Ncert Solutions Chemistry Class 11th

 $\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium ; $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165*1000}{-550}K$

$\begin{array}{l}=3*100K\\ =300K\end{array}$

Chapter 2 is Structure of Atom, aka the OG Science breakdown of how literally everything exists. You'll learn stuff like Bohr's model, orbitals, quantum numbers — all the nerdy magic that makes atoms behave. Sounds heavy, but once you get the hang of it, it lowkey feels like unlocking the Matrix. Trust the process, and maybe don't sleep through this one.

NCERT solutions for class 11 Chemistry chapter 9 hydrocarbon are crucial for understanding the fundamental principle of organic chemistry including the classification nomenclature and reaction of alkanes and alkyne and their vital role in daily life as fuse and in the manufacturing of polymer and other products

Ka for C₃H₇COOH = 2 * 10^-5

$pKa=-\mathcal{l}og\left(2*10^{-5}\right)=5-\mathcal{l}og2$                

=5 – 0.3 = 4.7

pH of 0.2 (M) solution =

    $pH=\frac{pKa-\mathcal{l}ogC}{2}$            

$=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

$pH=27*10^{-1}$          

$\therefore$ Ans 27

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2*0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}*1\right)}^2}{\left(\frac{0.5}{1.5}*1\right)}$

$=\frac{4}{3}=1.33$

$=-710.15J/mol$

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium ; $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165*1000}{-550}K$

$\begin{array}{l}=3*100K\\ =300K\end{array}$

Here are some of the best NCERT Class 11 Chemistry Chapter 11 exercise solutions you can download and study .

1. LearnCBSE

Provides detailed, step by step solutions covering every question—from Intext to Exercise problems.

Available in English, per the latest NCERT syllabus for 2023–24 & 2025- 2026

Ortho product has intramolecular H- bonding & para product has intermolecular H- bonding. Thus it can be separated by steam distillation due to difference in B.

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