Science Answer the questions.[Che-5M]

$\because$ 28g of N₂ has number of molecules = 6.022 × 10²³

$\therefore$ 100g of N₂ has number of molecules $=\frac{6.022\times10^{23}\times100}{28}=2.1\times10^{24}$

Atoms in 100g of N₂ = 2.1 × 10²⁴ × 2

= 42 × 10²⁴ atoms

Similarly, molar mass of NH₃ = 14 + 3 × 1 = 17g

$\because$ 17g NH₃ has number of molecules = 6.022 × 10²³

$\therefore$ 100g NH₃ has number of molecules $=\frac{6.022\times10^{23}\times100}{17}=3.54\times10^{24}$

Atoms in 100g of NH3 = 3.54 × 10²⁴ × 4 = 1.416 × 10²⁵

Thus, 100g of NH₃ has more number of atoms.

1. 1 mole = molar mass of a substance.

$\therefore$ 1 mole of carbon atoms weigh = 12g

$\therefore$ 5 moles of carbon atoms will weigh = 12 × 5 = 60g

Hence, Raunak’s container has weigh = 60g In the same way,

$\therefore$ 1 mole of sodium atoms weigh = 23g

$\therefore$ 5 moles of sodium atoms will weigh = 23 × 5 = 115g

Hence, Krish’s container has weigh = 115g

Thus, Krish’s container is heavier than Raunak’s container.

As, 1 mole = 6.022 × 10²³ atoms

2. Here, both containers have 5 moles of each carbon and sodium, therefore,both containers have equal number of atoms, e., 5 × 6.022 × 10²³ atoms, or 3.011 × 10²⁴ atoms in each.

1. Solid P: Calcium Carbonate (CaCO₃)

Gas Q: Carbon dioxide (CO₂)

Solid R: Calcium oxide (CaO)

2. Total mass of Q and R = 22gm + 28gm = 50gm
3. Total mass of Q and R (50gm) is equal to mass of reactant (50gm).
4. The law of conservation of mass is followed, i.e. total mass of product is equal to mass of reactant.
5. Law of conservation of mass is illustrated by the example.

$\ \ 2\text{Cu} \ \ \ \ \ \ +\ \ \ \ \ \ \ \ \ \ \ \text{O}_2\ \ \ \ \ \ \ \ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ }\ \ \ \ 2\text{CuO}\\ ^{(3.92\text{gm})} \ \ \ \ \ \ \ \ ^{(4.90-3.92\ =\ 0.98\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(4.90\text{gm})}$

$\frac{3.92}{3.92}=1,\ \frac{0.98}{3.92}=0.25,\ \frac{4.90}{3.92}=1.25$

So, 1 equivalent of Cu reacts with 0.25 equivalent of O₂ to from 1.25 equivalent of copper oxide.

Reaction 2:

$2\text{CuO}\ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ }\ \ \ \ 2\text{Cu}\ \ \ +\ \ \ \ \ \ \ \ \ \ \text{O}_2\\ ^{(4.55\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(3.64\text{gm})}\ \ \ \ \ \ ^{(4.55-3.64\ =\ 0.91\text{gm})}$

$\frac{4.55}{3.64}=1.25,\ \frac{3.64}{3.64}=1,\ \frac{0.91}{3.64}=0.25$

Here again, one can see that 1.25 equivalent of CuO decomposed to form 1 equivalent of Cu and 0.25 equivalent of oxygen.

Hence, law of constant proportion is verified.

Number of moles in 1g of sulphur dioxide:

Number of moles in 1g of oxygen:

$\frac{1}{32}$ mole of oxygen will contain $=\frac{\text{x}\ \times\ 64}{32}=2\text{x}$

1. $\ \text{Element}:\ \ \ \ \ \text{Ca}\ \ \ \ \ \ \ \ \ \ \text{F}\\ {\text{Valencies}}:\ \ +2\ \ \ \ \ \ -2$

Thus, the resulting compound is CaF₂

2. $\ \text{Element}:\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \text{S}\\ {\text{Valencies}}:\ \ +1\ \ \ \ \ -2$

Thus, the resulting compund is H₂S

3. $\ \text{Element}:\ \ \ \ \ \ \ \text{N}\ \ \ \ \ \ \ \ \ \ \text{H}\\ {\text{Valencies}}:\ \ -3\ \ \ \ \ \ +1$

Thus, the resulting compound is NH₃

4. $\ \text{Element}:\ \ \ \ \ \ \text{C}\ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\ {\text{Valencies}}:\ \ +4\ \ \ \ \ \ -1$

Thus, the resulting compound is CCl₄

5. $\ \text{Element}:\ \ \ \ \ \text{Na}\ \ \ \ \ \ \ \ \ \ \text{O}\\ {\text{Valencies}}:\ \ +1\ \ \ \ \ \ -2$

Thus, the resulting compound is Na₂O

6. $\ \text{Element}:\ \ \ \ \ \ \ \text{C}\ \ \ \ \ \ \ \ \ \ \text{O}\\ {\text{Valencies}}:\ \ +4\ \ \ \ \ \ -2$

Thus, the resulting compound is CO₂

= 6.022 × 10²⁶Mg atoms

10³ moles of Mg²⁺ ions = 10³ × 6.022 × 10²³

= 6.022 × 10²⁶Mg²⁺ ions

One Mg²⁺ ion is formed from one Mg atom by loss of 2 electrons,

As Mg → Mg²⁺ + 2e^- i.e.

$\therefore$ Difference in the mass of 6.022 × 10²⁶Mg atom and Mg²⁺ ions

= mass of 2 × 6.022 × 10²⁶ electrons

= 2 × 6.022 × 1026 × 9.1 × 10^-31kg.

(as mass of an electron = 9.1 × 10^-31kg)

= 109.6004 × 10^-5kg = 1.096 × 10^-3kg = 1.096g

1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16

= 102g

$\text{i.e.},102\text{g} \ \text{of} \ \text{AI}_2\text{O}_3=6.022\times10^{23} \ \text{molecules} \ \text{of} \ \text{AI}_2\text{O}_3$

$\text{Then}, \ 0.051\text{g} \ \text{of} \ \text{AI}_2\text{O}_3 \ \text{contains}=\frac{6.022\times10^{23}}{102}\times0.051 \ \text{molecules}$

$=3.011\times10^{20} \ \text{molecules} \ \text{of} \ \text{AI}_2\text{O}_3$

The number of aluminium ions (Al³⁺) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al³⁺) present in 3.011 × 10²⁰ molecules

(0.051 g ) of aluminium oxide (Al₂O₃) = 2 × 3.011 × 10²⁰

= 6.022 × 10²⁰.

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{X}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ 2$

To stabilize the compound with hydrogen, two atoms of X are required. Therefore, the valency of X in this compound will be two.

CX₂:

 $\text{Element/ Ion}\ \ \ \ \ \ \ \text{C}\ \ \ \ \ \ \text{X}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ 2$

To stabilize the compound with hydrogen, two atoms of X are required. Thus, the valency of X in this compound is two.

XO₂:

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{X}\ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ -2$

To stabilize this compound with oxygen, four atoms of X are required. Thus the valency of X in this compound if four.

XO₃:

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{X}\ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\ \ \ \ \ \ -2$

To stabilize this compound with oxygen, six atoms of X are required. Thus, the valency of X in this compound is six.

1. Liquid X: Water.

Gas Y: Oxygen.

Gas Z: Hydrogen.

2. $\frac{\text{mass of Z}}{\text{mass of Y}}=\frac{22\text{gm}}{178\text{gm}}=1\ :\ 8$

3. Law of constant proportion is illustrated by this example.
4. Two sources of liquid X– Sea, Well.
5. Gas Y(oxygen) is necessary for breathing.