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Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of H ₂ gas. Statement II : Four g of propyne reacts with NaNH ₂ to liberate NH ₃ gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:

Options

  1. AStatement I is incorrect but Statement II is correct
  2. BBoth Statement I and Statement II are correct
  3. CStatement I is correct but Statement II is incorrect
  4. DBoth Statement I and Statement II are incorrect

Correct answer

C. Statement I is correct but Statement II is incorrect

Step-by-step solution

CH ₃- C CH + Na CH ₃- C C ⁻ Na ⁺+ 1 2 H ₂ Statement-l is correct. Moles of C ₃ H ₄= 4 40 =0.1 mole CH ₃- C CH + NaNH ₂ CH ₃- C C ⁻ Na ⁺+ NH ₃ 0.1 mole 0.1 mole Volume of NH ₃=(0.1)(22.4)=2.24 ~L Statement-II is incorrect.

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