Chemistry

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) Cr                  (d)  [Ar] 3d⁵4s²

(ii) Fe²⁺             (c) [Ar]3d⁶4s⁰

(iii)Ni²⁺               (a) [Ar]3d⁸4s⁰

(iv)Cu  (b) [Ar]3d¹⁰4s¹

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C)

The empirical formula mass of CH₂O is 30 g

The relation between empirical and molecular formula is given as,

n = $\frac{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{E}\mathrm{m}\mathrm{p}\mathrm{i}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}}$  …. (1)

On substituting the values in equation (1)

n = $\frac{180\mathrm{}\mathrm{g}}{30\mathrm{}\mathrm{g}\mathrm{}\mathrm{}}$  = 6

Thus, the molecular formula of the compound will be,

Molecular formula of the compound = (CH₂O)₆ = C₆H₁₂O₆

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) Photon  (d) Exhibits both momentum and wavelength

(ii) Electron  (d) Exhibits both momentum and wavelength

(iii) ψ 2 (b) Probability density

                                          (c) Always positive value

(iv)  Principal quantum  (a) Value is 4 for N shell 

        number n  (c) Always positive value

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

The molar mass of carbon is 44 g mol^-1

44 g of carbon dioxide contains 12 g of carbon

The percentage composition is given as, % composition

= $\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{}}$ $*100$

On substituting the values in the above equation,

% of carbon = $\frac{12\mathrm{}\mathrm{g}}{44\mathrm{}\mathrm{g}\mathrm{}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}\mathrm{}\mathrm{}}$ $*100$

= 27.27%

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) X-rays  (d) ν = 10¹⁸Hz

(ii) UV  (c) ν = 10¹⁶Hz

(iii) Long radio waves (a) ν = 10 -10⁴ Hz

(iv)Microwave  (b) ν = 10¹⁰ Hz

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