Ncert Solutions Chemistry Class 11th

This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (A)

The molarity (M) is given by the formula:

M $=$ $\frac{\mathrm{n}}{{\mathrm{V}}_{\left(\mathrm{L}\right)}\mathrm{}}$

On substituting the values in the above equation:

0.02M = $\frac{\mathrm{n}*\mathbf{}1000\mathrm{L}}{100\mathrm{}}$

n = 0.002 mol

The number of molecules can be calculated as, number of moles

= $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}}$

On substituting the values in the above equation:

0.002 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{6.022*\mathrm{}{10}^{23}\mathrm{}}$

number of molecules = 0.002 $*$ 6.022 $*$ 10²³

= 12.044 $*$  10²⁰

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) Principal quantum number (b) energy and size of orbital

(ii) Azimuthal quantum number (d) shape of the orbital

(iii) Magnetic quantum number (a) orientation of the orbital

(iv) Spin quantum number (c) spin of electron

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (D)

The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$  …. (1)

The number of moles of HCl is calculated by using equation (1) as follows,

Moles of HCl $\mathrm{}=\frac{18.25\mathrm{}\mathrm{g}}{36.5\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}}$ 0.5 mol

The molality (m) is given by the formula:

m = $\frac{\mathrm{n}}{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}{\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}}^{\mathrm{k}\mathrm{g}}}$

On substituting the values in the above equation:

Molality = $\frac{0.5\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}1000\mathrm{}\mathrm{k}\mathrm{g}}{500}$ = 1m

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) Cu              (c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

(ii) Cu²⁺            (d) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹

(iii) Zn²⁺           (a)1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

(iv) Cr³⁺          (e) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C)

The molar mass of glucose is  180 g mol^- . The molarity (M) is given by the formula:

M = $\frac{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$

On substituting the value in the above equation, the mol can be calculated as

M  = $\frac{0.90\mathrm{}\mathrm{g}\mathrm{}{\mathrm{L}}^{-1}}{180\mathrm{}\mathrm{g}\mathrm{}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}$ = 0.005M

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  option (i)& (iv)

1. Principal quantum number (n): It represents the size and energy of orbitals.
2. Azimuthal quantum number (l): It represents the subshell and shape of orbitals.
3. Magnetic quantum number (ml): It represents the orientation of the orbitals.
4. Spin quantum number (ms): It represents the direction of spin of electrons in the orbitals.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

The relation between molarity and volume is given as,

M₁ V₁= M₂ V₂

On substituting the value in the above equation, the political can be calculated as

5M* 500 mL = M₂*1500 mL 

M =1.66M

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  option (i) & (iii)

Isoelectronic species are those species that possess the same number of electrons. Here both Na⁺, Mg²⁺ possess 10 electrons and both Na⁺, O^2– possess 10 electrons.