Q6 / 75 · JEE Main 2025 (22 Jan Shift 1) · Chemistry

Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of $\mathrm{H}_2$ gas. Statement II : Four g of propyne reacts with $\mathrm{NaNH}_2$ to liberate $\mathrm{NH}_3$ gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:

  • A
    Statement I is incorrect but Statement II is correct
  • B
    Both Statement I and Statement II are correct
  • Correct answer
    C
    Statement I is correct but Statement II is incorrect
  • D
    Both Statement I and Statement II are incorrect

Solution

$\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}+\mathrm{Na} \rightarrow \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+}+\frac{1}{2} \mathrm{H}_2$ Statement-l is correct. Moles of $\mathrm{C}_3 \mathrm{H}_4=\frac{4}{40}=0.1$ mole $\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}+\mathrm{NaNH}_2 \rightarrow \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+}+\mathrm{NH}_3$ 0.1 mole 0.1 mole Volume of $\mathrm{NH}_3=(0.1)(22.4)=2.24 \mathrm{~L}$ Statement-II is incorrect.

Hence, the correct answer is option: 'C'.

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