Chemistry NCERT Exemplar Solutions Class 11th Chapter One

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 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 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 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 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

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

Molecular mass of Ca₃ (PO₄)₂ = 310.18 g / mole 

Given mass of calcium 4 x 30= 120 g

Given mass of phosphorous =  31 x 2 =62 g

Given mass of oxygen = 16 x 8 = 128 g

Mass percent of Calcium = $\frac{120}{310.18\mathrm{}}$ x 100 = 38.71%

Mass percent of Phosphorous = $\frac{62}{310.18\mathrm{}}$  x 100 = 20%

Mass percent of Oxygen = $\frac{128}{310.18\mathrm{}}$ x 100 = 41.29 %

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

SI unit of the mole is mol. The amount of a substance that contains as many particles or entities as there are atoms in exactly

12 g (0.012 kg) of the C-12  isotope is defined as a mole. One mole is defined as follows:

1 mole =  6.023 x 10²³