Mole Concept and Stoichiometry | Numericals on Gay Lussacs Law ICSE Class 10 |

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Numericals on Gay Lussacs Law of Chemistry ICSE Class 10 | numericals on avagadros law class 10 icse | ‪@sirtarunrupani‬
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Mole Concept and Stoichiometry Class 10 ICSE | Numericals on Gay Lussacs Law ICSE Class 10 |‪@sirtarunrupani‬
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Syllabus of Mole Concept and Stoichiometry:
(i) Gay Lussac’s Law of Combining Volumes; Avogadro’s Law.
• Idea of mole - a number just as a dozen, a gross (Avogadro’s number).
• Avogadro’s Law - statement and explanation.
• Gay Lussac’s Law of Combining Volumes.
- Statement and explanation.
• Understanding molar volume- “the mass of 22.4 litres of any gas at S.T.P. is equal
to its molar mass”. (Questions will not be set on formal proof but may be taught for
clear understanding).
• Simple calculations based on the molar volume and Gay Lussac’s law.
(ii) Refer to the atomicity of hydrogen, oxygen, nitrogen and chlorine (proof not required). The explanation can be given using equations for the formation of HCl, NH3, and NO.
(iii)Vapour Density and its relation to relative molecular mass:
• Molecular mass = 2×vapour density
(formal proof not required)
• Deduction of simple (empirical) and molecular formula from:
(a) the percentage composition of a compound.
(b) the masses of combining elements.
(iv) Mole and its relation to mass.
• Relating mole and atomic mass; arriving at gram atomic mass and then gram
atom; atomic mass is a number dealing with one atom; gram atomic mass is the
mass of one mole of atoms.
• Relating mole and molecular mass arriving at gram molecular mass and gram molecule - molecular mass is a number dealing with a molecule, gram molecular mass is the mass of one mole of molecules.
• Simple calculations based on relation of mole to mass, volume and Avogadro’s
number.
(v) Simple calculations based on chemical equations
Related to weight and/or volumes of both reactants and products.
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27:35 volume of water is 300 cm 3 because volume of H2 is 200 and volume of used O2 is 100 when they combine +

[@chrisop2641]

It will be 0 as its a liquid and gay lusacc law does not apply to liquids

[@realkrsnafan7243]

​@@chrisop2641what if h20 is in water vapour form?

[@vijaybahadurojha4487]

It not mention that it is in vapour form so it will be 0

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57:24 - 3×20=60cm³

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According to gay lussac's law, agar solid or liquid ka volume consider nahi kiya jata hai, to H2O bhi consider nahi kiya jayega. Agar H2O water na hoke water vapour hua to?

[@ranju235]

Water Vapour ko consider kiya jata hai kyuki vo toh gaseous state mein hai
Water in liquid form is having 0 vol

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​@@ranju235 correct ✅ yes it is considered.

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57.38. Ans is 60 cc

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57:20 (sir iska answer hai 60cm3 of co2) will be formed

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57:31 60 cm3

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27:22 volumes of water is 200 cm3

[@amrit_srivastava]

Gay Lussac Law is only for gases, in the equation it is 2H2O so it won't be applied here

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Volume will be zero bcoz Gay lussacs law does not support with solid and liquid

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57:30 3x20 = 60

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8 no. Co2 volume will be 60cc

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12:59 sir but you said that there will be 6.023 x 10²³ but in this equation you put 6.022 insted of 6.023 sir can you explain why is it 🤔

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57:33 vol of CO2 is 60 cc

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Volume of water = 2 volume

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27:25 volume of water will be Nil

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Volume of 3Carbon dioxide is 20×3=60

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57:07 Ans =60cm³

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Volume of water is 0

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Zero vol of water

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27:46 200

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6.023 x10 power -3

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CH4+2O2= Co2+2H2o

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57:29 60

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27:12 200cm3

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No in gay lussac, it only applied in gas not with solid or liquid
2H20 is liquid so there is no volume

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Vol of water is 2

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60cm3

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Gay lussac law apply only for gases

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volume of water is 2

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60

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57:35 60 cm3

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60