JEE Main202522 Jan 2025Morning ShiftChemistryHydrocarbonsActual
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
- AStatement I is incorrect but Statement II is correct
- BBoth Statement I and Statement II are correct
- CStatement I is correct but Statement II is incorrect
- 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.