Atoms & Molecules

Que 1. A colorless liquid is thought to be a pure compound. Analysis of three samples of the material yield the following results.

	Mass of Sample	Mass of carbon	Mass of Hydrogen
Sample 1	1.0 g	0.862 g	0.138 g
Sample 2	1.549 g	1.335 g	0.214 g
Sample 3	0.988 g	0.852 g	0.136 g

 
Could the material be a pure compound?

Ans. Analysis

	Mass of Carbon	+	Mass of Hydrogen	=	Mass of Sample
Sample 1	0.862 g	+	0.138 g	=	1.0 g
Sample 2	1.335 g	+	2.214 g	=	1.549 g
Sample 3	0.852 g	+	0.136 g	=	0.988 g

 
Yes, the material is a pure compound as all the three samples have the same composition.

Que 2. A big drop has volume 1.0 mL. How many molecules of water are there is this drop, If the density of water is 1g/mL?

Ans. Volume of drop of water = 1.0 mL

          Density of water = 1.0 g/mL

∴                Mass of drop of water = Volume ×  Diabesity = 1.0 g

                  Molecular mass of H₂O = 2× 1 u + 1 × 16 u = 18 u

                  Gram molecular mass of water = 18 g/mol

                  18 g of water contains = 6.022 × 10²³ molecular

∴                1 g of water contains =  

       =    

Que 3. What is the fraction of the mass of water due to neutrons?

Ans. Mass of one mole (Avogadro Number) of neutrons ~  1 g

         Mass of the one neutrons =  

         Mass of 8 molecule of water =  

         There are 8 neurons of water =   

         Mass of one molecule of water =  

         Fraction off mass of the water due to neutrons  ~ 

Que 4. You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?

Ans. On heating the power, it will char if it is a sugar.
Alternatively, the powder may be dissolved in water and checked for its conduction of electricity. If it conducts, it is a salt.

Que 5. Calculate the number of electrons present in 15.4 of carbon tetrachloride (CCI₄ ).

Ans. Number of moles of CCI_4 =   

∵                                                 = 0.1 mole

              1 mole of CCI₄  = 6.022× 10²³ molecules of CCI₄

∴                       0.1 mole of CCI₄ = 0.1× 6.022 ×  10²³ moles of CCI₄

                                                        = 6.022× 10²² molecules of CCI₄

We know that one atom of carbon has 6 electrons and one atom of chlorine has 17 electrons.
Therefore, one molecule of CCI₄  will contain 6 + (4× 17) = 74 electrons.

∴ Number of electrons in 6.022× 10²² molecules of CCI₄ 

                                                    = 74× 6.022 × 10²² electrons

                                                    = 445.6× 10²² electrons

                                                    = 4.456× 10²⁴ electrons