MHT CET · Chemistry · Some Basic Concepts of Chemistry

What is the number of moles and total number of atoms respectively present in \(5 \cdot 6 \mathrm{~cm}^{3}\) of ammonia gas at STP?

A \(2 \cdot 50 \times 10^{-3} \mathrm{~mol}\) and \(1 \cdot 5 \times 10^{20}\) atoms
B \(1.505 \mathrm{~mol}\) and \(6.022 \times 10^{20}\) atoms
C \(2 \cdot 05 \mathrm{~mol}\) and \(1 \cdot 50 \times 10^{20}\) atoms
D \(2 \cdot 50 \times 10^{-4} \mathrm{~mol}\) and \(6 \cdot 022 \times 10^{20}\) atoms

Verified Solution

Answer & Solution

Correct Answer

(D) \(2 \cdot 50 \times 10^{-4} \mathrm{~mol}\) and \(6 \cdot 022 \times 10^{20}\) atoms

Step-by-step Solution

Detailed explanation

Vol. of \(\mathrm{NH}_{3}\) gas at \(\mathrm{STP}=5.6 \mathrm{~cm}^{3}=5.6 \times 10^{-3} \mathrm{dm}^{3}\)
Now \(22.4 \mathrm{dm}^{3}\) of \(\mathrm{NH}_{3}=1\) mole of \(\mathrm{NH}_{3}\) at \(\mathrm{STP}\)
\(\therefore 5.6 \times 10^{-3} \mathrm{dm}^{3}\) of \(\mathrm{NH}_{3}=\frac{5.6 \times 10^{-3}}{22.4}=2.5 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{NH}_{3}\)
Now, No. of atoms \(=2.5 \times 10^{-4} \times 6.022 \times 10^{21} \times 4=6.022 \times 10^{20}\) atoms

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app