Chemistry M.C.Q (1 Marks)

($2$) $4 \mathrm{~g}$ of Helium $=\frac{4 \mathrm{~g}}{4 \mathrm{~g}}$ mole $=1 \mathrm{~mole}=\mathrm{NA}_{\mathrm{Ae}}$ atom

($3$) $2.2710982 \text { of He at STP } $$ =\frac{2.271}{22.710982} \text { mole } $

$ =0.1 \mathrm{~mole} $

$ =0.1 \mathrm{~N}_{\mathrm{A}} \mathrm{He} \text { atom }$

($4$) $4 \mathrm{~mol}$ of $\mathrm{He}=4 \mathrm{~N} \mathrm{Ne}$ atoms

So, empirical formula$*=A=1:B=1:C=3$

$\therefore$ The correct empirical formula of compound $\mathrm{X}$ is $\mathrm{ABC}_3$

$ \mathrm{W}=\frac{\mathrm{M} \times \mathrm{M}_2 \times \mathrm{V}(\text { in } \mathrm{mL})}{1000}=\frac{0.75 \times 36.5 \times 25}{1000}$

$ =0.684 \mathrm{~g} \text { (Mass of } \mathrm{HCl}) $

$ \underset{36.5 \mathrm{~g}}{\mathrm{HCl}}+\underset{40 \mathrm{~g}}{\mathrm{NaOH}} \longrightarrow \mathrm{HCl}+\mathrm{NaOH} $

$36.5 \mathrm{~g} \mathrm{HCl}$ reacts with $\mathrm{NaOH}=40 \mathrm{~g}$

$0.684 \mathrm{~g} \mathrm{HCl} \text { reacts with } \mathrm{NaOH}=\frac{40}{36.5} \times 0.684=0.750 \mathrm{~g}$

Amount of $\mathrm{NaOH}$ left $=1 \mathrm{~g}-0.750 \mathrm{~g}=0.250 \mathrm{~g}=250 \mathrm{mg}$

Weight of pure limestone $\left( CaCO _3\right)=20 \%$ of $20\,g$

$=\frac{20}{100} \times 20$

$=4\,g$

$n _{ CaCO _3}=\frac{4}{100}=0.04$

$CaCO _3 \rightarrow CaO + CO _2$

$n =0.04 \quad n =0.04$

$n _{ CO _2}=0.04$

$W _{ CO _2}=0.04 \times 44$

$=1.76\,g$

$1=\frac{0.5}{\text { Weight of solvent }( g )} \times 1000$

Weight of solvent $( g )=500 \,g$

no. of moles of $CaCO _{3}$ (pure) $=\frac{1}{2} \times$ mole of $HCl$

$\quad \text { [Mole }=\text { molarity } \times \text { volume(in Itr. }) \text { ] }$

$=\frac{1}{2} \times 0.5 \times \frac{50}{1000}=0.0125$

weight of $CaCO _{3}$ (pure) $=$ mole $\times$ mol. wt

$=0.0125 \times 100=1.25 \,g$

$\%$ purity $=\frac{\text { wt. of pure substance }}{\text { wt. of impure sample }} \times 100$

$95=\frac{1.25}{\text { wt. of impure sample }} \times 100$

wt. of impure sample $=\frac{1.25 \times 100}{95}=1.32 \,g$

$H \quad 22 \quad \frac{22}{1}=22 \quad \frac{22}{6.5}=3.38=3$

Emperical formula $=\mathrm{CH}_{3}$

$=\frac{w}{\text { molar mass }} \times N _{ A } \times$ atomicity

$(1)$ $\frac{1}{7} \times N _{ A } \times 1$

$(2)$ $\frac{1}{108} \times N _{ A } \times 1$

$(3)$ $\frac{1}{24} \times N _{ A } \times 1$

$(4)$ $\frac{1}{32} \times N _{ A } \times 2$

mass of solution $=$ volume of solution $\times$ density $=1000 \times 1.28$

$=1280 \mathrm{g}$

mass of solvent $=$ mass of solution $-$ mass of solute $=1280-80$

$=1200 \mathrm{g}$

molality $=\frac{2}{1200} \times 1000=\frac{20}{12}=\frac{10}{6}=\frac{5}{3}=1.67 \mathrm{m}$

(moles)$_{1}=\frac{2.3}{46}=\frac{1}{20} \quad \quad 0 \quad \quad \quad 0$

(moles)$_f 0\quad \quad \quad \quad \quad \quad \frac{1}{20}\quad \quad \quad \frac{1}{20}$

$\quad \quad \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{CO}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$

$\left[\mathrm{H}_{2} \mathrm{O} \text { absorbed by } \mathrm{H}_{2} \mathrm{SO}_{4}\right]$

(moles)$_1\quad \frac{4.5}{90}=\frac{1}{20} 0 \quad \quad \quad 0 \quad \quad 0$

(moles)$_f\quad 0\quad \quad \quad \ \frac{1}{20} \quad \quad \frac{1}{20} \quad \quad \frac{1}{20}$

$\mathrm{CO}_{2}$ is absorbed by KOH. So the remaning product is only CO. moles of CO formed from both reactions

$=\frac{1}{20}+\frac{1}{20}=\frac{1}{10}$

Left mass of $\mathrm{CO}=$ moles $\times$ molar mass

$=\frac{1}{10} \times 28$

$={2.8 \mathrm{g}}$

$\Rightarrow 42 mg =3.018 \times 10^{20} $

$\Rightarrow 84 mg =6.02 \times 10^{20}$

$84 g =6.02 \times 10^{23}$

$\therefore 85.7 \%=\frac{85.7}{100} \times 84=72 g $

$1 mole$ of $C =12 g$

6 moles of $C =12 \times 6=72 g$

$\therefore$ There are 6 carbons in each molecule

$84-72=12 g$ of $H =12 H$ atoms in each molecule

$\therefore$ Molecular formula $= C _6 H _{12}$

$\because 0.1 \mathrm{mole}$

$X Y_{2} \equiv 10 g \therefore 1 \mathrm{mole}$

$X Y_{2} \equiv 100 g=X+2 Y=100 \ldots (1)$

For $X_{3} Y_{2}$

$\because 0.05$ mole

$X_{3} Y_{2}$

$\because 9 g$

$\therefore 1$ mole $X_{3} Y_{2} \equiv 180 g$

$3X+2Y=180\dots (2)$

On solving $(1)$ and $(2)$, $X=40$ And $Y=30$

If Avogadro Number $(NA)$ is changed than mass of $1 mol\;( 6.022 \times 10^{20}$ atom) of carbon. $=\frac{12 \times 6.022 \times 10^{20}}{6.022 \times 10^{23}}=12 \times 10^{-3} g$

Therefore the mass of $1 \;mol$ of carbon is changed

$=0.05 \mathrm{mole}$

Moles of $\mathrm{NaCl}=50 \times 5.8 / 100 \times 58.5$

$=0.05 \mathrm{mole}$

$\mathrm{AgNO} 3+\mathrm{NaCl} \rightarrow \mathrm{AgCl}+\mathrm{NaNO} 3$

mass of $\mathrm{AgCl}=$ mole $\times$ molar mass

${=0.05 \times 143.5}$

${=7.16 \mathrm{g}}$

$\ln 18 \;\mathrm{g}=1 \mathrm{moles}$

$1\; mole$ contain $6.022 \times 10^{23}$ molecules of water

so $18\; moles$ contain maximum number of molecules

$\quad \quad \mathrm{H}_{2}: \mathrm{O}_{2}$

mass $2: 32$

i.e. $1: 16$

therefore, Molar ratio $=\frac{n_{\mathrm{H}_{12}}}{n_{0_{2}}}=\frac{x / 2}{4 x / 32}=\frac{(x) \times 32}{2 \times 4 x}=\frac{4}{1}=4: 1$

$v_{H_{2}}: v_{O_{2}}: v_{C H_{2}}=n_{H_{12}}: n_{O_{2}}: n_{C H_{4}}$

$\Rightarrow v_{H_{2}}: v_{O_{2}}: v_{C H_{4}}=\frac{m_{H_{2}}}{M_{H_{2}}}: \frac{m_{O_{2}}}{M_{O_{2}}}: \frac{m_{C H_{4}}}{M_{C H_{4}}}$

But $m_{H_{2}}=m_{O_{2}}=m_{C H_{4}}=m\left[\therefore n=\frac{\text { mass }}{\text { molar massd }}\right]$

Thus, $\quad v_{H_{2}}: v_{O_{2}}: v_{C H_{4}}=\frac{m}{2}=\frac{m}{1}=\frac{m}{16}=16: 1: 2$

$\quad \quad \quad \quad H_{2}(g)+C l_{2}(g) \rightarrow 2 H C l(g)$

Initial vol. $22.4 L \;\;11. 2 L\;\;\;\;\;\;\;\;\;2 m o l$

$\therefore 22.4 \mathrm{L}$ volume at $\mathrm{STP}$ is occupied by

$C l_{2}=1 \text { mole }$

$\therefore 11.2 \mathrm{L}$ volume will be occupled by

$C l_{2}=\frac{1 \times 11.2}{22.4} \mathrm{mol}=0.5 \mathrm{mol}$

$22.4 \mathrm{L}$ volume at $\mathrm{STP}$ is occupied by $H_{2}=1 \mathrm{mol}$

Thus, $H_{2}(g)+C l_{2}(g) \rightarrow 2 H C l(g)$

$\quad \quad 1 \mathrm{mol} \quad \quad 1\mathrm{mol}\quad \quad 0.5 \mathrm{mol}$

since, $C l_{2}$ possesses minimum number of moles,

thus it is the limiting reagent.

As per equation,

$1\; mole$ of $C l_{2}=2$ moles of $\mathrm{HCl}$

$\therefore 0.5$ mole of $C l_{2}=2 \times 0.5$ mole of $\mathrm{HCl}$

$=1.0$ mole of $\mathrm{HCl}$

Hence, $1.0 \;mole$ of $\mathrm{HCl}(\mathrm{g})$ is produced by $0.5\; mole$ of $C l_{2}[\text { or } 11.2 \mathrm{L}]$

$M g+\frac{1}{2} O_{2} \rightarrow M g O$

Moles: $\frac{1.0}{24} ; \frac{0.56}{32}$

${\frac{0.5}{12} ; \frac{0.07}{4}}$

${\frac{0.5}{12}-x ; \frac{0.07}{4}-\frac{x}{2}}$

Oxygen is limiting reagent so, $\frac{0.07}{4}-\frac{x}{2}=0$

$x=\frac{0.07}{2}$

Excess $M g=\frac{0.5}{12}-\frac{0.07}{2} \;\mathrm{mol}$

Mass of $M g$ is $=1-0.7 \times 12=0.16\; \mathrm{g}$

Thus, when $1.0 \;g$ of magnesium is burnt with $0.56 \;\mathrm{g} O_{2}$ in a closed vessel, $0.16\; \mathrm{g}$ magnesium is left in excess.

Molar conc $=\frac{n \times 1000}{V_{\text {solution }(\mathrm{mL})}}=\frac{10^{-3} \times 1000}{100}$

Molar conc. $=0.01 \mathrm{M}$

$a A(g)+b B(g) \rightarrow c C(g)+d D(g)$

Here, '$a$' moles of $A(g)$ reacts with '$b$' moles of $B(g)$ to give '$c$' mole of $C(g)$ and '$d$' moles of $D(g)$

No. of moles of $\mathrm{AgNO}_{3}=10^{-3} \mathrm{mol}$

No. of moles the chloride $=0.5 \times 10^{-3} \mathrm{mol}$

Suppose the formula for the chloride is $\mathrm{XCl}_{\mathrm{n}}$ then moles of chloride ion $=\mathrm{n} \times 0.5 \times 10^{-3}$

Reaction goes as follows:

$A g^{+}+C l^{-} \rightarrow A g C l$

Then, going by stoichiometry we get

$n \times 0.5 \times 10^{-3}=10^{-3}$

$\Rightarrow n=2$

Therefore, formula is $\mathrm{XCl}_{2}$

Among these elements, metals are in the maximum number. There are $95$ metals and seventeen non-metals. Rest are metalloids such as Silicon, Arsenic, Boron, etc.

Diamond and graphite are purely made up of carbon i.e $C$, an element.

Ozone is trioxygen molecule i.e $O _3$, an element.

Dimension of Joule i.e. work $ = F \times L$ $ = ML{T^{ - 2}} \times L$

$ = \left[ {M{L^2}{T^{ - 2}}} \right]$

$\frac{1}{{Pa}} = \frac{1}{{{\rm{Pressure}}}} = \frac{1}{{\frac{F}{A}}} = \frac{{1 \times A}}{F} = \left[ {ML{T^{ - 1}}} \right]$

So, $J \,P{a^{ - 1}}$ $ = \left[ {M{L^2}{T^2}} \right]$ $ = \left[ {{L^2} \times L} \right]\; = \left[ {{L^3}} \right]$.

$1 \,g =1000\, mg$

Now, convert all option value in mg

$16 \,g =16000\, mg$

$16.4\, g =16400\, mg$

$16.428\, g =16428\, mg$

$16.4284 \,g =16428.4\, mg$

from above in 16428 all digits are significant and nearest to milligram.

$1$ erg energy $=10^{-7}\, J$

$1$ Joule $=1 \,J$

$1$ Calorie energy $=4.18 \,J$

Hence, answer is option $D$.

Volume of solution $=1.5 \mathrm{mL}$

Mass of solution = Volume $\times$ Density

$=1.5 \mathrm{mL} \times 3.12 \mathrm{g} \mathrm{mL}^{-1}$

$=4.68 \mathrm{g}=4.7 \mathrm{g}$ (up to $2$ significant figures)

Hence, $CuO$ and $Cu _2 O$ form two different compounds in a ratio of $1: 2$ with the fixed weight of oxygen.

For example: In $CO : 12$ parts by mass of carbon combine with $16$ part by mass of oxygen.

In $CO _2, 12$ parts by mass of carbon combine with $32$ parts by mass of oxygen. The ratio of masses of oxygen which combine with a fixed mass of carbon in these compounds is $16: 32$ or $1: 2$, which is a simple whole number ratio.

$12 \,g +32\, g =44 \,g$

Mass of reactants and products should be equal to each other, in accordance with the law of conservation of mass.

Compound $A \Rightarrow 1 \,g$ of $N _2+0.57\, g$ of $O _2$

Compound $B \Rightarrow 2 \,g$ of $N _2+2.24 \,g$ of $O _2$

$\therefore 1 \,g$ of $N _2$ will require $\frac{1 \times 2.24}{2}=1.12\, g ^{\text {of }} O _2$

Compound $C \Rightarrow 3\, g$ of $N _2+5.11\, g _2$ of $O _2$

$\therefore 1 \,g$ of $N _2$ will require $\frac{1 \times 5.11}{3}=1.70 \,g$ of $O _2$

The ratio of $O _2$ in $A , B$ and $C =0.57: 1.12: 1.7 \approx 1: 2: 3$

Therefore, the given results obey the law of multiple proportions.

In $CH _4$, ratio of masses of $C$ and $H =6: 2$

According to law of reciprocal proportions,

The ratio of masses of $C$ and $O=6: 16$

Hence, $SnCl _2$ and $SnCl _4$ illustrates the law of multiple proportions.

$118.7 \mathrm{g}$ of tin combine with $71 \mathrm{g}$ of chlorine to form one mole of $S n C l_{2} .$

The ratio of the masses of chlorine that combine with same mass of tin is $142$

$71=2: 1$

This is a simple ratio. It indicates the law of multiple proportions.

Similarly other ratio of the reactants are $2: 1,2: 5$ in case of fixing mass of $N_{2}$ as $28 \mathrm{g}$ and varing mass of $O_{2}$ as $32 \mathrm{g}$ and $80 \mathrm{g}$ respectively.

Here two elements combine to form more than one compound and the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. Thus they follow the law of multiple proportions. thus option $C$ is correct.

                 $12:$     $16$

In $CO_2$    $12:$     $32$

$(b)$ $6 \times {10^{23}}$ atoms of $C$ has mass $ = 12\,gm$

$3 \times {10^{23}}$ atoms of $C$ has mass $ = \frac{{12 \times 3 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 6\,gm$

$(c)$ $1$ mole of $S$ has mass $= 32\,gm$

$(d)$ $7.0\,gm$ of $Ag$

So, lowest mass = $6\,gm$ of $C$.

$\therefore$ $1.12$ $ \times {10^{ - 7}}$ of gas at $STP$ has $\frac{{6 \times {{10}^{23}} \times 1.12 \times {{10}^{ - 7}}}}{{22400}}$ $= .03 \times {10^{14}} = 3 \times {10^{12}}$.

According to the mole concept

$17\,gm$ $N{H_3}$ has molecules $ = 6.02 \times {10^{23}}$

$\therefore \,\,1\,gm$ $N{H_3}$ has molecules $ = \frac{{6.02 \times {{10}^{23}}}}{{17}}$

$\therefore $ $4.25\,gm$ $N{H_3}$ has molecules $ = \frac{{6.02 \times {{10}^{23}} \times 4.25}}{{17}} = 1.5 \times {10^{23}}\,molecule$

$14\,g$ ${N_2}$ has no. of moles $ = \frac{{14}}{{28}} = \frac{1}{2}$

No. of moles are same, so no. of molecules are same.

Hence the total no. of electrons in one mole of $H _2$ are $12.046 \times 10^{23}$.

$18\,gm$ ${H_2}O$ =$6.023 \times {10^{23}}$ molecule $\therefore $ $34\,gm$ ${H_2}O$ = $\frac{{6.023 \times {{10}^{23}}}}{{18}} \times 34$ $ = 11.37 \times {10^{23}}$ mole

$(b)$ $28\,gm$ of $C{O_2}$ $44\,gm$ $C{O_2}$$ = 6 \times {10^{23}}$ molecules

$28\,gm$ $C{O_2}$ $ = \frac{{6 \times {{10}^{23}}}}{{44}} \times 28 = 3.8 \times {10^{23}}$

$(c)$ $46\,gm$ of $C{H_3}OH$ $32\,gm$ $C{H_3}OH\; = 6 \times {10^{23}}$ molecules

$46\,gm$ $C{H_3}OH\; = \frac{{6 \times {{10}^{23}}}}{{32}} \times 46 = 8.625 \times {10^{23}}$

$(d)$ $108\,gm$ of ${N_2}{O_5} = 6 \times {10^{23}}$ molecules

$54\,gm$ of ${N_2}{O_5}\; = \frac{{6 \times {{10}^{23}}}}{{108}} \times 54 = 3 \times {10^{23}}$ molecules.

Molecular weight $= 46 +16 = 62$

$62\,gm$ of $N{a_2}O$ = $1$ $mole$

$620\,gm$ of $N{a_2}O$= $10$ $mole$.

also $4\,g$ of sulphur $ = \frac{4}{{32}} = \frac{1}{8}$ $mole.$

Hence, number of $Na$ atoms = No. of moles $ \times $ number of atom $ \times $ Avogadro’s number

$2 \times 4 \times 6.023 \times {10^{23}} = 48 \times {10^{23}}$

No.  of atoms $ = 3 \times 6.023 \times {10^{21}}$=$18.069 \times {10^{21}}$ atoms

(b) $22.4\,L$ of $C{O_2}$

No. of atoms =$6.023 \times {10^{23}} \times 3$$ = 18.069 \times {10^{23}}$ atoms

(c) $0.44\,gm$ of $C{O_2}$

No. of moles $ = \frac{{0.44}}{{44}} = \frac{1}{{100}} \times 6.023 \times {10^{23}}$ moles

$= 6.023 \times {10^{21}}$ moles $ = 3 \times 6.023 \times {10^{21}}$ atoms $18.069 \times {10^{21}}$ atoms

$\because 48\,g\,{C_2}{H_5}OH$ has $H$ atom $ = 6 \times {N_A}$

$\therefore $ $0.046\,g\,\,{C_2}{H_5}OH$ has $H$ atoms $ = \frac{{6 \times 6.02 \times {{10}^{23}} \times 0.046}}{{46}}$$ = 3.6 \times {10^{21}}$

Here, $N_A$ represents Avogadro's number. $M$ represents the molar mass of compound.

Number of molecules present in $36 \,g$ of water $=\frac{36}{18} \times N _{ A }=2\, N _{ A }$

Number of molecules present in $28\, g$ of $CO =\frac{46}{46} \times N _{ A }= N _{ A }$

Number of molecules present in $46 \,g$ of $C _2 H _5 OH =\frac{46}{46} \times N _{ A }= N _{ A }$

Number of molecules present in $54 \,g$ of $N _2 O _5=\frac{54}{108} \times N _{ A }=0.5\, N _{ A }$

no. of molecule $=10 \times 6.02 \times 10^{23}$

$=6.02 \times 10^{24} $

no. of $\mathrm{NH}_{3}$ molecules $=2 \times \mathrm{N}_{\mathrm{A}}$

one $\mathrm{NH}_{3}$ molecule contains $10$ electron total no. of electrons $=2 \times \mathrm{N}_{\mathrm{A}} \times 10=20\, \mathrm{N}_{\mathrm{A}}$

$=720+122 =842 $

wt of a molecule $=842 \times 1.67 \times 10^{-24}$

$=1.406 \times 10^{-21}\, \mathrm{g}$

$\frac{64}{64}=1\, \mathrm{mol}\, \mathrm{SO}_{2}$

$\frac{44}{1}=1\, \mathrm{mol}$ of $\mathrm{CO}_{2}$

$\frac{48}{48}=1\, \mathrm{mol}$ of $\mathrm{O}_{3}$

$\frac{8}{2}=4\, \mathrm{mol}$ of $\mathrm{H}_{2}$

$=224\, \mathrm{L}$

Moles $=\frac{224}{22.4}=10$

No. of atoms $=10 \times 2 \times 6.02 \times 10^{23}$

$=1.2 \times 10^{25}$

${{\text{n}}_{{{\text{H}}_2}{\text{O}}}} = \frac{{36}}{{18}} = 2$

$\mathrm{n}_{\mathrm{CO}}=\frac{28}{28}=1$

${n_{{C_2}{H_5}OH}} = \frac{{46}}{{46}} = 1$

${n_{{N_2}{O_5}}} = \frac{{54}}{{108}} = 0.5$

$ \therefore \text { No. of atoms } =4 \times \frac{4.25}{17} \times 6 \times 10^{23} $

$=6 \times 10^{23}$

No. of molecules of $\mathrm{CO}_{2}=\frac{1}{2} \times \mathrm{N}_{\mathrm{A}}$

No. of atoms $=3 \times \frac{1}{2} \times \mathrm{N}_{\mathrm{A}}=1.5 \mathrm{N}_{\mathrm{A}}$

No. of $\mathrm{NH}_{4}^{+}$ ions in $1.8\, \mathrm{g}=\frac{1.8}{18} \times \mathrm{N}_{\mathrm{A}}=0.1\, \mathrm{N}_{\mathrm{A}}$

$\text { No. of protons in } 1.8\, \mathrm{g} \,\mathrm{NH}_{4}^{+} =11 \times 0.1\, \mathrm{N}_{\mathrm{A}}$

$=1.1 \,\mathrm{N}_{\mathrm{A}}$

in $8\, \mathrm{g}\, \mathrm{O}_{2}$ No. of atom

$\frac{8}{16}=\frac{N}{N_{A}} \Rightarrow N=\frac{N_{A}}{2}$

At condition $\mathrm{B}=0.5$ mole

At condition $\mathrm{C}=0.5$ mole

At condition $\mathrm{D}=0.1$ mole

i.e. at condition $D$ we have smallest number of mole so we also have smallest number of molecule in condition $D$.

$\therefore$ No. of mole of $\mathrm{O}$ atom $=4 \times 2=8$

$ \therefore \text { No. of } \mathrm{O} \text { atom } =8 \times \mathrm{N}_{\mathrm{A}}$

$=8 \times 6.023 \times 10^{23} =48.184 \times 10^{23} $

$=4.8184 \times 10^{24}$

Butane and isobutane have same formula.

$\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \Rightarrow \frac{40.2}{134} \times \mathrm{N}_{\mathrm{A}} \times 2=0.6\, \mathrm{N}_{\mathrm{A}} \,\mathrm{C}$ -atoms

$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \Rightarrow \frac{72}{180} \times \mathrm{N}_{\mathrm{A}} \times 6=2.4\, \mathrm{N}_{\mathrm{A}} \,\mathrm{C}$ -atoms

$\mathrm{C}_{5} \mathrm{H}_{10} \Rightarrow \frac{35}{70} \times \mathrm{N}_{\mathrm{A}} \times 5=2.5\,\mathrm{N}_{\mathrm{A}}\, \mathrm{C}$ -atoms

$ \text { Moles of } \mathrm{CaCO}_{3} \text { in } 10 \,\mathrm{g} =\frac{10}{100} $

$=0.1\, \mathrm{mol}=0.1 \,\mathrm{g} \,\text { atom }$

$=0.1 \times 6.02 \times 10^{23} \times 3 $

$=1.806 \times 10^{23}$

$\mathrm{H}_{2} \Rightarrow \frac{2 \mathrm{g}}{2} \times \mathrm{N}_{\mathrm{A}} \times 2=2 \mathrm{N}_{\mathrm{A}}$ atoms

$\mathrm{C} \Rightarrow 2 \times \mathrm{N}_{\mathrm{A}}$ atoms

$\mathrm{SO}_{2} \Rightarrow 3 \times \mathrm{N}_{\mathrm{A}} \times 3=9 \mathrm{N}_{\mathrm{A}}$ atoms

$1 \mathrm{mg}$ of $N_{2}=\frac{2 N \times 10^{-3}}{28}$ atoms

$1 \mathrm{mg}$ of $N a=\frac{N \times 10^{-3}}{23}$ atoms

$1 \mathrm{mL}=1 \mathrm{g} H_{2} O=\frac{3 N}{18}$ atoms

$8$ moles of oxygen is contained by $1$ mole of $Mg_3(PO_4)_2$.

$0.50$ moles of oxygen is contained by $\frac{{0.50 \times 1}}{8}$ $=0.0625$ moles of $Mg_3(PO_4)_2$.

So, your answer is $0.0625$ moles

Number of moles $=\frac{1000}{22.4}=44.6$

$\text { No. of atom }=\text { Mole } \times \mathrm{N}_{\mathrm{A}}=6\, \mathrm{N}_{\mathrm{A}}$

No. of $O-$atoms $=4$

$\therefore$ so, ratio of no. of atoms $=$ ratio of no. of moles

$\frac{12}{4}=\frac{6}{n_{o}}$

$n_{o}=2$

$0.1 \times 3=0.3\, \mathrm{mol}$ atom

$\mathrm{CO}_{2} \Rightarrow \frac{22}{44}=0.5\, \mathrm{mol}: 0.5 \times 3=1.5\,\mathrm{mol}$ atoms

$\mathrm{H}_{2} \mathrm{SO}_{4} \Rightarrow \frac{2.45}{98}=0.025\, \mathrm{mol}: 0.025 \times 7=0.175 \,\mathrm{mol\,atoms}$

$C_2H_5OH = 1\,mole, N_2O_5 = 1/2\,mole$

Moles of oxygen $=0.125 \times 2=0.25\, \mathrm{mol}$

$6.022 \times 10^{23}$ atom have $1 \;mol$ of $NH_3$ 

then $2.4 \times 10^{24}$ atom have $4\;mol$ of $NH_3$ 

In $SC{l_2}$ valency of sulphur = $2$ 

So equivalent mass of sulphur $ = \frac{{32}}{2} = 16$.

wt. of chlorine = $35.5 $

$\therefore $ wt. of metal $ = 74.5 - 35.5 = 39$

Equivalent weight of metal $ = \frac{{{\rm{weight of metal}}}}{{{\rm{weight of chlorine}}}} \times 35.5$

$ = \frac{{39}}{{35.5}} \times 35.5 = 39$

$\therefore $ $22.4\,L$ of gas has mass =$\frac{{7.5}}{{5.8}} \times 22.4 = 28.96$

So molecular weight = $29$

So, molecular formula of compound is $NO$

$\therefore $ $22.4\, L$ of gas at $S.T.P$. weight $ = 22.4 \times 1.16$ $ = 25.984 \approx 26$

This molecular weight indicates that given compound is ${C_2}{H_2}$.

$22.4\,gm$ of gas occupies $22.4\,L$ at $S.T.P.$

$\therefore $ $11.2\,gm$ of gas occupies $\frac{{22.4}}{{22.4}} \times 11.2 = 11.2\,L$.

Molecular weight of $\begin{array}{*{20}{c}}{COOH}\\{\mathop C\limits^| OOH}\end{array} \cdot 2{H_2}O = \frac{{126}}{2} = 63$.

Molecular weight of ${H_3}P{O_3}$$ = 3 + 31 + 48 = 82$

$\therefore $ Equivalent weight $ = \frac{{{\rm{Molecular weight}}}}{{{\rm{Basicity}}}} = \frac{{82}}{2}= 41.$

$194\,gm$ caffeine has = $\frac{{28.9}}{{100}} \times 194 = 56.06\,gm$

$\therefore $ No. of atoms in caffeine $ = \frac{{56.06}}{{14}} \approx 4$.

$6.023 \times {10^{23}}$atoms has mass =$10.86 \times {10^{ - 23}} \times 6.023 \times {10^{23}}$ = $65.40\,gm$

This is the atomic weight of $Zn.$

$\therefore $ $0.3$ mole ${(COOH)_2}\;.\;2{H_2}O$ has $96 \times 0.3 = 28.8\,gm$

$\therefore $ No. of gram atoms of oxygen $ = \frac{{28.8}}{{16}} = 1.8$.

wt. of metal $= 53\,gm$

wt. of oxygen $= 47\,gm$

Equivalent weight of oxygen $ = \frac{{{\rm{wt}}{\rm{. of  metal}}}}{{{\rm{wt}}{\rm{. of  oxygen}}}} \times 8$ $ = \frac{{53}}{{47}} \times 8 = 9.02$

Valency $ = \frac{{2 \times V.D}}{{E + 35.5}} = \frac{{2 \times 66}}{{9 + 35.5}} = \frac{{132}}{{44.5}} = 2.96 \approx 3$

$\therefore$ Atomic Weight = Equivalent Weight $\times$ Valancy $ = 9.02 \times 3 = 27.06$

So, equivalent weight of bromine $= 80\,gm$

$4\,gm$ of bromine combines with $1\,gm$ of $Ca$

$\therefore $ $80\,gm$ of bromine combines with = $\frac{1}{4} \times 80 = 20$.

$24\,g \,Mg $ evolves $22.4\,L$ ${H_2}$ at $STP$

$\therefore $ $12\,g$ $Mg$ evolves ${H_2}$ at $STP$ $\frac{{22.4}}{{24}} \times 12$ $=11.2\,L$ at $STP.$

$\therefore $ $22.4\,L$ of gas has mass $ = \frac{{4.4}}{{2.24}} \times 22.4 = 44$

So given gas is $C{O_2}$ because $C{O_2}$ has molecular mass $=44.$

$\therefore $ $8.96\,L$ of a gas at $STP$ has no. of molecules $ = \frac{{6.02 \times {{10}^{23}} \times 8.96}}{{22.4}}$

$ = 2.408 \times {10^{23}}$$ = 24.08 \times {10^{22}}$.

Vapour density of metal chloride $= 59.25$

$\therefore $ molecular weight of metal chloride

$ = 2 \times V.D = 2 \times 59.25 = 118.5$

$\therefore $ valency of metal $ = \frac{{{\rm{molecular \,weight \,of \,metal \,chloride}}}}{{{\rm{equivalnet\, weight\, of \,metal }} + {\rm{35}}{\rm{.5}}}}$

Valency of metal $ = \frac{{118.5}}{{9 + 35.5}} = \frac{{118.5}}{{44.5}} = 2.66$

Therefore atomic weight of the metal = equivalent weight $ \times $ valency $ = 9 \times 2.66 = 23.9$

$\therefore $ Atomic weight of metal $ = 37.2 \times 2 = 74.4$

$\therefore $ Formula of chloride $ = MC{l_2}$

Hence, molecular weight of chloride

$(MC{l_2}) = 74.4 + 2 \times 35.5 = 145.4$

$4.4\,g$ $C{O_2}$ occupies $ = \frac{{22.4}}{{44}} \times 4.4$= $2.24\,L$.

$ = 8.9 \times 3 = 26.7$ $\left( {{\rm{Valency}} = \frac{{{\rm{26}}{\rm{.89}}}}{{{\rm{8}}{\rm{.9}}}} \approx 3} \right)$.

Twice $16 \times 2 = 32$

Change in $0.5$ per atom $ = 7 - 6 = 1$

$\therefore $ Equivalent weight of $KMn{O_4}$ $ = \frac{{{\text{Molecular weight of }}KMn{O_4}}}{{{\text{Change of 0}}{\text{.5 per atom}}}}$$ = \frac{M}{1} = M$.

$22400 = 44$

$1120 = \frac {1120}{22400} \times 44 = 2.2\ g$

$=\frac{10 \times 0.2+11 \times 0.8}{0.2+0.8}=10.8$

since mixture is equi-molecular,

$0.1 \,\mathrm{mol}$  $0.05\,{\text{mol}}\,{\text{C}}{{\text{H}}_4}\xrightarrow{{ \times 16}}0.8\,{\mkern 1mu} {\text{gm}}$

${\text{0}}{\text{.1}}\,{\text{mol}}\,0.05\,{\text{mol}}\,{{\text{C}}_2}{{\text{H}}_6}\xrightarrow{{ \times 30}}1.5\,{\mkern 1mu} {\text{gm}}$

$\therefore $ Mass of mixture $=2.3\, \mathrm{gm}$

$12 / 16=X / 100$

$X=75$

So, the correct option is $C$

$=0.988 \times 12+0.0118 \times 13+0.0002 \times 14$

$=12.0122$

$(II) $ $0.5\, \mathrm{g}$ atom of oxygen $=8\, \mathrm{g}$

$(III)$ $\frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \times 32=16 \,\mathrm{g}\, \mathrm{O}_{2}$

$(IV)$ $\frac{5.6}{22.4} \times 44\, \mathrm{g} \,\mathrm{CO}_{2}=11 \,\mathrm{g}\, \mathrm{CO}_{2}$

$\mathrm{E}_{\mathrm{x}_{2} \mathrm{y}_{3}}=\frac{\mathrm{x}}{3}+\frac{\mathrm{Y}}{2}=18,2 \mathrm{X}+3 \mathrm{Y}=108$

On solving,

$\mathrm{X}=30, \mathrm{Y}=16$

$1$ $mol$ $\mathrm{XY}_{2}=\frac{5}{0.05}=100$

$\mathrm{X}+2 \mathrm{Y}=100$

$3.01 \times 10^{23}$ molecule $\mathrm{X}_{2} \mathrm{Y}_{3}=85\, \mathrm{gm}$

$1\, \mathrm{mol}\, \mathrm{X}_{2} \mathrm{Y}_{3}=\mathrm{N}_{\mathrm{A}}$ molecule $=170\, \mathrm{gm}$

$2 \mathrm{X}+3 \mathrm{Y}=170.........(ii)$

On solving.

$\mathrm{X}=40 . \mathrm{Y}=30$

$\frac{8}{m}=\frac{5.6}{22.4}$

$\therefore $ $\mathrm{m}=32 ; \text { vapour density }=16$

$=\frac{78.4 \times 100}{0.5}$

$=1.568 \times 10^{4}$

$H_3PO_4 + Ca(OH)_2 \to  CaHPO_4 + 2H_2O$

In this reaction

$H_{3} P O_{4} \rightarrow H P O_{4}^{2-}+2 H^{+}$

Orthophosphoric acid leaves $2$ hydrogen ions.

Hence, its valency factor is $=2$

Equivalent weight $=\frac{\text { Molecular Weight }}{2}$

Molecular weight $=1 \cdot 3+31+4 \cdot 16=98$

Therefore,

The equivalent weight is $=\frac{98}{2}=49$

                              $=\frac {35\times 25 +37\times 75}{25+75}$

$H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$ 

In this reaction

$H_{3} P O_{4} \rightarrow H P O_{4}^{2-}+2 H^{+}$

Orthophosphoric acid leaves $2$ hydrogen ions.

Hence, its valency factor is $=2$

Equivalent weight $=\frac{\text { Molecular Weight }}{2}$

Molecular weight $=1 \cdot 3+31+4 \cdot 16=98$

Therefore,

The equivalent weight is $=\frac{98}{2}=49$

$=\frac{\text { weight of metal }}{\text { weight of chlorine }} \times$ eq. wt. of $\mathrm{Cl}$

$\mathrm{E}_{\mathrm{M}}=\frac{(74.5-35.5)}{35.5} \times 35.5=39.0$

Eq. wt. of $\mathrm{KIO}_{3}=\frac{214}{4}=53.5$

$\mathrm{Ew}=\frac{\mathrm{M}}{3 / 2}=\frac{2 \mathrm{M}}{3}$

$\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow \mathrm{CaHPO}_{4}+2 \mathrm{H}_{2} \mathrm{O}$

In this reaction

$H_{3} P O_{4} \rightarrow H P O_{4}^{-2}+2 H^{+}$

Orthophosphoric acid leaves $2$ hydrogen ions.

Hence, its valency factor is $=2$

Equivalent weight $=\frac{\text { Molecular Weight }}{2}$

Molecular weight $=1 \cdot 3+31+4 \cdot 16=98$

Therefore,

The equivalent weight is $=\frac{98}{2}=49$

$(II)$ $0.5$ $g$ atoms of oxygen $=8\, \mathrm{g}$

$(III)$ $\frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \times 32=16 \,\mathrm{g}\, \mathrm{O}_{2}$

$(IV)$ $\frac{5.6}{22.4} \times 44\, \mathrm{g} \,\mathrm{CO}_{2}=11 \,\mathrm{g}\, \mathrm{CO}_{2}$

$Fe$ present in $100\,mg$ of ${(CHCOO)_2}Fe$

$ = \frac{{56}}{{170}} \times 100\,mg = 32.9\,mg$

This is present in $400\,mg$ of capsule

$\%$ of $Fe$ in capsule $ = \frac{{32.9}}{{400}} \times 100 = 8.2$.

$\therefore $ $100\,gm$ of $NaOH$ contains $\frac{{16}}{{40}} \times 100$ = $40\%$ oxygen.

$60\,gm$ of urea contains $28\,gm$ of nitrogen

$\therefore $ $100\,gm$ of urea contains $\frac{{28}}{{60}} \times 100 = 46.66$.

For example, $CH _2 O$ and $C _6 H _{12} O _6$ have the same empirical formula but different molecular formula, they have different molecular weights.

So, Molecular formula $ = {C_2}{H_4}{O_2}$

So, Empirical formula $ = C{H_2}O$ (Simplest formula).

Ratio of $C, H$ and $O$ $ = 1:2:1$

In acetic acid $\mathop {C{H_3} - \mathop {{\text{ }}C}\limits_{||}  - O - H}\limits_{O\,\,\,{\kern 1pt} {\kern 1pt} } $

Ratio of $C, H$ and $O$ $1:2:1$.

$\%$ of ${P_2}{O_5} = \frac{{{\rm{wt}}{\rm{. of }}{P_2}{O_5}}}{{{\rm{wt \,of \,salt}}}} \times 100$ $ = \frac{{142}}{{264}} \times 100 = 53.78\% $.

$\%$ of ${H_2}O = \frac{{3 \times 18}}{{284}} \times 100 = 19\% $.

simplest formula = $X_2Y$

$3=\frac{1000 \,\mathrm{M}}{1000 \times 1.1-\mathrm{M} \times 40}$

$=4.34$

moles of nitrogen $=30.5 / 14$

$=2.18$

ratio of moles of nitrogen and oxygen $=2.18: 4.34$

$=1: 2$

empirical formula $ = NO_2$

molecular formula $=(\mathrm{NO}_ 2) \mathrm{n}$

where $\mathrm{n}=$ molecular mass/ empirical mass

$=92 / 46$

$=2$

therefore the formula of the compound is $\mathrm{N}_ 2 \mathrm{O} _4 .$

$C=70.6 g=\frac{70.6}{12}$ moles $=5.88$ moles

$H=4.2 g=\frac{4.2}{1}$ moles $=4.2$ moles

$N=11.8 g=\frac{11.8}{14}$ moles $=0.84$ moles

$O=13.4 g=\frac{13.4}{16}$ moles $=0.84$ moles

$C: H: N: O=5.88: 4.2: 0.84: 0.84$

$C: H: N: O=7: 5: 1: 1$

Therefore, empirical formula of Kelvar is $C_{7} H_{5} N O$

$ = \frac{{1.15}}{{23}}$  : $\frac{{3.01 \times {{10}^{22}}}}{{6.02 \times {{10}^{23}}}}$  : $0.1$

$=$ $0.05$     :    $0.05$     :      $0.1$ 

$=$ $1$        :      $1$        :       $2$

$E.F.=NaCO_2$

$N$                 $2.8$        $\frac {2.8}{14}=0.2=1$

$O$                $3.2\,g$      $\frac {3.2}{16}=0.2=1$

Simplest formula of the compound $=NO$

$C=3 / 12=\frac{1}{4}=1$

$\mathrm{O}=4 / 16=\frac{1}{4}=1$

$\mathrm{N}=7 / 14=\frac{1}{2}=2$

Empirical formula $=$ $NO_2$ and molecular formula $=N_2O_4$ 

$C=70.6 g=\frac{70.6}{12}$ moles  $=5.88$ moles

$H=4.2 g=\frac{4.2}{1}$ moles $=4.2$ moles

$N=11.8 g=\frac{11.8}{14}$ moles $=0.84$ moles

$O=13.4 g=\frac{13.4}{16}$ moles $=0.84$ moles

$C: H: N: O=5.88: 4.2: 0.84: 0.84$

$C: H: N: O=7: 5: 1: 1$

Therefore, empirical formula of Kelvar is $C_{7} H_{5} N O$

$\therefore n = \frac{{78}}{{13}} = 6$

$13.5 \,\mathrm{g} \text { contains } \frac{9}{12} \,\mathrm{mole}$

$108 \,\mathrm{g}$ contains $\frac{9 \times 108}{12 \times 13.5}=6$ mole carbon

i.e., $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{N}_{2}$

Increase in volume $ = 150\,ml - 100\,ml = 50\,ml$ increase.

$2 \times 276\,gm$ of $A{g_2}C{O_3}$ gives $4 \times 108\,gm$

$\therefore $ $1\,gm$ of $A{g_2}C{O_3}$ gives $ = \frac{{4 \times 108}}{{2 \times 276}}$

$\therefore $ $2.76\,gm$ of  $2A{g_2}C{O_3}$ gives $\frac{{4 \times 108 \times 2.76}}{{2 \times 276}} = 2.16\,gm$

According to equation $\frac{3}{2}$ mole of ${O_2}$ combines with $2$ mole $Al$.

$2$ mole $Al$ $= 54\,gm$

mole wt of ${H_3}P{O_4}= 3 + 31 + 64 = 98$

$\frac{{98}}{2} = 49$

Atomic wt. of $BaC{O_3}$ =$137 + 12 + 16 \times 3= 197$

No. of mole $ = \frac{{{\rm{wt}}{\rm{.\, of \,substance}}}}{{{\rm{mol wt}}{\rm{.}}}}$

$1$ mole of $Ba{(OH)_2}$ gives $1$ mole of $BaC{O_3}$

$\therefore $ $205$ mole of $Ba{(OH)_2}$ will give $0 .205$ mole of $BaC{O_3}$

$\therefore $ wt. of $0.205$ mole of $BaC{O_3}$ will be $.205 \times 197 = 40.385\,gm \approx \,40.5\,gm$

$2$ mole $ \to $ $1$ mole [$100 \times 1 =100$ millimole]

$\therefore $ $100$ miliimole $ \to $ $50$ millimole ${H_2}S$ required

$CuS{O_4} \equiv C{u^{ + 2}} + \mathop {{S^{2 - }}}\limits_{({H_2}S)} \to CuS$

$1$ mole $ \to $ $1$ mole [$100 \times 1 =100$ millimole]

$\therefore $ $100$ millimole $ \to $ $100$ millimole ${H_2}S$ required

Ratio $\frac{{50}}{{100}} = \frac{1}{2}$.

$n = \frac{{12\,gm}}{{24\,gm}} = \frac{1}{2}$ mole of ${H_2}$

$0.5$ mole of oxygen react with $1$ mole of $Mg$

$1.5$ mole of oxygen react with $\frac{{1.5}}{{0.5}} = 3$

mole $24 \times 3 = 72\,gm$.

$100\, g$ $CaC{O_3}$ with $2 \,N\, HCl$ gives $44\, g$ $C{O_2}$

$100\, g$ $CaC{O_3}$ with $1\,N \,HCl$ gives $22\, g$ $C{O_2}$

at $t = 0$      $56\,gm$    $8\,gm$      $0\,gm$

                   $=$ $2\,mole$    $4\,mole$   $0\,mole$

at equilibrium  $2 - 1$       $4 - 3$         $34\,gm$

                  $=$ $1\,mole$   $=$ $1\,mole$  $=$ $2\,mole$

According to eq. $(i)$ $2$ $mole$ of ammonia are present $\&$ to produce $2$ $mole$ of $N{H_3}$, we need $1$ mole of ${N_2}$ and $3$ $mole$ of ${H_2}$ hence $2 - 1 = 1$ $mole$ of ${N_2}$ and $4 - 3 = 1$ mole of ${H_2}$ are present at equilibrium in vessel.

$0.02 = \frac{{0.02 \times 22.4}}{2} = 0.224$.

Equivalent weight $=\frac{M}{6}$.

$MnO _4^{-}+ e ^{-} \rightarrow MnO _4^{2-}$

The number of electrons gained per molecule of $KMnO _4$ is $1$.

So, the equivalent mass is $158$.

as transfer of $5{e^ - }$ takes place when $KMn{O_4}$ acts as oxidant in acidic medium.

$2KMn{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O$$ + 5O$

 Equivalent weight $= \frac{{158}}{5}$ $= 31.6$ 

No. of electrons lossed $= 12 -6 = 6$ 

Equivalent weight $=\frac{M}{6} = \frac{{294}}{6} = 49$ .

$2KMn{O_4} + {H_2}O\,\xrightarrow{{ + 3{e^ - }}}\,2KOH + 2Mn{O_2} + 3[O]$

 Equivalent weight = $\frac{M}{3}$

change in vol. $=+75\, \mathrm{mL}$

$=\frac{1}{2} \times 0.01=0.005$

$ \text { vol. at } \,\mathrm{N} \cdot \mathrm{T} \cdot \mathrm{P} =0.005 \times 22400 $

$=112\, \mathrm{ml}$

Acc. to $L.R.$                            $0.1$

$\mathrm{n}_{\mathrm{Cl}_{2}}=\frac{11.2}{22.4}=0.5\, \mathrm{mole}$

                          ${H_{2(g)}} + C{l_{2(g)}} \to 2HC{l_{(g)}}$

initially               $1$ mol       $0.5$ mole      $0$

after reaction   $(1-0.5)$                $0.5 \times 2$

                          $=0.5$ mole $0$             $=1$ mole

$1$ $vol.$ $\mathrm{C}_{2} \mathrm{H}_{2}=\frac{5}{2}\, \mathrm{vol} .$ of $\mathrm{O}_{2}$

$2 \,\mathrm{vol} . \mathrm{C}_{2} \mathrm{H}_{2}=5 \,\mathrm{vol}$ of $\mathrm{O}_{2}$

$1\, \mathrm{L} \quad 3 \,\mathrm{L}$

$10\, \mathrm{L} \quad 30 \,\mathrm{L}$

$\mathrm{V}_{\mathrm{air}}=\frac{30 \times 100}{20}=150\, \mathrm{L}$

$(L. R.)$

$72 \,g\, Mg$ will produce $\mathrm{Mg}_{3} \mathrm{N}_{2}=100 \,\mathrm{g}$

$48\, \mathrm{g} \,\mathrm{Mg} \text { will produce } \mathrm{Mg}_{3} \mathrm{N}_{2} =\frac{100}{72} \times 48 $

$=\frac{200}{3} \,\mathrm{g} $

$[\mathrm{L} \cdot \mathrm{R.}]$

$2 \,\mathrm{mol}\, \mathrm{H}_{2}$ reacts with $\mathrm{O}_{2}=1\, \mathrm{mol}$

$\therefore $ $3\, \mathrm{mol}\, \mathrm{H}_{2}$ will react with $\mathrm{O}_{2}=\frac{1}{2} \times 3=1.5 \,\mathrm{mol}$

remaining amount of $\mathrm{O}_{2}=0.5\, \mathrm{mol}=16\, \mathrm{g}$

$2 N_{2}+3 O_{2} \rightarrow 2 N_{2} O_{3}$

here we see that $2$ moles of $\mathrm{N} _2$ combine with $3$ moles of oxygen and give $2$ moles of $\mathrm{N} _2 \mathrm{O}_ 3$

$a/c$ to question,

$9$ mol of $O_2$ and $14$ mol of $n 2$ here allowed to react.

when $3 \mathrm{mol}$ of $O_2$ remain unreacted till then we have to find moles of $N_2O_3$ is formed.

at initial time, mole of $\mathrm{N}_ 2 \mathrm{O}_ 3=0$

mole of $\mathrm{O} _2=9$

mole of $\mathrm{N}_ 2=14$

after some time,

mole of $\mathrm{O} _2=9-3 x$

mole of $\mathrm{N}_ 2=14-2 \mathrm{x}$

mole of $\mathrm{N} _2 \mathrm{O}_ 3=2 \mathrm{x}$

$a/c$ to question, $9-3 x=3$

or, $x=2$

then, number of mole of $\mathrm{N} _2 \mathrm{O}_ 3$ is formed $=2 \times 2=4 \mathrm{mol} .$

$2 C_{4} H_{10}+13 O_{2} \rightarrow 8 C O_{2}+10 H_{2} O$

This means $2$ moles of this mixture requires $13$ moles of oxygen.

Molecular mass of butane $=58 g / m o l$

Molecular mass of oxygen $=32 g / m o l$

Moles of butane in $10 k g=\frac{10 \times 1000}{58}=\frac{10000\times 13}{58\times 2}$ moles of $O_{2}=\frac{130000\times 32}{116} g$

$O_{2}=\frac{130 \times 32}{116}\, k g\, O_{2}$

Therefore, $35.862 \,\mathrm{kg}$ oxygen is used.

$\mathrm{nC}_{2} \mathrm{H}_{2}=\frac{7.8}{26}=0.3, \mathrm{nO}_{2}\, \mathrm{req}=0.3 \times \frac{5}{2}=0.75$

$\mathrm{V}_{\mathrm{O}_{2}}=0.75 \times 22.4=16.8\,\mathrm{L}$

       $(L.R.)$

and left air $=150 \times \frac {79}{100}$

$=270,000\, \mathrm{g} $

moles of $\mathrm{CaCO}_{3}=\frac{270,000}{100}=2700\, \mathrm{mol}$

$\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaO}+\mathrm{CO}_{2}$

$1$ mole               $1$ mole    $1$ mole

$2700$ mole $=2700$ mole $\mathrm{CaO}$

$\text { weight of } \mathrm{CaO} =\text { mole } \times \mathrm{M}_{\mathrm{w}} $

$=2700 \times \frac{56}{1000} \,\mathrm{kg}=151.2 \,\mathrm{kg} $

$58 \,\mathrm{g}\, \mathrm{C}_{4} \,\mathrm{H}_{10}$ requires $\mathrm{O}_{2}=6.5 \times 32\, \mathrm{g}$

$1000 \,\mathrm{g} \,\mathrm{C}_{4} \mathrm{H}_{10} \text { requires } \mathrm{O}_{2} =\frac{6.5 \times 32 \times 1000}{58}$

$=3.586\, \mathrm{kg}$

$100\, \mathrm{kg} \longrightarrow 44\,\mathrm{kg}$

$\frac{80}{100} \times 10\, \mathrm{kg} \longrightarrow \frac{44}{100} \times \frac{80}{100} \times 10=3.52 \,\mathrm{kg}$

$1$ mole             $\longrightarrow $ $2$ mole

$30 \,\mathrm{g} \quad \longrightarrow 2 \times 22.4\, \mathrm{L}$ at $STP$

$3\, \mathrm{g} \quad \longrightarrow \frac{2 \times 22.4}{30} \times 3=4.48 \,\mathrm{L}$

Hence for $6$ molecules of Oxygen, we need $2$ molecules $CS _2$

Hence option $D$ is the answer.

$X$ is $A B_{4}$

weight of $1$ mol of $X$ is weight of $1$ mol of $AB$ $_{4}$

weight of $1 \mathrm{mol}$ of $\mathrm{AB}_{4}=1 \times 12.01+4(35.5)$

$=154 \mathrm{g}$

weight of $1$ mol $\mathrm{X}=154 \mathrm{g}$

$2 \mathrm{N}^{5+}+14 \mathrm{e}^{-} \longrightarrow 2 \mathrm{N}^{2-}$

$\mathrm{N}^{5+}+7 \mathrm{e} \longrightarrow \frac{1}{2}\left[\mathrm{N}_{2} \mathrm{H}_{4}\right]$

$\mathrm{KClO}_{3} \Rightarrow 30.62 \times \frac{80}{100}=24.5\, \mathrm{g}$

no. of moles $=\frac{24.5}{122.5}=0.2\, \mathrm{mol}$

$2\, \mathrm{mol}\, \mathrm{KClO}_{3}$ produces $\mathrm{O}_{2}=3\, \mathrm{mol}$

$0.2\, \mathrm{mol} \,\mathrm{KClO}_{3}$ will produce $\mathrm{O}_{2}=\frac{3}{2} \times 0.2$

${=0.3 \,\mathrm{mol}}$

${=0.3 \times 22.4=6.72\, \mathrm{L}}$

$3 \mathrm{O}_{2} \rightarrow 2 \mathrm{O}_{3}$

$\therefore \text { Mole of Ozone formed } =\frac{2}{3} \times \frac{15}{100} \times 3$

$=0.3\, \mathrm{mole}$

$\therefore $ Mass of ozone formed $=0.3 \times 48\, \mathrm{g}=14.4\, \mathrm{g}$

$NO_3^ - = \frac{{0.1}}{2} = 0.05\,M$.