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

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**Solution 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’.**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:**

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**Solution 6:**

**Solution 7:**

**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.**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:**

- The empirical formula of C
_{6}H_{6}is: CH - The empirical formula of C
_{6}H_{12}O_{6}is: CH_{2}O. - The empirical formula of C
_{2}H_{2}is: CH - The empirical formula of CH
_{3}COOH is: CH_{2}O.

**Solution 11:**

Three pieces of information conveyed by the formula H_{2}O is that:

- It shows that there are 2 hydrogen atoms and 1oxygen atoms present in H
_{2}O. - The hydrogen and oxygen atoms are present in simplest whole number ratio of 2:1.
- It represents one molecule of compound water.

**Solution 12:**

**Solution 13:**

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**Solution 14:**

**Solution 15:**

- Na
_{2}SO_{4}.10H_{2}O. - C
_{6}H_{12}O_{6}.

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**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 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 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.