Class 11th

This is a Short Answer Type Questions as classified in NCERT Exemplar

Molar mass of NaOH = 40 g/ mole

Molar mass of water =18 g / mole

Mass of NaOH= 4 g

Number of moles of 4 g NaOH = $\frac{4\mathrm{}\mathrm{g}}{40\mathrm{}\mathrm{g}\mathrm{}}$ = 0.1 mol

Number of moles of H₂O= $\frac{36\mathrm{}\mathrm{g}}{18\mathrm{}\mathrm{g}\mathrm{}}$ =2 mol

Mole fraction of water

$\frac{26\mathrm{}\mathrm{g}}{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+\mathrm{}\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}{\mathrm{H}}_2\mathrm{O}}$

Mole fraction of NaOH = $\frac{2\mathrm{}\mathrm{g}}{0.1\mathrm{}+\mathrm{}2\mathrm{}\mathrm{g}\mathrm{}}$  = $\frac{2\mathrm{}\mathrm{g}}{2.1\mathrm{}\mathrm{g}\mathrm{}}$  = 0.95

Number of moles of= $\frac{2\mathrm{}\mathrm{g}}{0.1\mathrm{}\mathrm{g}\mathrm{}}$  = $\frac{0.1\mathrm{}\mathrm{g}}{2.1\mathrm{}\mathrm{g}\mathrm{}}$ = 0.047

Number of moles of NaOH

Mass of solution = Mass of solute + Mass of solvent

Mass of NaOH + Mass of water 4 g + 36 g = 40 g

Specific gravity of solution =1 g / ml

1 litre =1000 ml volume of solution = 40ml

40ml = 0.04 litre

Molarity = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}}$ = $\frac{0.1\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}}{0.04\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{}}$ = 2.5M

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i)

 “l” which is known as Azimuthal quantum number or orbital angular momentum or subsidiary quantum number depicts the three dimensional shape of the orbital.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Molality is the number of moles of substance (also known as the solute) found in a given mass of solvent (in Kg ) in which it is dissolved. Molality is calculated by using the formula.

Molality = $\frac{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{}}{\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{k}\mathrm{g}\mathrm{}}$

So, temperature has no effect on the molality of the solution because molality is expressed in mass.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  option (iii) 9

The total number of orbitals is given by n2, where n is the principal quantum number or the principal shell.

Thus for the third shell number of orbitals is 32 i.e. 9

This is a Short Answer Type Questions as classified in NCERT Exemplar

Mass of NaoH = 40 g

Mass of solvent =1000 g

Mass of solution = 40 x 3+1000

Density = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}$

$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}$ = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}{\mathrm{D}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}}$

$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}$ = $\frac{1120\mathrm{g}}{1.10\mathrm{g}/\mathrm{m}\mathrm{L}}$= 1009.0 mL

Molarity = $\frac{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}}$  = $\frac{3}{1.009\mathrm{}}=$  2.97M

1009.00 mL= 1.009 L

Hence, the molarity of the solution is 2.97M

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: (ii) Heisenberg's uncertainty principle.

According to Heisenberg's uncertainty principle it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Thus it implies that determining the trajectory of an electron is impossible as it requires exact position and velocity which is not possible as per the uncertainty principle.

This is a Short Answer Type Questions as classified in NCERT Exemplar

65.3 g of Zinc gives 22.7 litres of Hydrogen gas

32.65 g Zinc gives = 32.65g x 22.7 litres/65.3 = 11.35 L

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: option (iii) 2

The number of angular nodes is given by n-l

where n is principal quantum number, l is azimuthal quantum number

For 4d orbital, n=4 and l=2

Thus, the number of angular nodes

 = n-l

 = 4-2

 = 2

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  option (iv)

The number of radial nodes is given by n-l-1

where n is principal quantum number, l is azimuthal quantum number

For 3p orbital, n=3 and l=1

Thus, the number of radial nodes

= n-l-1 

= 3-1-1

 = 1

This is a Short Answer Type Questions as classified in NCERT Exemplar

(Natural abundance of 1H x molar mass ) + (Natural abundance of ²H x molar mass of 2H)

Natural abundance of  ¹H = 99.985

Natural abundance of ²H = 0.015

Average atomic mass = $\frac{99.985+0.030}{100\mathrm{}}$

$\frac{100.015}{100\mathrm{}}$ = 1.00015u