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Question

If 32 g of O2 contains 6.022 ×1023 molecules at NTP then 32 g of S, under the same conditions, will contain,


  1. 12.044 x 1023 S

  2. 1 x 1023 S

  3. 6.022 x 1023 S

  4. 3.011 x 1023 S


Solution

The correct option is C

6.022 x 1023 S


ns =3232 =1

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