(i) Assuming the density of water to be 1g//cm^(3), calculate the volume occupied by one molecule of water. (ii) Assuming the water molecule to be spherical, calculate the diameter of the water molecule. (iii) Assuming the oxygen atom occupied half of the volume occupied by the water molecule, calculate approximately the diameter of the oxygen atom.

(i) Assuming the density of water to be 1g//cm^(3), calculate the volu

1.	(i) Assuming the density of water to be 1g//cm^(3), calculate the volume occupied by one molecule of water. (ii) Assuming the water molecule to be spherical, calculate the diameter of the water molecule. (iii) Assuming the oxygen atom occupied half of the volume occupied by the water molecule, calculate approximately the diameter of the oxygen atom.
Answer» Solution :1 mole of `H_(2)O= 18g=18cm^(3)""(because "density of "H_(2)O=1g//cm^(3))` `=6.022xx10^(23)" MOLECULES of "H_(2)O` Thus, `6.022xx10^(23)" molecules of "H_(2)O" have VOLUME"=18cm^(3)` `therefore"1 molecule of "H_(2)O" willhave volume"=(18)/(6.022xx10^(23))cm^(3)=2.989xx10^(-23)cm^(3)` (ii) As water molecule is assumed to be spherical, if R is its radius, then its volume will be `(4)/(3)piR^(3)=2.989xx10^(-23)cm^(3)"or"R^(3)=7.133xx10^(-24)"or"R=(7.133)^(1//3)xx10^(-8)=1.925xx10^(-8)cm` Take `n=(7.133)^(1//3)""therefore""logn=(1)/(3)log7.133=(1)/(3)xx0.8533=0.2844` `n="Antilog 0.2844"=1.925` `therefore"Diameter of water molecule"=2xx1.925xx10^(-8)cm=3.85xx10^(-8)cm` (iii) As oxygen atom occupied half of the volume occupied by water molecule, hence if r is the radius of oxygen atom, then `(4)/(3)pir^(3)=(1)/(2)xx2.989xx10^(-23)cm^(3)"or"r^(3)=3.566xx10^(-24)" which GIVES r"=1.528xx10^(-8)cm` `therefore"Diameter of oxygen atom"=2xx1.528xx10^(-8)cm=3.056xx10^(-8)cm.`