Ncert Solutions Chemistry Class 12th

n-factor of KMnO₄ in acidic medium = 5

n-factor of Mohr's salt = 1

meq of KMnO₄ = meq of Mohr's salt

0.01 * 5 * V = 0.05 * 1 * 20

Volume of KMnO₄ used, V = 20 mL

So; Volume of KMnO₄ left in burette = 50 -20mL

= 30 mL

Moles of C = moles of CO₂

=  $\frac{0.793}{44}mol$

Mass of C =  $\frac{0.793}{44}*12g$

= 0.261g

Moles of H = 2 *  $\frac{0.442}{18}mol$

Mass of H =  $2*\frac{0.442}{18}*1g$

Total mass of compound = 0.492g (given)

So; mass of O = (0.492 – 0.216 – 0.049) g

= 0.227g

% of O = $\frac{0.227}{0.492}*100$  

= 46.14%

the nearest integer = 46

Bromination through free radical mechanism occurs at allylic carbon.

(a) $CoC{\mathcal{l}}_3.4N{H}_{3}is\left[CoC{l}_2{\left(N{H}_3\right)}_4\right]Cl$ will show cis- trans isomerism as;

(b) $CoC{\mathcal{l}}_3.5N{H}_{3}is\left[Co{\left(N{H}_3\right)}_{5}C\mathcal{l}\right]C{\mathcal{l}}_2$ will not show cis- trans isomerism.

(c) $CoC{\mathcal{l}}_3.6N{H}_{3}is\left[Co{\left(N{H}_3\right)}_6\right]C{\mathcal{l}}_3$ will not show cis- trans isomerism.

(d) $CoC\mathcal{l}{\left(N{O}_3\right)}_2.5N{H}_{3}is\left[Co{\left(N{H}_3\right)}_{6}C\mathcal{l}\right]{\left(N{O}_3\right)}_2$  will not show cis-trans isomerism.

$E_{S{n}^{2+}/Sn}^0=-0.140V=E_2^0$

$E_{S{n}^{4+}/Sn}^0=+0.010V=E_3^0$

Here, $\Delta G_3^0=\Delta G_1^0+\Delta G_2^0$

$-4F{E}_3^0=-2F{E}_1^0-2F{E}_2^0$

$E_3^0=\frac{E_1^0+E_2^0}{2}$

$+0.010=\frac{E_1^0+\left(-0.140\right)}{2}$

$E_1^0=E_{S{n}^{4+}/S{n}^{2+}}^0=+0.16$

= 16 * 10^-2 V

Compound X is sucrose as,

Here, glucose is A which on oxidation with HNO₃ gives saccharic acid as

B is b-D fructose which is laevarotatory.

Natural rubber is a linear polymer of isoprene called cis-polyisoprene and has elastic properties.

Ci-polyisoprene molecule consists of various chains held together by weak vander Waals interactions and has a coiled structure.