The no- of chlorine atoms that will be ionised displaystyle left - Clrightarrow Cl-e- right - by the energy released from the process displaystyle Cl-e-rightarrow Cl-for-6-023times 10-23- atom will be -IP- for- Cl - 1250 kJ displaystyle mole-1-and-EA- 350-kJ-mole-1-1-686times 10-23-1-543times 10-21-none of these1-456times 10-22-

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Toppr Admin · Accepted Answer

Since- 1250 kJ mole-x2212-1 energy is required to ionise 6-023-xD7-1023 atoms but 350 kJ mole-x2212-1 energy is released hence the no- of ionised atoms-xA0-6-023-xD7-1023-xD7-350kJmole-x2212-11250kJmole-x2212-1-1-686-xD7-1023

The no. of chlorine atoms that will be ionised (Cl→Cl++e−) by the energy released from the process Cl+e−→Cl−for6.023×1023 atom will be :(IPforCl=1250kJ mole−1andEA=350kJmole−1)1.686×10231.543×1021none of these1.456×1022

Since, 1250 kJ mole−1 energy is required to ionise 6.023×1023 atoms but 350 kJ mole−1 energy is released hence the no. of ionised atoms =6.023×1023×350kJmole−11250kJmole−1=1.686×1023

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