Vapours of Hg are present in the atomosphere form natural sources, such as volcanoes and from human activities. The cuurent level of Hg in atomsphere is 246.3 PPb by volume at 27^(@) C. [1 PPb by volume means 1 L of Hg for every 10^(9) L of air]. Calculate number of Hg atoms in atmosphere having volume of air 5 xx 10^(13) m^(3). Assume Hg vapour follow ideal gas behaviour. [Given : R = 0.0821 atm litre mole - K, N_(A) = 6 xx 10^(23) ] [Divide your answer by 10^(32)]

Vapours of Hg are present in the atomosphere form natural sources, suc

1.	Vapours of Hg are present in the atomosphere form natural sources, such as volcanoes and from human activities. The cuurent level of Hg in atomsphere is 246.3 PPb by volume at 27^(@) C. [1 PPb by volume means 1 L of Hg for every 10^(9) L of air]. Calculate number of Hg atoms in atmosphere having volume of air 5 xx 10^(13) m^(3). Assume Hg vapour follow ideal gas behaviour. [Given : R = 0.0821 atm litre mole - K, N_(A) = 6 xx 10^(23) ] [Divide your answer by 10^(32)]
Answer» SOLUTION :`PPB = (V_(Hg))/(V_("air")) xx 10^(9)` `246.3 = (V_(hg))/(5 xx 10^(13)) xx 10^(9)` `V_(Hg) = 246.3 xx 5 xx 10^(4)m^(3)` `V_(Hg) = 246.3 xx 5 xx 10^(7)` litre a 1 atm and 300 K for Hg vapour : (By volume data is given under SIMILAR P and T condition `PV_(Hg) = n_(Hg) RT` ` atm xx 246.3 xx 5 xx 10^(7) = n_(Hg) xx 0.0821 xx 300` `n_(Hg) = 5 xx 10^(8)` `N_(Hg) atm = 5 xx 10^(8) xx 6 xx 10^(23)` `= 3 xx 10^(32)` `= (3 xx 10^(32))/(10^(32)) = 3`