Chemistry NCERT Exemplar Solutions Class 11th Chapter One

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 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 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 Multiple Choice Questions as classified in NCERT Exemplar

Option (C)

The molar mass of glucose is  180 g mol^- . The molarity (M) is given by the formula:

M = $\frac{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$

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

M  = $\frac{0.90\mathrm{}\mathrm{g}\mathrm{}{\mathrm{L}}^{-1}}{180\mathrm{}\mathrm{g}\mathrm{}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}$ = 0.005M

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.