Question

The no. of chlorine atoms that will be ionised $(Cl\to C{l}^{+}+e^{-})$ by the energy released from the process $Cl+e^{-}\to C{l}^{-}$for$6.023\times {10}^{23}$ atom will be :

$(IPforCl=1250kJ$ $mol{e}^{-1}andEA=350kJmol{e}^{-1}$)

• A. $1.686\times {10}^{23}$
• B. $1.456\times {10}^{22}$
• C. $1.543\times {10}^{21}$
• D. none of these

Answer 1

The correct option is A $1.686\times {10}^{23}$

Since, 1250 kJ $mol{e}^{-1}$ energy is required to ionise $6.023\times {10}^{23}$ atoms but 350 kJ $mol{e}^{-1}$ energy is released hence the no. of ionised atoms $=\frac{6.023\times {10}^{23}\times 350kJmol{e}^{-1}}{1250kJmol{e}^{-1}}=1.686\times {10}^{23}$