Chemistry NCERT Exemplar Solutions Class 11th Chapter Nine

Acid = M (OH)2 Salt + H2O

M.E of Acid = me of Base

$\therefore 10*\left(0.1*nf\right)=\left(0.05*2\right)*30$

$\therefore n_f=3$

Thus basicity of acid = 3

Biuret test is given by proteins and dipeptide & compounds having following groups

Hence positive test given by tripeptide & biuret.

Monomer of novolac is

Molecular mass of monomer unit

              = 12 * 7 + 1 * 8 + 16 * 2

              = 84 + 8 + 32

              = 124 g

Molecular mass of polymer = 963 g

$\therefore$ n * mass of monomer = mass of polymer + (n – 1) * 18

(n – 1) unit of H₂O is removed during formation

$\therefore 124*n=963+18\left(n-1\right)$

106n = 945

 $\therefore n=\frac{945}{106}=8.91\approx 9$

FeCl₃. 3H₂O i.e. [Fe (H₂O)₃], K₃ [Fe (CN]₆ and [Co (NH₃)₆]Cl₃ the last two complex are inner-orbital complex due to presence of strong ligand.

$\therefore {\Delta }_{0}of{K}_3\left[Fe{\left(CN\right)}_6\right]>{\Delta }_{0}of\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3$

Because CN^- is strong ligand

$\therefore {\Delta }_0\propto \frac{1}{\lambda }$

More ${\Delta }_0$ smaller value of absorbed $\lambda$

$K_3\left[Fe{\left(CN\right)}_6\right]\to F{e}^{+3}=4s^{0}3d^{5}4p^0$

$\therefore u.e=1$

$\therefore \mu =\sqrt[]{1*\left(1+2\right)}B.M=\sqrt[]{3}B.M$

= 1.732 B.M

$\approx$ 2

Number of replaceable H is 2.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(i) Statements A and R both are correct and R is the correct explanation of A. Since, metals like Pd and Pt adsorbs a large volume of hydrogen, hence, these are used as a storage media for it. 

This is a assertion and reason answer type question as classified in NCERT Exemplar

(i) Statements A and R both are correct and R is the correct explanation of A.

MCl₂+Na₂CO₃→ MCO₃+2NaCl

MSO₄+Na₂CO₃→MCO₃+Na₂SO₄

This is a matching answer type question as classified in NCERT Exemplar

(i) → (b); (ii) → (c); (iii) → (a)