Some Basic Concepts of Chemistry

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

Option (C)

No. of mole given = $\frac{\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$

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

no, of mole =  $\frac{5.85\mathrm{}\mathrm{g}\mathrm{}}{58.5\mathrm{}\mathrm{g}\mathrm{}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}$  = 0.1 g

The molarity (M) is given by the formula:

M = $\frac{\mathrm{n}}{{\mathrm{V}}_1}$

On substituting the values in the above equation:

Molarity = $\frac{0.1\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}1000\mathrm{}\mathrm{L}}{500}$

= 0.2 mol L^-1

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

For the conversion from Fahrenheit to Celsius

$\mathrm{F}=\frac{9}{5}\mathrm{C}+32$

 On substituting the values in the above equation,

 200 = $\frac{9}{5}C+32$

         =93.3 $°$  C

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii)

Average of readings of student A,

Average of student A = $\frac{3.01+2.99}{2}=3.00$

Average of readings of student B,

Average of student B = $\frac{3.05+2.95}{2}=3.00$

The correct reading is 3. For both A and B, the average value is close to the correct value. Thus, readings of both are accurate.

The readings of student A are also very close to each other and close to the average value. Thus, readings are precise. Also, the readings of student B are also very close to each other and close to the average value.

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