Volume Of 1 Equivalent Of O2 At STP Is (A) 5.6L (B) 11.2L (C) 22.4L ...

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Sign InVolume of $ 1 $ equivalent of $ {{O}_{2}} $ at STP is(A) $ 5.6L $ (B) $ 11.2L $ (C) $ 22.4L $ (D) $ 20L $

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• Class 11

• Chemistry

• Volume of 1 equivalent of O2 a...

Last updated date: 07th Jul 2024•Total views: 343.2k•Views today: 7.43k AnswerStudy MaterialsNCERT Solutions For Class 11Important Questions for Class 11Revision Notes for Class 11NCERT BooksMaths Formula for Class 11Sample PapersClass 11 MathsClass 11 PhysicsClass 11 ChemistryClass 11 Business StudiesClass 11 EconomicsClass 11 AccountancySyllabusMaths SyllabusPhysics SyllabusChemistry SyllabusBiology SyllabusEnglish SyllabusHindi SyllabusAccountancy SyllabusBusiness Studies SyllabusEconomics SyllabusPolitical Science SyllabusPhysical Education SyllabusTextbook SolutionsClass 11 NCERT Exemplar SolutionsClass 11 HC Verma SolutionsClass 11 RD Sharma SolutionsClass 11 RS Aggarwal SolutionsClass 11 DK Goel SolutionsClass 11 TS Grewal SolutionsClass 11 Sandeep Garg SolutionsAnswerVerified343.2k+ viewsHint: We know that using the Avogadro’s law which states $ 1 $ mol of gas at NTP occupies 22.4 L of the gas, we can easily find the volume occupied by the oxygen atom at NTP but we have to find out the moles of oxygen from the moles of $ {{O}_{2}} $ and moles of $ {{O}_{2}} $ .Complete answer: This is based on Avogadro's law. So, let’s discuss Avogadro's law. This law states that any gas molecule at the standard conditions of temperature(i.e. $ {}^\circ C $ ) and pressure(i.e. $ 1 $ atm) contains Avogadro number of particles and $ 1 $ mol of the gas at STP or NTP occupies the volume of litres. Avogadro’s law is applicable only at the conditions of standard temperature and pressure and if the pressure and temperature are constant, then on increasing the amount of the gas, its volume also increases and vice-versaIt is already known that $ 1 $ mole of the gas( or $ 32g $ of $ {{O}_{2}} $ ) is equivalent to $ 22.4 $ Litres of the oxygen gas. For finding the gram equivalent volume of oxygen( O2), we equate the following: $ 32g $ of $ {{O}_{2}}\to 22.4L $ of $ {{O}_{2}} $ . So, $ 8g $ is equivalent to $ =\left( \dfrac{22.4}{32} \right)\times 8=5.6\text{ }L $ of the gas. Thus, the gram equivalent volume of oxygen is $ 5.6L. $ Therefore, the correct answer is option A.Note: Remember that STP is mostly used to calculate the gas density. The term which is similar to STP is NTP. The abbreviation of NTP is normal temperature and pressure. It defines temperature at and pressure at one atmosphere. NTP is set but mostly is considered as temperature. STP is mainly used to express the fluid flow.Recently Updated PagesWhy Are Noble Gases NonReactive class 11 chemistry CBSELet X and Y be the sets of all positive divisors of class 11 maths CBSELet x and y be 2 real numbers which satisfy the equations class 11 maths CBSELet x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSELet x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSELet x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSEWhy Are Noble Gases NonReactive class 11 chemistry CBSELet X and Y be the sets of all positive divisors of class 11 maths CBSELet x and y be 2 real numbers which satisfy the equations class 11 maths CBSELet x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSELet x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSELet x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSE

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