Some Basic Concepts of Chemistry

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%