O_(@)` at STP isA. 10mLB. 1mLC. 100mLD. 10L
The lead of the lead paintings becomes black due to formation of……… `H_(2)O_(2)` converts the black colour to white lead is…..A. `H_(2)S,PbSO_(4)`B. `PbS,PbSO_(4)`C. `PbSO_(4),PbS`D. `PbS,H_(2)S`
Molarity and volume strength of perhydrol respectivelyA. 100M and 8.9 VB. 8.9 M and 100VC. 8.9M and 30VD. 30M and 8.9 V

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