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20 volume H(2)O(2) means that one litre ...

20 volume `H_(2)O_(2)` means that one litre of the given sample will give ...... litres of `O_(2)` at S.T.P.

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To solve the problem, we need to determine how many liters of oxygen gas (O₂) can be produced from 1 liter of a 20 volume solution of hydrogen peroxide (H₂O₂) at standard temperature and pressure (STP). ### Step-by-Step Solution: 1. **Understand the Concept of Volume in H₂O₂**: - The term "20 volume" means that 1 liter of the hydrogen peroxide solution can produce 20 liters of oxygen gas when it decomposes. 2. **Decomposition Reaction of H₂O₂**: ...
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20 volume H_(2)O_(2) solution has a strength of about

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The strength of H_(2)O_(2) is expressed in many ways like molarity , normality , % strength and volume strengths But out of all these form of strengths , volume strength has great significance for chemical reactions . This decomposition of H_(2)O_(2) is shown as under H_(2)O_(2)(l) to H_(2)O(l) + (1)/(2)O_(2)(g) 'x' volume strength of H_(2)O_(2) means one volume (litre or ml) of H_(2)O_(2) releases x volume (litre or ml) of O_(2) at NTP . 1 litre H_(2)O_(2) release x litre of O_(2) at NTP =(x)/(22.4) moles of O_(2) From the equation , 1 mole of O_(2) produces from 2 moles of H_(2)O_(2) . (x)/(22.4) moles of O_(2) produces from 2xx(x)/(22.4) moles of H_(2)O_(2) =(x)/(11.2) moles of H_(2)O_(2) So, molarity of H_(2)O_(2)=((x)/(11.2))/(1)=(x)/(11.2)M Normality =n-factor xx molarity =2xx(x)/(11.2)=(x)/(5.6) N What volume of H_(2)O_(2) solution of "11.2 volume" strength is required to liberate 2240 ml of O_(2) at NTP?

The strength of H_(2)O_(2) is expressed in many ways like molarity , normality , % strength and volume strengths But out of all these form of strengths , volume strength has great significance for chemical reactions . This decomposition of H_(2)O_(2) is shown as under H_(2)O_(2)(l) to H_(2)O(l) + (1)/(2)O_(2)(g) 'x' volume strength of H_(2)O_(2) means one volume (litre or ml) of H_(2)O_(2) releases x volume (litre or ml) of O_(2) at NTP . 1 litre H_(2)O_(2) release x litre of O_(2) at NTP =(x)/(22.4) moles of O_(2) From the equation , 1 mole of O_(2) produces from 2 moles of H_(2)O_(2) . (x)/(22.4) moles of O_(2) produces from 2xx(x)/(22.4) moles of H_(2)O_(2) =(x)/(11.2) moles of H_(2)O_(2) So, molarity of H_(2)O_(2)=((x)/(11.2))/(1)=(x)/(11.2)M Normality =n-factor xx molarity =2xx(x)/(11.2)=(x)/(5.6) N What is the percentage strength of "15 volume" H_(2)O_(2) ?

The strength of H_(2)O_(2) is expressed in many ways like molarity , normality , % strength and volume strengths But out of all these form of strengths , volume strength has great significance for chemical reactions . This decomposition of H_(2)O_(2) is shown as under H_(2)O_(2)(l) to H_(2)O(l) + (1)/(2)O_(2)(g) 'x' volume strength of H_(2)O_(2) means one volume (litre or ml) of H_(2)O_(2) releases x volume (litre or ml) of O_(2) at NTP . 1 litre H_(2)O_(2) release x litre of O_(2) at NTP =(x)/(22.4) moles of O_(2) From the equation , 1 mole of O_(2) produces from 2 moles of H_(2)O_(2) . (x)/(22.4) moles of O_(2) produces from 2xx(x)/(22.4) moles of H_(2)O_(2) =(x)/(11.2) moles of H_(2)O_(2) So, molarity of H_(2)O_(2)=((x)/(11.2))/(1)=(x)/(11.2)M Normality =n-factor xx molarity =2xx(x)/(11.2)=(x)/(5.6) N 30 g Ba(MnO_(4))_(2) sample containing inert impurity is completely reacting with 100 ml of "28 volume" strength of H_(2)O_(2) in acidic medium then what will be the percentage purity of Ba(MnO_(4))_(2) in the sample ? (Ba=137, Mn=55, O=16)

The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. As oxidizing agent H_(2)O_(2) is converted into H_(2)O and as reducing agent H_(2)O_(2) is converted into O_(2) . For both cases its n-factor is 2. :. "Normality " "of" H_(2)O_(2) " solution " =2xx "molarity of" H_(2)O_(2) solution 40 g Ba(MnO_(4))_(2) (mol.mass=375) sample containing some inert impurities in acidic medium completely reacts with 125 mL of "33.6 V" of H_(2)O_(2) . What is the percentage purity of the sample ?

The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. As oxidizing agent H_(2)O_(2) is converted into H_(2)O and as reducing agent H_(2)O_(2) is converted into O_(2) . For both cases its n-factor is 2. :. "Normality " "of" H_(2)O_(2) "solution" =2xx "molarity of" H_(2)O_(2) solution 20mL of H_(2)O_(2) solution is reacted with 80 mL of 0.05 MKMnO_(4) "in acidic medium then what is the volume strength of" H_(2)O_(2) ?

Find the volume of oxygen at S. T.P. required. for the complete combustion of 2 litres of carbon monoxide at S.T.P.

The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. As oxidizing agent H_(2)O_(2) is converted into H_(2)O and as reducing agent H_(2)O_(2) is converted into O_(2) . For both cases its n-factor is 2. :. "Normality " "of" H_(2)O_(2) " solution " =2xx " molarity of" H_(2)O_(2) solution What is the molarity of "11.2 V" H_(2)O_(2) ?