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Q.

One end of a rope is tied to a 40 kg weight while the second end is coupled with an electric generator. A labourer lifts the weight up to 3 metre and then allows it to fall. As a result, the electric generator generates equal amount of electric work. The standard free energy change for the reaction Al2O3(s)+6Na(s)3Na2O(s)+2Al(s)is 451 kJ/mol of Al2O3 . The number of times the weight is lifted and dropped to produce 13.5 g aluminum will be if g=9.8 m/s2.

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a

288

b

28

c

432

d

144

answer is D.

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Detailed Solution

Al2O3(s)+6Na(s)3Na2O(s)+2Al(s)

2 x 27=54 g

          To produce 54 g Al, ΔG0=451 kJ

          To produce 54 g Al, ΔG0=45154×13.5=112.75 kJ

          Let Q=heat produced when 40 g falls through 3 m distance (h)

          But Q=mgh

=40 kg×9.8 m/s2×3 m=392 kg m2 s-2=392 J  kgm2 s-2=1 J

Hence, number of lifting and falling=ΔG0Q=112.75 kJ×1000 J1 kJ×1392 J=287.6288

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