Atoms and Molecules Worksheet-6

Atoms and Molecules Worksheet-6

 

  1. In a reaction 4.0 g of sodium carbonate were reacted with 10 g of hydrochloric acid. The product was a mixture of 2.5 g of carbon dioxide and 11.5 g of sodium chloride solution. Is this data in agreement with the law of conservation of mass?

 

  1. If 6.3 g of sodium bicarbonate are added to 15.0 g of ethanoic acid (or acetic acid) solution, the residue left is found to weigh 18.0 g. what mass of CO2 is released in the reaction?

 

  1. The percentage of the three elements calcium, carbon and oxygen in a sample of calcium carbonate is given as :

Calcium = 40.0%; Carbon = 12.0%; Oxygen = 48.0%

If the law of constant proportions is true, what weights of these elements will be present in 1.5 g of another sample of calcium carbonate?

 

  1. 0.24 g of a sample of a compound having boron and oxygen was found on analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound.

 

  1. Calculate the mass of 0.5 mole of O2 gas.

(Given : Gram atomic mass of oxygen = 16 g Gram molecular mass of oxygen (O2) = 32 g)

 

  1. Calculate the mass of 0.5 mole of O atoms.

(Given : Gram atomic mass of oxygen = 16 g, Gram molecular mass of oxygen (O2) = 32 g)

 

  1. Calculate the mass of 3.011 × 1023 atoms of O.

(Given : Gram atomic mass of oxygen = 16 g, Gram molecular mass of oxygen (O2) = 32 g)

 

  1. A sample contains 14 g of Nitrogen, 32g of oxygen and 4g of hydrogen. How many molecules of each is present in the sample?

 

  1. How many atoms of N and H are present in 0.5 mole of ammonia (NH3)?

 

  1. What is the weight of 1.20 × 1024 atoms of Na (Atomic mass of Na = 23)?

 

Answer:

  1. The chemical reaction leading to products is :

Mass of reactants = (4.0 + 10.0) = 14.0 g

Mass of products = (11.5 + 2.5) = 14.0 g

The mass of the reactants is equal to the mass of the products formed. Thus the data is in agreement with the law of conservation of mass.

 

  1. The chemical reaction leading to products is :

Mass of reactants = (6.3 + 15.0) = 21.3 g

Mass of products = Mass of solution + Mass of carbon dioxide released.

                   = 18.0 g + x g

According to law of conservation of mass,

Mass of reactants = Mass of products

          21.3 g = (18.0 + x)g

or      x = 21.3 – 18.0 = 3.3 g

∴ Mass of carbon dioxide released = 3.3 g

 

  1. A pure chemical compound always consists of the same elements combined together in a fixed proportion by mass.

In 1.5 g of another sample of Calcium carbonate 0.6 g of calcium, 0.18 g of carbon, and 0.72 g of oxygen is present.

 

  1. Mass of the compound = 0.24 g

Mass of boron in the compound = 0.096 g

Mass of oxygen in the compound = 0.144          

 

  1. 0.5 mole of O2 gas

∴ mass of O2 in grams (m) = no. of moles × M = 0.5 × (32 g) = 16 g

 

  1. 0.5 mole of oxygen (O) atoms

Mass of oxygen (O) in grams (m) = No. of moles × M = 0.5 × (16g) = 8g

 

  1. Explanation: 3.011 × 1023 atoms of oxygen (O)

Step I : Calculation of number of gram atoms of oxygen

         

Step II : Calculation of mass of oxygen (O) atoms

Mass of oxygen (O) atoms = Gram atomic mass of oxygen × No. Of gram atoms of oxygen

           = (16g) × 0.5 = 8g

 

  1. One Gram molecular weight of a compound = 1 mole = 6.022 × 1023 molecules

(1) 28 g N2 = 1 mole = 6.022 × 1023 molecules

14 g N2 = 0.5 mole = 6.022 × 1023 × 0.5 = 3.011 × 1023 molecules

(2) 32 g of O2 = 1 mole = 6.022 × 1023 molecules

(3) 2 g of H2 = 1mole = 6.022 × 1023 molecules

 

  1. 1 mole of ammonia contains 1 × 6.022 × 1023 atoms and 3 × 6.022 × 1023 atoms

∴ 0.5 mole of NH3 will contain

3.012 × 1023 atoms of nitrogen and 9.035 × 1023 atoms of hydrogen

 

  1. 6.022 × 1023 atoms of sodium = 23 g

1.20 × 1024 atoms of sodium

45.82 g