Frank ICSE Class 10 Chemistry Solutions -Mole Concept And Stoichiometry

Frank ICSE Solutions for Class 10 Chemistry – Mole Concept And Stoichiometry

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Solution 1:

1. Gay-Lussac’s law: It states that ‘when gases react, they do so in volumes which bear a simple ratio to one another, and also to the volume of the gaseous product, provided all the volumes are measured at the same temperature and pressure’.
2. Avogadro’s law : It states that ‘Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules’.

Solution 2:

Solution 3:
When stating the volume of a gas, the pressure and temperature should also be given because the volume of a gas is highly susceptible to slight change in pressure and temperature of the gas.

Solution 4:

Solution 5:

Solution 6:

Solution 7:

1. Gram atom: “The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element”.
   For example: The gram atomic mass of hydrogen is 1g. So, 1g of hydrogen is 1 gram atom of hydrogen.
2. Gram mole: “A sample of substance with its mass equal to its gram molecular mass is called one gram molecule of this substance or one gram mole”.
   For example: Gram molecular mass of oxygen is 32 g. So One gram mole of oxygen is 32g.

Solution 8:

Solution 9:
Empirical formula:“Empirical formula of a compound is the formula which gives the number of atoms of different elements present in one molecule of the compound, in the simplest numerical ratio”.
Molecular formula: “Molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of the compound”.

Solution 10:

1. The empirical formula of C₆H₆ is: CH
2. The empirical formula of C₆H₁₂O₆ is: CH₂O.
3. The empirical formula of C₂H₂ is: CH
4. The empirical formula of CH₃COOH is: CH₂O.

Solution 11:
Three pieces of information conveyed by the formula H₂O is that:

1. It shows that there are 2 hydrogen atoms and 1oxygen atoms present in H₂O.
2. The hydrogen and oxygen atoms are present in simplest whole number ratio of 2:1.
3. It represents one molecule of compound water.

Solution 12:

Solution 13:

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Solution 14:

Solution 15:

1. Na₂SO₄.10H₂O.
2. C₆H₁₂O₆.

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Solution 1996-1:

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Solution 1998-1:

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Solution 1999-1:

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Solution 1999-2:

Solution 1999-3:

Solution 2000-2:

Solution 2000-1:

Solution 2000-1:
Gay – Lussac proposed this law.

Solution 2001-2:
Molecular mass of ethane = 30
According to Gay-Lussac’s law:
2 vol. of C2H6 requires= 7 vol. of oxygen
Vol. of C2H6 = 2 vol. = 100 L
Vol. of oxygen required = 7 vol. =350 L

Solution 2001-3:

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Solution 2001-4:
The term is vapour density.

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Solution 2007-1:

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Solution 2008-1:
The gas laws which relates the volume of a gas to the number of molecules of the gas is avogadro’s law

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Solution 2008-2:

Solution 2008-3:

Solution 2009-2:

Solution 2009-3:
The correct statement is that equal volumes of all gases under identical conditions contain the same number of molecules.

Solution 2009-4:

Solution 2009-1:
The relative molecular mass of the gas is 10.

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