Some Basic Concepts of Chemistry

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

In a, c, and d options given reactions, all the elements are balanced and hence, following the law of conservation of mass.

In reaction (B), C₃H₈+ O₂→ CO₂+H₂0  elements are not balanced. Hence product are different. It

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A)

According to the law of conservation of mass, matter can neither be created nor be destroyed. In the given reaction,

4Fe + 3O₂→ 2Fe₂O₃

The mass of reactant is equal to the mass of product. Hence, it follows the law of conser

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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

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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.

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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 compo

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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 = $\frac{\mathrm{n}\times \mathbf{}1000\mathrm{L}}{100\mathrm{}}$

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

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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

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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 (D)

(A) The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}\dots$ ….(1)

The number of moles of He is calculated by using equation (1) as follows

Moles of O₂ = $\frac{4\mathrm{}\mathrm{g}}{4\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}}$  =1 mol

The number of atoms can be calculated as, number o