Some Basic Concepts of Chemistry

On heating, we have XCO₃ $\to$ XO + CO₂

The loss of mass (= 4.08 g $-$ 3.64 g = 0.44 g) is due to the removal of CO₂. Thus

Amount of CO₂ released = ${{0.44g} \over {44\,g\,mo{l^{ - 1}}}}$ = 10^$-$2 mol

Hence, Amount of XO = 10^$-$2 mol

Now, mass of BaO in the mixture after heating = 3.64 g $-$ (10^$-$2 mol) (M_XO)

= 3.64 g $-$ (10^$-$2 mol) (M_X + 16 g mol^$-$1)

= 3.48 g $-$ (10^$-$2 mol) (M_X)

From the dissolution reactions

BaO + 2HCl $\to$ BaCl₂ + H₂O

XO + 2HCl $\to$ XCl₂ + H₂O

We conclude that

Amount of HCl consumed for the dissolution process = $2\left[ {{{3.48g - ({{10}^{ - 2}}mol){M_X}} \over {{M_{BaO}}}} + {{10}^{ - 2}}mol} \right]$

Amount of HCl taken for the dissolution process = (100 mL) (0.1 M) = $\left( {{{100} \over {1000}}L} \right)$ (1 mol L^$-$1) = 0.1 mol

Amount of remaining HCl = 0.1 mol $-$ $2\left[ {{{3.48g - ({{10}^{ - 2}}mol){M_X}} \over {154g\,mo{l^{ - 1}}}} + {{10}^{ - 2}}mol} \right]$

Since the remaining HCl required 16 mL of 2.5 M NaOH $\left( { = {{16} \over {1000}} \times 2.5\,mol} \right)$ for complete neutralization, we would have

${{{16 \times 2.5} \over {1000}}}$ mol = 0.1 mol $-$ 2 $\left[ {{{3.48g - ({{10}^{ - 2}}mol){M_X}} \over {154g\,mo{l^{ - 1}}}} + {{10}^{ - 2}}mol} \right]$

This gives ${M_X} = {{\left( {{{16 \times 2.5} \over {1000}} - 0.1 + 2 \times {{10}^{ - 2}}} \right)\left( {{{154} \over 2}} \right)g + 3.48g} \over {({{10}^{ - 2}}\,mol)}}$ = 40 g mol^$-$1

Hence, the element X is Ca.

3 M Na₂S₂O₃ solution means 3 moles of Na₂S₂O₃ present in 1 litre of solution.

$\therefore$ Volume of solution = 1 litre = 1000 ml

Density of solution = 1.25 g/ml

$\therefore$ Mass of solution = 1000 $\times$ 1.25 = 1250 gm

Molar mass of Na₂S₂O₃ = 158 gm

$\therefore$ Weight of solute Na₂S₂O₃ = 3 $\times$ 158 = 474 gm

$\therefore$ Weight of water = 1250 $-$ 474 = 776 gm

(i) % weight of $N{a_2}{S_2}{O_3} = {{Weight\,of\,N{a_2}{S_2}{O_3}} \over {Weight\,of\,solution}} \times 100$

$ = {{474} \over {1250}} \times 100$

$ = 37.92$

(ii) Mole fraction of $N{a_2}{S_2}{O_3} = {{Moles\,of\,N{a_2}{S_2}{O_3}} \over {Moles\,of\,N{a_2}{S_2}{O_3} + Moles\,of\,{H_2}O}}$

$ = {3 \over {3 + {{776} \over {18}}}}$

$ = 0.065$

(iii) Molality of solution (m) $ = {{Moles\,of\,solute} \over {Weight\,of\,solvent\,in\,kg}}$

$\therefore$ $m = {3 \over {{{776} \over {1000}}}}$

$ = 3.865$

$N{a_2}{S_2}{O_3}\buildrel {} \over \longrightarrow 2N{a^ + } + {S_2}O_3^{ - 2}$

$\therefore$ Molality of $N{a^ + } = 2 \times 3.865 = 7.732$

and Molality of ${S_2}O_3^{ - 2} = 3.865$

To find the molality of the solution, we'll first understand the definition. Molality ($m$) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is:

$m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

Given that the mass of the salt is 3 g and its molecular weight is 30 g/mol, we can calculate the moles of solute (salt) using the formula:

$ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molecular weight of solute}} $

$ \text{moles of salt} = \frac{3 \, \text{g}}{30 \, \text{g/mol}} $

$ \text{moles of salt} = 0.1 \, \text{mol} $

We also have 250 g of water as the solvent. Since molality requires the mass of the solvent to be in kilograms, we convert the mass of water to kilograms:

$ 250 \, \text{g} = 0.250 \, \text{kg} $

Now, we can substitute the values into the molality formula:

$ m = \frac{0.1 \, \text{mol}}{0.250 \, \text{kg}} $

$ m = 0.4 \, \text{mol/kg} $

Therefore, the molality of the solution is $0.4 \, \text{mol/kg}$.

From the given chemical equations, we find that

2 mol of NH₂OH $\equiv$ 4 mol Fe²⁺ and 1 mol MnO$_4^ - $ $\equiv$ 5 mol Fe²⁺

Amount of MnO$_4^ - $ consumed in the oxidation of iron (H)

V M = (12 mL) (0.02 M) = $\left( {{{12} \over {1000}}L} \right)$ (0.02 mol L^$-$1) = ${{12 \times 0.02} \over {1000}}$ mol

Since 1 mol MnO$_4^ - $ $\equiv$ 5 mol Fe²⁺, we have

Amount of Fe²⁺ formed by the reduction of Fe³⁺ by NH₂OH = (5) $\left( {{{12 \times 0.02} \over {1000}}} \right)$ mol

Now since 2 mol NH₂OH $\equiv$ 4 mol Fe²⁺, we have

Amount of NH₂OH present in 50 mL of diluted solution = $\left( {{2 \over 4}} \right)(5)\left( {{{12 \times 0.02} \over {1000}}} \right)$ mol

Amount of NH₂OH present in 1 L of diluted solution = $\left( {{{1000} \over {50}}} \right)\left( {{2 \over 4}} \right)(5)\left( {{{12 \times 0.02} \over {1000}}} \right)$ mol

Amount of NH₂OH present in 1 L of undiluted solution = $\left( {{{1000} \over {10}}} \right)\left( {{{1000} \over {50}}} \right)\left( {{2 \over 4}} \right)(5)\left( {{{12 \times 0.02} \over {1000}}} \right)$ mol = 1.2 mol

Mass of NH₂OH present in 1 L of undiluted solution = (1.2 mol) (33 g mol^$-$1) = 39.6 g.

The equations involved are

$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O] \times 2$

${H_2}{O_2} \to {O_2} + 2{H^ + } + 2{e^ - }] \times 5$

$2MnO_4^ - + 5{H_2}{O_2} + 6{H^ + } \to 2M{n^{2 + }} + 5{O_2} + 8{H_2}O$

From this equation, we find that

Equivalent mass of ${H_2}{O_2},\,{M_{eq}} = {{Molar\,mass\,of\,{H_2}{O_2}} \over {2\,eq\,mo{l^{ - 1}}}} = {{34g\,mo{l^{ - 1}}} \over {2\,eq\,mo{l^{ - 1}}}} = 17\,g\,e{q^{ - 1}}$

Mass of H₂O₂ in the given 1.0 g sample, ${m_{eq}} = {X \over {100}} \times 1.0\,g$

Amount (in equivalents) of H₂O₂ in the given 1.0 g sample is

${n_{eq}} = {m \over {{M_{eq}}}} = {{(X/100)g} \over {17\,g\,e{q^{ - 1}}}} = {X \over {17 \times 100}}eq$

If N_KMnO4 is the normality of KMnO₄, the amount (in equivalents) of KMnO₄ consumed is

$n{'_{eq}} = {V_{KMn{O_4}}}\,{N_{KMn{O_4}}} = \left( {{X \over {1000}}L} \right){N_{KMn{O_4}}}$

Equating Eqs (1) and (2), we get

$\left( {{X \over {1000}}L} \right){N_{KMn{O_4}}} = {X \over {17 \times 100}}eq$

or, ${N_{KMn{O_4}}} = {{10} \over {17}}eq\,{L^{ - 1}} = 0.588\,eq\,{L^{ - 1}}$

(a) Volume of mixture of CO and CO₂ = 1 L

Let volume of $C{O_2} = V\,L$

$\therefore$ Volume of $CO = (1 - V)\,L$

Now mixture is passed through red hot charcoal.

$\therefore$ Reaction is :

Volume only count for gases for the reaction involving gases. That is why volume of solid C is not counted.

After reaction completed volume becomes = 1.6 L

$\therefore$ $1 + V = 1.6$

$ \Rightarrow V = 0.6$

$\therefore$ Volume of CO₂ in the mixture = 0.6 L and volume of CO = $ = 1 - 0.6 = 0.4\,L$.

$\therefore$ 60% CO₂ and 40% CO.

(b) We know,

% of any element $ = {{Weight\,of\,element} \over {Molar\,mass\,of\,compound}} \times 100$

Let, atomic mass of metal M = A

Given compound contains 28% nitrogen.

And 3 atom of metal combine with 2 atoms of nitrogen.

$\therefore$ $28 = {{14 \times 2} \over {3A + 14 \times 2}} \times 100$

$ \Rightarrow 3A + 28 = 100$

$ \Rightarrow 3A = 72$

$ \Rightarrow A = 24$

$\therefore$ Atomic weight of metal = 24

$\therefore$ Metal is = Mg

$3Mg + {N_2} \to M{g_3}{N_2}$

111 mg CaCl₂ will give CaCO₃ = 100 mg

1 mg CaCl₂ will give CaCO₃ = ${{100} \over {111}}$ mg

Similarly,

1 mg MgCl₂ will give $CaC{O_3} = {{100} \over {95}}$ mg

Total CaCO₃ formed due to 1 mg CaCl₂ and 1 mg of MgCl₂

$ = {{100} \over {111}} + {{100} \over {95}}$ mg = 2 mg

CaCO₃ per litre of water = 2 mg

Weight of 1 ml of water = 1 g = 10³ mg

Weight of 1000 ml of water = 10³ $\times$ 10³ mg = 10⁶ mg

2 mg of CaCO₃ is present in 10⁶ mg of water.

or 2 parts of CaCO₃ in 10⁶ part of water.

(b) $C{a^{2 + }} + N{a_2}C{O_3} \to CaC{O_3} + 2N{a^ + }$

Equivalent weight of $C{a^{2 + }} = {{40} \over 2} = 20$

1 milliequivalent of Ca²⁺ = 20 mg

1 milliequi. of Ca²⁺ = 1 millieq. of Na₂CO₃

$\therefore$ 1 milliequivalent of Na₂CO₃ is required to soften 1 litre of hard water.

(c) Magnesium burns in O₂ to form MgO. The reaction involved is:

$\mathop {2\,mg}\limits_{2 \times 24 = 48\,g} + \mathop {{O_2}}\limits_{32\,g} \to \mathop {2MgO}\limits_{2[24 + 16] = 80\,g} $

(i) According to the above equation, 48 g Mg requires O₂ = 32g

1 g Mg requires O₂ = ${{32} \over {48}}$ g = 0.66 g

Since the oxygen available is only 0.50 g, so whole of magnesium will not burn. Thus magnesium is in excess.

(ii) Again from the balanced equation, 32 g O₂ reacts with magnesium = 48 g

0.5 g O₂ reacts with magnesium

$ = {{48} \over {32}} \times 0.5 = 0.75$ g

Hence, weight of magnesium in excess

= (1.0 $-$ 0.75) = 0.25 g

(iii) Unused magnesium dissolves in H₂SO₄ according to the following equation:

$\mathop {Mg}\limits_{24\,g} + \mathop {{H_2}S{O_4}}\limits_{98\,g} \to MgS{O_4} + {H_2}$

$\therefore$ H₂SO₄ required to dissolve 0.25 g Mg

$ = {{98} \over {24}} \times 0.25 = 1.021\,g$

MgO formed dissolves in H₂SO₄ according to the following equation:

$\mathop {MgO}\limits_{24 + 16 = 40\,g} + \mathop {{H_2}S{O_4}}\limits_{98\,g} \to MgS{O_4} + {H_2}O$

MgO formed from 0.75

$Mg = {{80} \over {48}} \times 0.75 = 1.25\,g$

H₂SO₄ required to dissolve 1.25 g MgO

$ = {{98} \over {40}} \times 1.25 = 3.062\,g$

Total weight of H₂SO₄ required to dissolve the residue = 1.021 + 3.062 = 4.083 g

Strength of given

H₂SO₄ = Normality $\times$ Eq. wt.

= 0.5 $\times$ 49 = 24.5 g L^$-$1

Hence, 24.5 g H₂SO₄ is present in 1000 mL 4.083 g H₂SO₄ is present in

$ = {{1000} \over {24.5}} \times 4.083$ mL = 166.65 mL

To find the amount of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ present in the sample, we can follow the steps below, applying stoichiometry and the principle of equivalence in titration. Let's analyze each step to solve this problem:

When $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ is strongly heated, it decomposes, and the end product mentioned is $ \text{Mn}_3\text{O}_4 $. The chemical equation for the decomposition isn't provided directly, but it's crucial to know that we end up with $ \text{Mn}_3\text{O}_4 $ which indicates the reduction of the sulfate.

The $ \text{Mn}_3\text{O}_4 $ residue reacts with $ \text{FeSO}_4 $ in the presence of dilute $ \text{H}_2\text{SO}_4 $. Here, $ \text{Fe}^{2+} $ acts as a reducing agent.

The solution from step (ii) completely reacts with $ \text{KMnO}_4 $, using 50 ml of $ \text{KMnO}_4 $ solution.

Here, we learn that 25 ml of the $ \text{KMnO}_4 $ solution require 30 ml of 0.1 N $ \text{FeSO}_4 $ for complete reaction. This gives us a direct relationship to calculate the normality (and thus the molarity) of the $ \text{KMnO}_4 $ solution.

To find the normality of the $ \text{KMnO}_4 $ solution, we use:

$ V_1N_1 = V_2N_2 $

where,

• $ V_1 $ is the volume of $ \text{FeSO}_4 $ (30 ml),


• $ N_1 $ is the normality of $ \text{FeSO}_4 $ (0.1 N),


• $ V_2 $ is the volume of $ \text{KMnO}_4 $ (25 ml),


• $ N_2 $ is the normality of $ \text{KMnO}_4 $, which we will calculate.

$ 30 \times 0.1 = 25 \times N_2 $

$ N_2 = \frac{30 \times 0.1}{25} = \frac{3}{25} = 0.12\, \text{N} $

Now, knowing that 50 ml of $ \text{KMnO}_4 $ reacts with the $ \text{Mn}_3\text{O}_4 $ formed from the original $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $, and since 25 ml of $ \text{KMnO}_4 $ is 0.12 N, 50 ml of $ \text{KMnO}_4 $ is equivalent to 60 ml of 0.1 N $ \text{FeSO}_4 $. This equivalence helps us understand the amount of $ \text{Fe}^{2+} $ that would react with the $ \text{Mn}^{2+} $ from the original sample.

Given that 60 ml of 0.1 N $ \text{FeSO}_4 $ is used in total, and knowing that 1 mole of $ \text{MnSO}_4 $ reacts to eventually form $ \text{Mn}_3\text{O}_4 $ and is equivalent to the reaction of $ \text{Fe}^{2+} $ with $ \text{KMnO}_4 $, we can calculate the moles of $ \text{MnSO}_4 $ initially present.

$ \text{Moles of } \text{Fe}^{2+} = V \times N = 60 \times 0.1 \times 10^{-3} \text{ L} \times \text{N} = 6 \times 10^{-3} \text{ moles} $

Given the molar ratio of $ \text{Fe}^{2+} $ to $ \text{MnSO}_4 $ in the reactions (which can vary based on the specific reactions involved but typically could be 1:1 in redox reactions involving manganate and iron(II) ions in acidic solution), we can assume that the moles of $ \text{Fe}^{2+} $ directly give us the moles of $ \text{MnSO}_4 $ initially present because each $ \text{Fe}^{2+} $ ion reduces a $ \text{MnO}_4^- $ ion, and the $ \text{Mn}_3\text{O}_4 $ would be linked back to the $ \text{MnSO}_4 $ based on the stoichiometry not given.

Thus, the amount of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ present initially can be approximated if we know its molecular weight:

$ \text{MW of } \text{MnSO}_4\cdot4\text{H}_2\text{O} = 151.00 + (4 \times 18.015) = 151.00 + 72.06 = 223.06 \, \text{g/mol} $

Therefore, the mass of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ is:

$ \text{Mass} = 6 \times 10^{-3} \text{ moles} \times 223.06 \, \text{g/mol} = 1.33836 \, \text{g} $

In this way, we find that the sample originally contains approximately $ 1.34 $ g of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $. This calculation is based on an understanding of the stoichiometry of the reactions involved and the titration data. Some assumptions had to be made due to missing precise chemical equations, especially in the conversion between $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ to $ \text{Mn}_3\text{O}_4 $, and how it reacts with $ \text{KMnO}_4 $ and $ \text{FeSO}_4 $.

To find the total number of electrons in 18 ml of water, we need to follow a series of steps involving concepts from chemistry, particularly Avogadro's number and the composition of water molecules.

First, let's determine the number of moles of water in 18 ml. The density of water is approximately 1 g/ml, so 18 ml of water would have a mass of 18 grams.

The molar mass of water (H₂O) is 18 g/mol (approximately 1 g/mol for Hydrogen and 16 g/mol for Oxygen, hence 2*1 + 16 = 18 g/mol). To calculate the number of moles of water, we use the formula:

$\text{moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}}$

Substituting the given values:

$\text{moles of water} = \frac{18 \text{ g}}{18 \text{ g/mol}} = 1 \text{ mol}$

Now, each water molecule contains 10 electrons (2 in each hydrogen atom and 6 in the oxygen atom).

To find the total number of electrons in all the water molecules, we multiply the number of moles of water by Avogadro's number (which is approximately $6.022 \times 10^{23}$ mol⁻¹) to get the number of molecules, then multiply that by the number of electrons per molecule:

$\text{Total electrons} = \text{moles of water} \times \text{Avogadro's number} \times \text{electrons per molecule}$

Substituting the values we know:

$\text{Total electrons} = 1 \text{ mol} \times 6.022 \times 10^{23} \text{ mol}^{-1} \times 10$

$\text{Total electrons} = 6.022 \times 10^{24}$

Therefore, there are $6.022 \times 10^{24}$ electrons in 18 ml of water.

The modern atomic unit is based on the mass of carbon-12. Specifically, it is defined as one twelfth of the mass of a carbon-12 atom in its ground state. This unit of mass is known as the unified atomic mass unit (u) or dalton (Da). Thus, the atomic mass of carbon-12 is exactly 12 u by definition. This standard provides a convenient way for scientists to compare the masses of different atoms and molecules, as it is based on a specific, widely available atom.

Volume of CO₂ evolved = 1336 mL, Temperature = 273 + 27 = 300 K, Pressure = 700 mm

Converting the volume of CO₂ to volume at STP by using the gas equation,

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$ ; ${{700 \times 1336} \over {300}} = {{760 \times V} \over {273}}$

V = 1120 mL

Where V is the volume of CO₂ at STP.

Now, 22400 mL of CO₂ at STP weigh = Gram mol. wt. of CO₂ = 44 g

$\therefore$ 1120 mL of CO₂ at STP weigh

$ = {{44} \over {22400}} \times 1120 = 2.20$ g

Weight of metal carbonate = 4.215 g

Weight of CO₂ evolved = 2.20 g

Weight of metal oxide left after evolution of CO₂ gas = 4.215 $-$ 2.20 = 2.015 g

If equivalent weight of metal is E then using the relation:

${{Weight\,of\,metal\,carbonate} \over {weight\,of\,metal\,oxide}}$

$ = {{E + Eq.\,wt.\,of\,C{O_3}^{2 - }} \over {E + Eq.\,wt.\,of\,{O^{2 - }}}}$

Substituting values ${{4.215} \over {2.015}} = {{E + 30} \over {E + 8}}$ [$\because$ Eq. wt. of $C{O_3}^{2 - } = {{60} \over 2} = 30$]

On solving, E = 12.15

Let,

We have x gm of $N{a_2}O$ and y gm of ${K_2}O$ in 0.5 gm of feldspar.

$\therefore$ Total NaCl obtained $ = {x \over {62}} \times 2 \times 58.5$ gm

and total KCl obtained $ = {y \over {94}} \times 2 \times 74.5$

According to the question,

${x \over {31}} \times 58.5 + {y \over {47}} \times 74.5 = 0.118$

$ \Rightarrow 1.887x + 1.585y = 0.118$ ....... (1)

Now, 2nd part of the reaction happens

Similarly,

$\therefore$ Total moles of AgCl produced

$ = \left( {{x \over {31}} + {y \over {47}}} \right)$ moles

Molecular weight of AgCl = 108 + 35.5 = 143.5

Weight of $AgCl = \left( {{x \over {31}} + {y \over {47}}} \right) \times 143.5$

According to the question,

$\left( {{x \over {31}} + {y \over {47}}} \right) \times 143.5 = 0.2451$

$ \Rightarrow 4.63x + 3.053y = 0.2451$ ....... (2)

Solving equation (1) and (2), we get

$x = 0.0179$ gm

and $y = 0.0531$ gm

$\therefore$ % of $N{a_2}O$ in 0.5 g of feldspar $ = {{0.0179} \over {0.5}} \times 100 = 3.58\% $

and % of ${K_2}O$ in 0.5 of feldspar $ = {{0.0531} \over {0.5}} \times 100 = 10.62\% $

After explosion volume of mixed remain gas of ${V_{{O_2}}} + {V_{C{O_2}}} = 25$ ml

$\therefore$ $30 - 5\left( {x + {y \over 4}} \right) + 5x = 25$

$ \Rightarrow 30 - 5x - {{5y} \over 4} + 5x = 25$

$ \Rightarrow 30 - {{5y} \over 4} = 25$

$ \Rightarrow {{5y} \over 4} = 5$

$ \Rightarrow y = 4$

Now $KOH$ is added to the mixture. As $KOH$ is basic nature so it will absorb $C{O_2}$. After adding $KOH$ volume reduce to 15 ml from 25 ml.

$\therefore$ Absorbed $C{O_2}$ by $KOH$ $ = 25 - 15 = 10$ ml

$\therefore$ $5x = 10$

$ \Rightarrow x = 2$

$\therefore$ Molecular Formula of hydrocarbon gas $ = {C_2}{H_4}$

We know,

Vapour density of single was $ = {{\mathrm{Molar\,Mass}} \over 2}$

and Vapour density (VD) of mixture of gas $ = {{\mathrm{Average\,Molar\,Mass\,({M_{avg}}})} \over 2}$

Here mixture of NO₂ and N₂O₄ gas present.

$\therefore$ $\mathrm{VD} = {{{\mathrm{M_{avg}}}} \over 2}$

Given, $VD = 38.3$

$\therefore$ $38.3 = {{{\mathrm{M_{avg}}}} \over 2}$

$ \Rightarrow {\mathrm{M_{avg}}} = 76.6$

Now let in 100 gm of mixture there are x gm of NO₂.

And molecular mass of $N{O_2} = 14 + 32 = 46$

And let weight of N₂O₄ in 100 gm of mixture $ = 100 - x$

And molecular mass of ${N_2}{O_4} = 14 \times 2 + 16 \times = 92$

$\therefore$ Molos of $N{O_2} = {x \over {46}}$

and moles of ${N_2}{O_4} = {{100 - x} \over {92}}$

We know,

Number of moles $(n) = {{\mathrm{Given\,Mass\,(W)}} \over {\mathrm{Molar\,Mass\,(M)}}}$

$ \Rightarrow M = {W \over n}$

Given, $M = 76.6$

$W = 100$ gm

$n = {x \over {46}} + {{100 - x} \over {92}}$

$ \Rightarrow 76.6 = {{100} \over {{x \over {46}} + {{100 - x} \over {92}}}}$

$ \Rightarrow {x \over {46}} + {{100 - x} \over {92}} = {{100} \over {76.6}}$

$ \Rightarrow {{2x + 100 - x} \over {92}} = {{100} \over {76.6}}$

$ \Rightarrow x + 100 = {{9200} \over {76.6}}$

$ \Rightarrow 76.6x + 7660 = 9200$

$ \Rightarrow 76.6x = 1540$

$ \Rightarrow x = {{1540} \over {76.6}} = 20.1$

$\therefore$ Weight of $N{O_2} = 20.1$ g

and weight of ${N_2}{O_4} = 79.9$ g

$\therefore$ Moles of $N{O_2} = {{20.1} \over {46}} = 0.437$ moles

Most of the naturally occurring elements occurs in the form of isotopes and if different isotopes present in the element are significant in quantity then by calculating their weighted average we get fractional atomic weight.

Note that in question it is given that most of the element's atomic weight is fractional not all elements. Because for example atomic weight of Cl = 35.5 which is fractional but atomic weight of C = 12 which is integer. That is why we can't say all element's atomic weight is fractional. The reason is explained below $\to$

Naturally available two isotopes of Cl are 75% ${}_{17}C{l^{35}}$ and 25% ${}_{17}C{l^{37}}$ . Here both isotopes are significantly present inside Cl. So their weighted average depends on both the isotopes. And by calculating we find atomic weight of Cl = 35.5 (It is a fraction). There are 3 main naturally available isotopes of Carbon (1) 98.9% ${}_6{C^{12}}$ (2) 1.1% ${}_6{C^{13}}$ and (3) 0.0001% ${}_6{C^{14}}$ . As ${}_6{C^{13}}$ and ${}_6{C^{14}}$ are negligible compared to ${}_6{C^{12}}$ . So their weighted average depends on only ${}_6{C^{12}}$ . By calculating we find atomic weight of C = 12.0107 $\simeq$ 12 (integer)

To determine the weight of AgCl that will be precipitated when a solution containing 4.77 g of NaCl is mixed with a solution of 5.77 g of AgNO₃, we employ stoichiometry. The reaction between NaCl and AgNO₃ is as follows:

$ \text{NaCl}_{(aq)} + \text{AgNO}_{3_{(aq)}} \rightarrow \text{AgCl}_{(s)} + \text{NaNO}_{3_{(aq)}} $

From the equation, you can see the reaction proceeds on a 1:1 molar basis for both NaCl and AgNO₃ to AgCl.

To start, we determine the moles of NaCl and AgNO₃ using their respective molar masses (NaCl = 58.44 g/mol, AgNO₃ = 169.87 g/mol):

$ \text{Moles of NaCl} = \frac{4.77}{58.44} \approx 0.0816 \text{ mol} $

$ \text{Moles of AgNO}_{3} = \frac{5.77}{169.87} \approx 0.0340 \text{ mol} $

Since NaCl is in excess (0.0816 mol > 0.0340 mol of AgNO₃), the limiting reagent is AgNO₃. So, the amount of AgCl formed is directly proportional to the amount of AgNO₃ present.

Now, we calculate the mass of AgCl formed. AgCl has a molar mass of approximately 143.32 g/mol (Ag = 107.87 g/mol + Cl = 35.45 g/mol). Given that 0.0340 mol of AgNO₃ will react to form the same amount of AgCl (because of the 1:1 mole ratio), the mass of AgCl produced is:

$ \text{Mass of AgCl} = 0.0340 \, \text{mol} \times 143.32 \, \text{g/mol} \approx 4.873 \, \text{g} $

Therefore, approximately 4.873 g of AgCl will be precipitated when a solution containing 4.77 g of NaCl is added to a solution of 5.77 g of AgNO₃.

1st Part : Volume of 100 g of 13% H₂SO₄ solution = ${{100} \over {1.02}}$ mL = 98.04 mL

13 g of H₂SO₄ $\equiv$ ${{13} \over {98}}$ $\equiv$ 0.1326 mol of H₂SO₄

$\therefore$ 98.04 mL solution contains 0.1326 mol of H₂SO₄.

Hence, molarity of the solution = ${{0.1326} \over {98.04}} \times 1000$ = 1.35 mol L^$-$1

Again, molality (m) = ${{Mass\,of\,the\,solute \div \,Molar\,mass} \over {Mass\,of\,the\,solvent}} \times 1000$

Now, 13% solution of H₂SO₄ means that 13g of H₂SO₄ is dissolved in 87g of solvent.

Thus, molality = ${{13/98} \over {87}} \times 1000 = {{13 \times 1000} \over {87 \times 98}} = 1.52$

$\therefore$ Molality of 13% H₂SO₄ solution = 1.52 (m).

2nd Part : We know, Normality = Molarity $\times$ basicity of the acid. Here, basicity of H₂SO₄ = 2.

$\therefore$ N = 1.35 $\times$ 2 = 2.70 For dilution, N₁V₁ = N₂V₂

$\therefore$ 100 $\times$ 2.70 N = 1.5 N $\times$ V₂ or, V₂ = ${{100 \times 2.70} \over {1.5}}$ or, V₂ = 180

$\therefore$ The sulphuric acid sample should be diluted upto 180 mL to prepare 1.5 N solution.

Let, percentage of isotope with atomic weight $10.01 = x$

and percentage of isotope with atomic weight $11.01 = 100 - x$

We know, average atomic weight formula,

${A_{avg}} = {{\sum {\left( {\% \,\mathrm{of}\,{I_i} \times \mathrm{Atomic\,Weight}} \right)} } \over {\sum {\left( {\% \,\mathrm{of}\,{I_i}} \right)} }}$

Here ${I_i}$ = i th isotope of an element.

Here given,

${A_{avg}} = 10.81$

$\therefore$ $10.81 = {{10.01 \times x + 11.01 \times (100 - x)} \over {x + 100 - x}}$

$ \Rightarrow 10.81 = {{10.01x + 11.01(100 - x)} \over {100}}$

$ \Rightarrow 1081 = 10.01x + 1101 - 11.01x$

$ \Rightarrow x = 1101 - 1081$

$ \Rightarrow x = 20$

$\therefore$ Isotope with weight 10.01 present = 20% and isotope with weight 11.01 present = 80%