Given below are two statements:
Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas.
Statement II: Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP.
In the light of the above statements, choose the most appropriate answer from the options given below:
Show Hint
- Statement I is correct but Statement II is incorrect.
- Both Statement I and Statement II are incorrect.
- Statement I is incorrect but Statement II is correct.
- Both Statement I and Statement II are correct.
The Correct Option is A
Solution and Explanation
To determine the accuracy of Statements I and II, let's analyze each one:
Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas.
The reaction of propyne (C₃H₄) with sodium (Na) involves the acidic hydrogen present in the terminal alkyne. The reaction is:
C₃H₃ - H + Na → C₃H₃Na + 1/2 H₂
In this reaction, one mole of propyne reacts with sodium to form sodium propyne and release half a mole of hydrogen gas. Therefore, Statement I is correct.
Statement II: Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP.
First, calculate the amount of propyne:
When propyne reacts with NaNH₂, the hydrogen atom is replaced, and NH₃ gas is formed. According to stoichiometry, 1 mole of NaNH₂ liberates 1 mole of NH₃:
C₃H₄ + NaNH₂ → C₃H₃Na + NH₃
The molar volume of gas at STP is 22.4 L/mol. Hence, the volume occupied by 0.1 mol NH₃ at STP is:
The calculated volume of NH₃ is 2.24 L, not 224 mL. Hence, Statement II is incorrect.
In conclusion, the correct choice is: Statement I is correct but Statement II is incorrect.
Learn with videos:
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Organic Chemistry Questions
- Example of vinylic halide is
- Structure of 4-Methylpent-2-enal is
- In the given reactions identify the reagent A and reagent B
- Given below are two statement one is labeled as Assertion (A) and the other is labeled as Reason (R).
Assertion (A): CH2 = CH – CH2 – Cl is an example of allyl halide
Reason (R): Allyl halides are the compounds in which the halogen atom is attached to sp2 hybridised carbon atom.
In the light of the two above statements, choose the most appropriate answer from the options given below - Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B)
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,