Class 11th

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: option (ii)

The black body is the ideal body that absorbs and emits radiation of all frequencies. The radiation that it emits is said to be black body radiation. With an increase of temperature, the frequency of the black body radiation increases.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: option (i)

 As the chemical properties depend upon the number of electrons present in the atom thus the isotopes resemble the same chemical properties as they have the same number of electrons i.e. atomic number.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C)

(A) The molecule of a compound is obtained when two or more atoms of different elements are combined together.

(B) A compound can be separated into its constituent elements by physical methods of separation.

(C) A compound does not retains the physical and chemical properties of its constituent elements.

(D) When two elements react to form two or more than two chemical compounds, the ratio between different masses of one of the elements combining with a fixed mass of the other.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A)

The density (d ) is given by the formula:

d = $\frac{\mathrm{m}}{\mathrm{V}\mathrm{}\mathrm{}}$

Here, m is the mass and v is the volume

On substituting the values in the above equation:

3.12 g mL^-1 = $\frac{\mathrm{m}}{1.5\mathrm{}\mathrm{m}\mathrm{L}\mathrm{}\mathrm{}}$ M = 4.68 g

The mass will be 4.7 upto 2 significant figure.

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