JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry

\(2.8 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CO}_2\) is left after removing \(10^{21}\) molecules from its ' \(x\) ' mg sample. The mass of \(\mathrm{CO}_2\) taken initially is
Given: \(\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\)

A 98 .3 mg
B 48.2 mg
C 196 .2 mg
D 150 .4 mg

Verified Solution

Answer & Solution

Correct Answer

(C) 196 .2 mg

Step-by-step Solution

Detailed explanation

Moles of removed \(\mathrm{CO}_2=\frac{10^{21}}{6.02 \times 10^{23}} \mathrm{~mol}\) \(=1.66 \times 10^{-3} \mathrm{~mol}\) mole of \(\mathrm{CO}_2\) left \(=2.8 \times 10^{-3} \mathrm{moles}\) total moles of \(\mathrm{CO}_2\) taken initially…

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

Match \(List-I\) with \(List-II\).

\(List-I\)	\(List-II\)
\((A)\) \(\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }\)	\((I)\) Dipole moment
\((B)\) \(\mu=Q \times I\)	\((II)\) Bonding molecular orbital
\((C)\) \(\frac{N_{b}-N_{a}}{2}\)	\((III)\) Anti-bonding molecualr orbital
\((D)\) \(\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }\)	\((IV)\) Bond order

JEE Mains 2022 Medium

Explore more questions on app

From JEE Mains

The sum of all rational terms in the expansion of \(\left(1+2^{1 / 3}+3^{1 / 2}\right)^6\) is equal to

JEE Mains 2025 Easy

Given

Reaction	Energy Change (in \(kJ\) )
\(Li(s) \to Li(g)\)	\(161\)
\(Li(g) \to Li^+(g)\)	\(520\)
\(\frac {1}{2}F_2(g)\,\to F(g)\)	\(77\)
\(F(g) + e^- \to F^-(g)\)	(Electron gain enthalpy)
\(Li^+ (g) + F^-(g) \to LiF(s)\)	\(-1047\)
\(Li (s) + \frac {1}{2}F_2(g)\to LiF(s)\)	\(-617\)

Based on data provided, the value of electron gain enthalpy of fluorine would be.....\(kJ\,mol^{-1}\)

JEE Mains 2013 Hard

Explore more questions on app

\(2.8 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CO}_2\) is left after removing \(10^{21}\) molecules from its ' \(x\) ' mg sample. The mass of \(\mathrm{CO}_2\) taken initially isGiven: \(\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\)

Step-by-step Solution

\(2.8 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CO}_2\) is left after removing \(10^{21}\) molecules from its ' \(x\) ' mg sample. The mass of \(\mathrm{CO}_2\) taken initially is
Given: \(\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\)