1. Find the number of atoms of each type present in 90 grams of glucose.
2. Calculate the volume occupied at STP by
   a) 64 gram of oxygen gas
   b) 6.02 X 10²² molecules of CH₄
   c) 5 mole of nitrogen gas
3. How many atoms and molecules of sulphur are present in 100 grams of S₈
4. Calculate the number of electrons, protons and neutrons present in 3.2 gram of oxygen gas.
5. 9.07 X 10¹⁷ atoms of iron wiegh as much as 1 cc of hydrogen gas at STP. What is the atomic mass of iron ?
6. Calculate the percentage composition of each element in cane sugar.
7. Calculate the amount of
   a) Hydrogen gas
   b) Oxygen gas
   produced by the electrolysis of 64 gram of water.
8. calculate the volume of hydrogen and nitrogen gas obtained by the decomposition of 100 litre of ammonia gas at STP.
9. calculate the volume of carbondioxide released on thermal decomposition of 50g of lime stone.
10. 24g of Mg ribbon is burnt in 11.2 litre of oxygen gas. FInd
    a) limiting reagent and excess reagent
    b) mass of MgO formed
11. Calculate the mass of
    a) one atom of Ca
    b) one mole of Ca
    c) one litre of water
12. The density of 5 molal solution of KOH solution is 1.15 g/ml. Calculate the molarity of the solution.
    

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A-1) The molar mass of glucose is 180g. This means that there are 6.022 X 10²³ molecules of glucose present in 180g of glucose. Thus

180g of glucose = 6.022 X 10²³ molecules of glucose

So, 1g of glucose = (6.022 X 10²³ / 180) molecules of glucose

and 90g of glucose = (6.022 X 10²³ X 90 / 180) molecules of glucose

= 3.011 X 10²³ molecules of glucose

Now the molecular formula of glucose is C₆H₁₂O₆. So, 

number of carbon atoms in one molecule of glucose = 6

and number of carbon atoms in 3.011 X 10²³ molecules of glucose = 6 X 3.011 X 10²³ = 1.807 X 10²⁴

Similarly, number of oxygen atoms in 3.011 X 10²³ molecules of glucose = 6 X 3.011 X 10²³ = 1.807 X 10²⁴

and number of Hydrogen atoms in 3.011 X 10²³ molecules of glucose = 12 X 3.011 X 10²³ = 3.612 X 10²⁴

A-7) A balanced chemical equation showing the electrolysis of water is as follows

 2H₂O → 2H₂ + O₂

according to the above equation, 36g of water gives 4g of hydrogen and 32g of oxygen. So, 

36g of H₂O → 4g of H₂

1g of H₂O → (4 / 36) g of H₂

and so 64g of H₂O → (4 X 64 / 36) g of H₂

= 7.111 g of H₂ 

Similarly, 

36g of H₂O → 32g of O₂

1g of H₂O → (32 / 36) g of O₂

and so 64g of H₂O → (32 X 64 / 36) g of O₂

= 56.889g of O₂

A-12) 5 molal KOH solution means that 5 moles of KOH is present in 1Kg of solvent. 

Mass of this solution will be = mass of 5 moles of KOH + 1000g of water

= 5 X 56 + 1000 = 1280g

The molarity of a solution is given by 

molarity = number of moles of the solute / Volume of solution (in litres) 

The volume of the solution can be calculated from the density using the following relation

Density = mass / Volume

Volume = mass / density = 1280 / 1.15 = 1113.043 ml = 1.113 L

So molarity = 5 / 1.113 = 4.492 M