Atomic, Molecular and Equivalent masses Questions in English

(D) The oxidation and reduction half-reactions are:
Oxidation: $N_2H_4 \rightarrow N_2 + 4H^+ + 4e^-$
Reduction: $IO_3^- + 6H^+ + Cl^- + 4e^- \rightarrow ICl + 3H_2O$
The equivalent mass is the molecular mass divided by the number of electrons gained or lost per molecule.
For $N_2H_4$: Molecular mass = $32 \ g/mol$. Number of electrons lost = $4$.
Equivalent mass of $N_2H_4 = \frac{32}{4} = 8$.
For $KIO_3$: Molecular mass = $214 \ g/mol$. Number of electrons gained = $4$.
Equivalent mass of $KIO_3 = \frac{214}{4} = 53.5$.

(A) Elements have non-integral atomic masses because they exist as a mixture of isotopes with different masses,which occur in nature in specific proportions.
For example,Chlorine $(Cl)$ has two isotopes with mass numbers $35$ and $37$,present in a ratio of $3:1$.
Therefore,the average atomic mass is calculated as: $\frac{3 \times 35 + 1 \times 37}{4} = 35.5 \ u$.
Thus,the presence of isotopes with different masses leads to non-integral atomic masses.

(C) Atomic weight (or relative atomic mass) is a dimensionless quantity defined as the ratio of the average mass of atoms of an element to $1/12$th of the mass of a carbon-$12$ atom.
However,when expressed in terms of mass,the unit used is the atomic mass unit,denoted as $amu$ or $u$ (unified mass).

(B) The atomic weight of an element is defined as the mass of one atom of that element expressed in atomic mass units $(amu)$.
By definition,$1 \ amu = \frac{1}{12}$ of the mass of one $C-12$ atom.
Therefore,if the atomic weight is $x$,the mass of one atom is $x \ amu$.

(C) In the current system,the atomic mass of carbon is $12$ and the atomic mass of oxygen is $16.$
To find the new atomic weight of carbon $(X)$ when oxygen is assigned a value of $100,$ we use the ratio:
$\frac{\text{Mass of Carbon}}{\text{Mass of Oxygen}} = \frac{12}{16} = 0.75$
Setting up the proportion:
$\frac{X}{100} = 0.75$
$X = 0.75 \times 100 = 75$
Therefore,the atomic weight of carbon would be $75.$
The correct option is $C$.

(D) The non-integral atomic masses of elements arise due to the existence of isotopes of that element,which possess different atomic masses.
These isotopes occur in nature in varying natural abundances.
Therefore,the observed atomic mass is the weighted average of the masses of these isotopes.
Thus,both $A$ and $C$ are correct.

(A) The average atomic mass of carbon is calculated as follows:
Average atomic mass of $C = (0.988 \times 12) + (0.0118 \times 13.1) + (0.0002 \times 14.1)$
$= 11.856 + 0.15458 + 0.00282 = 12.0134 \ u$
The molecular mass of $CH_4$ is given by the sum of the average atomic mass of carbon and $4$ times the atomic mass of hydrogen:
Molecular mass of $CH_4 = 12.0134 + (4 \times 1.008)$
$= 12.0134 + 4.032 = 16.0454 \ u$
Rounding to the nearest provided option,the correct value is $16.0452 \ u$.

(B) $CO_2$ molecule consists of one carbon atom and two oxygen atoms.
Let the isotopes of carbon be $C_1, C_2$ (total $2$) and isotopes of oxygen be $O_1, O_2, O_3$ (total $3$).
The structure of $CO_2$ is $O-C-O$.
The number of ways to choose the carbon atom is $2$.
The number of ways to choose the two oxygen atoms (where order does not matter as the two positions are equivalent) is given by the combination with repetition formula: $\frac{n(n+1)}{2}$,where $n=3$.
Number of ways to choose oxygen atoms = $\frac{3(3+1)}{2} = \frac{12}{2} = 6$.
Total number of $CO_2$ molecules = (Ways to choose $C$) $\times$ (Ways to choose $O$ pair) = $2 \times 6 = 12$.

(B) The equivalent weight $E$ is defined as the molar mass divided by the total valence change or total charge.
For $X_2Y$,the valency of $Y$ is $2$ and $X$ is $1$. The equivalent weight is $E_{X_2Y} = \frac{M_{X_2Y}}{n-factor}$.
Assuming $X$ has valency $1$ and $Y$ has valency $2$,$E_{X_2Y} = \frac{2X + Y}{2} = 38 \Rightarrow 2X + Y = 76$ (Equation $1$).
For $X_2Y_3$,the valency of $Y$ is $2$ and $X$ is $3$. The equivalent weight is $E_{X_2Y_3} = \frac{2X + 3Y}{6} = 18 \Rightarrow 2X + 3Y = 108$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(2X + 3Y) - (2X + Y) = 108 - 76$
$2Y = 32 \Rightarrow Y = 16$.
Substituting $Y = 16$ into Equation $1$:
$2X + 16 = 76$ $\Rightarrow 2X = 60$ $\Rightarrow X = 30$.
Thus,the atomic masses are $X = 30$ and $Y = 16$.

(C) In the reaction $Al(OH)_3 + 2H_3PO_3 \to Al(OH)(H_2PO_3)_2 + 2H_2O$:
For $Al(OH)_3$: Two $OH^-$ groups are replaced by $H_2PO_3^-$ groups. Thus,the acidity (n-factor) is $2$.
Equivalent weight $E_{Al(OH)_3} = \frac{M_1}{2}$.
For $H_3PO_3$: Each $H_3PO_3$ molecule loses one $H^+$ ion to form $H_2PO_3^-$. Thus,the basicity (n-factor) is $1$.
Equivalent weight $E_{H_3PO_3} = \frac{M_2}{1}$.

(C) For $XY_{2}$:
$0.05 \ mol = 5 \ g$
$1 \ mol = \frac{5}{0.05} = 100 \ g/mol$
So,$X + 2Y = 100$ --- $(i)$
For $X_{2}Y_{3}$:
$3.011 \times 10^{23}$ molecules is $0.5 \ mol$ (since $1 \ mol = 6.022 \times 10^{23}$ molecules).
$0.5 \ mol = 85 \ g$
$1 \ mol = 85 \times 2 = 170 \ g/mol$
So,$2X + 3Y = 170$ --- $(ii)$
Multiply equation $(i)$ by $2$:
$2X + 4Y = 200$ --- $(iii)$
Subtract $(ii)$ from $(iii)$:
$(2X + 4Y) - (2X + 3Y) = 200 - 170$
$Y = 30$
Substitute $Y = 30$ into $(i)$:
$X + 2(30) = 100$
$X + 60 = 100$
$X = 40$
Therefore,the atomic masses of $X$ and $Y$ are $40$ and $30$ respectively.

(C) The average atomic weight is calculated by the weighted average of the isotopes:
$\text{Average atomic weight} = \frac{(\text{Mass of isotope } 1 \times \text{Percentage } 1) + (\text{Mass of isotope } 2 \times \text{Percentage } 2)}{100}$
$\text{Average atomic weight} = \frac{(10 \times 20) + (11 \times 80)}{100}$
$\text{Average atomic weight} = \frac{200 + 880}{100} = \frac{1080}{100} = 10.80$

(A) The molar mass of $C_{60}H_{122} = (60 \times 12) + (122 \times 1) = 720 + 122 = 842 \ g/mol$.
The weight of one molecule is calculated by dividing the molar mass by Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$.
Weight of one molecule $= \frac{842}{6.022 \times 10^{23}} \approx 1.398 \times 10^{-21} \ g \approx 1.4 \times 10^{-21} \ g$.

(C) The average atomic mass is calculated using the formula: $\text{Average atomic mass} = \frac{m_1x_1 + m_2x_2}{x_1 + x_2}$
Given: $m_1 = 35, x_1 = 25\%$ and $m_2 = 37, x_2 = 75\%$.
Substituting the values: $\text{Average atomic mass} = \frac{35 \times 25 + 37 \times 75}{25 + 75}$
$= \frac{875 + 2775}{100}$
$= \frac{3650}{100} = 36.5$.

(B) The given reaction is $H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$.
In this reaction,$H_3PO_4$ loses $2$ hydrogen ions to form $HPO_4^{2-}$.
Thus,the valency factor ($n$-factor) of $H_3PO_4$ is $2$.
The molecular weight of $H_3PO_4 = (3 \times 1) + 31 + (4 \times 16) = 3 + 31 + 64 = 98$.
The equivalent weight is calculated as $\frac{\text{Molecular Weight}}{n\text{-factor}} = \frac{98}{2} = 49$.

(C) Given: Mass of metallic chloride = $74.5 \ g$,Mass of chlorine = $35.5 \ g$.
Mass of metal = $74.5 \ g - 35.5 \ g = 39.0 \ g$.
Equivalent weight of chlorine = $35.5$.
Using the formula: $\frac{\text{Mass of metal}}{\text{Equivalent weight of metal}} = \frac{\text{Mass of chlorine}}{\text{Equivalent weight of chlorine}}$.
$\frac{39.0}{E_M} = \frac{35.5}{35.5}$.
$E_M = 39.0$.

(A) Milli-equivalents (Meq.) of $Ca(OH)_2 = \frac{w}{E} \times 1000$
$= \frac{74}{74 / 2} \times 1000 = 2000$
$\left( \text{Since } E_{Ca(OH)_2} = \frac{74}{2} = 37 \right)$
$(b)$ Meq. of $NaOH = \frac{w}{E} \times 1000$
$= \frac{20}{40} \times 1000 = 500$
$\left( \text{Since } E_{NaOH} = 40 \right)$

(C) Molar mass is the mass of $1 \, mol$ of a substance.
For $AB_2$: Molar mass $= \frac{125 \times 10^{-3} \, kg}{5 \, mol} = 25 \times 10^{-3} \, kg \, mol^{-1}$.
So,$M_A + 2M_B = 25 \times 10^{-3} \quad (i)$.
For $A_2B_2$: Molar mass $= \frac{300 \times 10^{-3} \, kg}{10 \, mol} = 30 \times 10^{-3} \, kg \, mol^{-1}$.
So,$2M_A + 2M_B = 30 \times 10^{-3} \quad (ii)$.
Subtracting equation $(i)$ from $(ii)$:
$(2M_A + 2M_B) - (M_A + 2M_B) = (30 - 25) \times 10^{-3}$.
$M_A = 5 \times 10^{-3} \, kg \, mol^{-1}$.
Substituting $M_A$ in equation $(i)$:
$5 \times 10^{-3} + 2M_B = 25 \times 10^{-3}$.
$2M_B = 20 \times 10^{-3}$.
$M_B = 10 \times 10^{-3} \, kg \, mol^{-1}$.

(B) The molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \, g/mol$.
Moles of $CaCO_3$ in $10 \, g = \frac{10 \, g}{100 \, g/mol} = 0.1 \, mol$.
Since $1 \, \text{mole}$ of $CaCO_3$ contains $1 \, \text{mole}$ of $Ca$ atoms (or $1 \, \text{g-atom}$ of $Ca$),$0.1 \, \text{mole}$ of $CaCO_3$ contains $0.1 \, \text{g-atom}$ of $Ca$.

(A) At $STP$,$1 \ mol$ of $O_2$ occupies $22400 \ mL$ and provides $4 \ equivalents$ of oxygen (since $O_2 + 4e^- \rightarrow 2O^{2-}$).
$22400 \ mL$ of $O_2$ corresponds to $4 \ equivalents$ of oxygen.
$46.6 \ mL$ of $O_2$ corresponds to $\frac{4}{22400} \times 46.6 \approx 0.00832 \ equivalents$ of oxygen.
According to the law of equivalence,the number of equivalents of metal equals the number of equivalents of oxygen.
$\frac{\text{Weight of metal}}{\text{Equivalent weight of metal}} = 0.00832$
$\frac{0.1}{E} = 0.00832$
$E = \frac{0.1}{0.00832} \approx 12.02 \approx 12$.

(A) The molecular weight $(Mw)$ of $H_3PO_4$ is calculated as: $(3 \times 1) + 31 + (4 \times 16) = 3 + 31 + 64 = 98 \ g/mol$.
In the reaction $Ca(OH)_2 + H_3PO_4 \to CaHPO_4 + 2H_2O$,two hydrogen atoms of $H_3PO_4$ are replaced by one $Ca^{2+}$ ion.
Therefore,the $n$-factor for $H_3PO_4$ is $2$.
The equivalent weight $(Ew)$ is given by the formula: $Ew = \frac{Mw}{n-factor} = \frac{98}{2} = 49$.

(B) The reaction is $I_2 + 2S_2O_3^{2-} \longrightarrow 2I^{-} + S_4O_6^{2-}$.
In this reaction,the oxidation state of $I$ changes from $0$ in $I_2$ to $-1$ in $I^-$.
Total change in oxidation state for one molecule of $I_2$ is $2 \times |0 - (-1)| = 2$.
Thus,the valence factor $(v.f.)$ is $2$.
The molar mass of $I_2$ is $2 \times 127 = 254 \ g/mol$.
The equivalent mass is calculated as $\frac{\text{Molar mass}}{v.f.} = \frac{254}{2} = 127$.

(D) In the given reaction,$NaOH + H_3PO_4 \to NaH_2PO_4 + H_2O$,$H_3PO_4$ reacts with $NaOH$ to form $NaH_2PO_4$.
This indicates that $H_3PO_4$ loses only one $H^+$ ion.
Therefore,the basicity (n-factor) of $H_3PO_4$ in this reaction is $1$.
Equivalent weight = $\frac{\text{Molecular weight}}{\text{n-factor}}$.
Molecular weight of $H_3PO_4 = (3 \times 1) + 31 + (4 \times 16) = 98 \ g/mol$.
Equivalent weight = $\frac{98}{1} = 98$.

(C) Since noble gases are chemically inert,the concept of equivalent mass is not applicable to them.
Noble gases exist as monoatomic molecules,therefore their atomic mass is equal to their molecular mass.
The atomic mass of noble gases can be determined from their vapor density,which is easily measurable.
The relationship is given by: $Atomic \ mass = 2 \times Vapor \ density = Molecular \ mass$.

(D) Given: Equivalent mass of metal $= 29.73$.
Vapour density of metal chloride $= 130.4$.
Molecular weight of metal chloride $= 2 \times \text{Vapour density} = 2 \times 130.4 = 260.8$.
Let the valency of the metal be $n$. The formula of the metal chloride is $MCl_n$.
Molecular weight $= \text{Atomic weight of metal} + n \times 35.5$.
Since $\text{Atomic weight} = n \times \text{Equivalent mass}$,we have:
$260.8 = n \times 29.73 + n \times 35.5$.
$260.8 = n(29.73 + 35.5) = n(65.23)$.
$n = \frac{260.8}{65.23} \approx 4$.
Atomic weight of metal $= n \times \text{Equivalent mass} = 4 \times 29.73 = 118.92$.

(B) The reaction is $Ca(OH)_2 + H_3PO_4 \rightarrow CaHPO_4 + 2H_2O$.
In this reaction,$H_3PO_4$ loses $2$ $H^+$ ions to form $HPO_4^{2-}$.
Thus,the n-factor (valency factor) of $H_3PO_4$ is $2$.
The molecular weight of $H_3PO_4 = (3 \times 1) + 31 + (4 \times 16) = 98 \ g/mol$.
Equivalent weight $= \frac{\text{Molecular Weight}}{\text{n-factor}} = \frac{98}{2} = 49$.

(A) In the compound $X_2O_3$,the valency of $X$ can be determined from the formula.
Since $O$ has a valency of $2$,in $X_2O_3$,$X$ must have a valency of $3$.
Equivalent mass is defined as $\frac{\text{Atomic mass}}{\text{Valency}}$.
Given atomic mass of $X = 91.5$.
Equivalent mass of $X = \frac{91.5}{3} = 30.5$.

(B) Let the metal be $M$ and its valency be $x$. The formula of the metal chloride is $MCl_x$.
The molecular weight of $MCl_x$ is given by: $\text{Atomic weight of } M + x \times \text{Atomic weight of } Cl = 80$.
We know that $\text{Atomic weight of } M = \text{Equivalent weight} \times \text{Valency} = 4.5x$.
Substituting this into the molecular weight equation:
$4.5x + x(35.5) = 80$
$4.5x + 35.5x = 80$
$40x = 80$
$x = 2$
Therefore,the atomic weight of the metal is $4.5 \times 2 = 9$.

(A) Let the equivalent weight of the metal be $E$ and the equivalent weight of the nitrate radical $(NO_3^-)$ be $62$ and the sulphate radical $(SO_4^{2-})$ be $48$ (since $96/2 = 48$).
According to the law of equivalence,the number of equivalents of metal nitrate equals the number of equivalents of metal sulphate.
$\frac{1}{E + 62} = \frac{0.86}{E + 48}$
$E + 48 = 0.86(E + 62)$
$E + 48 = 0.86E + 53.32$
$E - 0.86E = 53.32 - 48$
$0.14E = 5.32$
$E = \frac{5.32}{0.14} = 38$
Thus,the equivalent weight of the metal is $38$.

(C) At $STP$,the volume of $1 \, mole$ of any gas is $22.4 \, L$ or $22400 \, cm^3$.
Given: Volume $V = 500 \, cm^3$,Mass $m = 0.581 \, g$.
Density $\rho = \frac{m}{V} = \frac{0.581 \, g}{500 \, cm^3} = 0.001162 \, g/cm^3$.
Molar mass $M = \rho \times 22400 \, cm^3/mol$.
$M = 0.001162 \times 22400 \approx 26.03 \, g/mol$.
The molar mass of $C_2H_2$ is $(2 \times 12) + (2 \times 1) = 26 \, g/mol$.
Therefore,the gas is $C_2H_2$.

(B) The molar mass of ethanol $(C_2H_5OH)$ is $2 \times 12 + 6 \times 1 + 16 = 46 \, g \, mol^{-1}$.
Density $(d)$ = $0.929 \, g \, cm^{-3}$.
Volume $(V)$ = $\frac{\text{Mass}}{\text{Density}} = \frac{46 \, g \, mol^{-1}}{0.929 \, g \, cm^{-3}} = 49.51 \, cm^3 \, mol^{-1}$.

(B) The molar mass of $C_6H_{12}O_6$ is $180 \ g/mol$.
The mass of one molecule is calculated by dividing the molar mass by Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$.
Mass of one molecule $= \frac{180 \ g/mol}{6.022 \times 10^{23} \ mol^{-1}} = 2.988 \times 10^{-22} \ g$.

(D) The molar mass of $O_2$ is $32 \ g/mol$.
The mass of one molecule of $O_2$ in grams is calculated as: $\frac{32 \ g/mol}{6.022 \times 10^{23} \ molecules/mol} \approx 5.31 \times 10^{-23} \ g$.
To convert this mass into kilograms,divide by $1000$ $(10^3)$:
$5.31 \times 10^{-23} \ g = \frac{5.31 \times 10^{-23}}{10^3} \ kg = 5.31 \times 10^{-26} \ kg$.

(D) The molecular formula of the compound is $C_{60}H_{122}$.
First,calculate the molar mass of $C_{60}H_{122}$:
$M = (60 \times 12.01) + (122 \times 1.008) = 720.6 + 122.976 = 843.576 \ g/mol$.
Using the Avogadro constant $(N_A = 6.022 \times 10^{23} \ mol^{-1})$,the mass of one molecule is given by:
$\text{Mass} = \frac{\text{Molar mass}}{N_A} = \frac{843.576}{6.022 \times 10^{23}} \approx 1.4008 \times 10^{-21} \ g$.
Thus,the mass of one molecule is approximately $1.4 \times 10^{-21} \ g$.

(C) Let the atomic mass of $X$ be $x$ and $Y$ be $y$.
For $X_2Y_3$:
$0.2 \ mol$ weighs $32.0 \ g$,so $1 \ mol$ weighs $32.0 / 0.2 = 160 \ g/mol$.
Thus,$2x + 3y = 160$ --- $(1)$
For $X_3Y_4$:
$0.4 \ mol$ weighs $92.8 \ g$,so $1 \ mol$ weighs $92.8 / 0.4 = 232 \ g/mol$.
Thus,$3x + 4y = 232$ --- $(2)$
Multiplying $(1)$ by $4$ and $(2)$ by $3$:
$8x + 12y = 640$
$9x + 12y = 696$
Subtracting the equations: $x = 56$.
Substituting $x = 56$ in $(1)$: $2(56) + 3y = 160 \implies 112 + 3y = 160 \implies 3y = 48 \implies y = 16$.
Therefore,the atomic masses are $X = 56$ and $Y = 16$.

(B) The reaction is: $\text{Compound} + O_2 \rightarrow \text{Oxide of the compound}$.
The mass of $O_2$ that combines with the compound is $(x_2 - x_1) \, g$.
By definition,the equivalent mass of a substance is the mass that combines with $8 \, g$ of oxygen.
Since $(x_2 - x_1) \, g$ of $O_2$ combines with $x_1 \, g$ of the compound,
Therefore,$8 \, g$ of $O_2$ will combine with: $\frac{x_1 \times 8}{x_2 - x_1} \, g$ of the compound.
Thus,the equivalent mass of the compound is $\frac{8x_1}{x_2 - x_1}$.

(A) The reaction is: $\text{Metal} + \text{Chlorine} \to \text{Metal Chloride}$.
The mass of chlorine combined with $A_1 \, g$ of metal is $(A_2 - A_1) \, g$.
By definition,the equivalent mass of an element is the mass that combines with $35.5 \, g$ of chlorine.
Since $(A_2 - A_1) \, g$ of chlorine combines with $A_1 \, g$ of metal,
Then $35.5 \, g$ of chlorine will combine with $\frac{A_1 \times 35.5}{A_2 - A_1} \, g$ of metal.
Therefore,the equivalent mass is $\frac{35.5 A_1}{A_2 - A_1}$.

(B) The mass of $4.6 \times 10^{22}$ atoms is $13.8 \, g$.
We know that $1 \, mol$ of an element contains $6.022 \times 10^{23}$ atoms (approximated as $6 \times 10^{23}$ atoms).
Atomic mass $= \frac{\text{Mass} \times N_A}{\text{Number of atoms}} = \frac{13.8 \, g \times 6 \times 10^{23}}{4.6 \times 10^{22}}$.
Atomic mass $= 3 \times 60 = 180 \, g \, mol^{-1}$.

(B) To find the equivalent weight,we determine the change in oxidation state of $FeS_2$ in the reaction $FeS_2 \to Fe_2O_3 + SO_2$.
In $FeS_2$,the oxidation state of $Fe$ is $+2$ and $S$ is $-1$ (as $S_2^{2-}$).
In $Fe_2O_3$,the oxidation state of $Fe$ is $+3$.
In $SO_2$,the oxidation state of $S$ is $+4$.
Change in oxidation state of $Fe$: $1 \times (+3 - (+2)) = +1$.
Change in oxidation state of $S$: $2 \times (+4 - (-1)) = 2 \times 5 = +10$.
Total change in oxidation state (n-factor) = $1 + 10 = 11$.
Equivalent weight = $\frac{\text{Molecular weight}}{\text{n-factor}} = \frac{M}{11}$.

(A) The molecular weight of the metal chloride $(MCl_3)$ is given by $2 \times \text{vapour density} = 2 \times 81.25 = 162.5$.
Let the equivalent weight of the metal be $E$.
The molecular weight of $MCl_3$ is $M + 3 \times 35.5 = 162.5$,where $M$ is the atomic weight of the metal.
Since the metal is trivalent,its atomic weight $M = 3 \times E$.
Substituting this into the equation: $3E + 106.5 = 162.5$.
$3E = 162.5 - 106.5 = 56$.
$E = \frac{56}{3} = 18.66$.

(A) Let $E$ be the equivalent weight of the metal.
The equivalent weight of oxygen $(O^{2-})$ is $8$ and the equivalent weight of chlorine $(Cl^{-})$ is $35.5$.
Using the principle of equivalence,the ratio of the weight of metal oxide to the weight of metal chloride is given by:
$\frac{\text{Weight of metal oxide}}{\text{Weight of metal chloride}} = \frac{E + E_{O^{2-}}}{E + E_{Cl^{-}}}$
Substituting the given values:
$\frac{8}{16.25} = \frac{E + 8}{E + 35.5}$
Cross-multiplying:
$8(E + 35.5) = 16.25(E + 8)$
$8E + 284 = 16.25E + 130$
$284 - 130 = 16.25E - 8E$
$154 = 8.25E$
$E = \frac{154}{8.25} \approx 18.66$

(A) The chemical reaction is given by: $4B + A \rightarrow X$.
Since $4 \ mol$ of $B$ and $1 \ mol$ of $A$ form $1 \ mol$ of $X$,the formula of $X$ is $AB_4$.
The molar mass of $X$ is the sum of the atomic masses of its constituent atoms:
$\text{Molar mass of } X = 1 \times (\text{Atomic mass of } A) + 4 \times (\text{Atomic mass of } B)$
$= 1 \times 12.01 + 4 \times 35.5$
$= 12.01 + 142$
$= 154.01 \ g/mol$.
Rounding to the nearest whole number as per the options,the weight of one mole of $X$ is $154 \ g$.

(A) Let the total mass of the metal oxide be $100 \ g$.
Mass of oxygen = $40 \ g$.
Mass of metal = $100 - 40 = 60 \ g$.
Equivalent weight of oxygen = $\frac{16}{2} = 8$.
Using the law of equivalence: $\frac{\text{Mass of metal}}{\text{Equivalent weight of metal}} = \frac{\text{Mass of oxygen}}{\text{Equivalent weight of oxygen}}$.
$\frac{60}{E} = \frac{40}{8}$.
$E = \frac{60 \times 8}{40} = \frac{480}{40} = 12$.
Therefore,the equivalent weight of the metal is $12$.

(B) The equivalent weight $(E.W.)$ of an acid is given by the formula: $E.W. = \frac{\text{Molecular Weight (M)}}{\text{Basicity (n-factor)}}$.
In the first reaction,$H_2SeO_4$ reacts with $1$ mole of $NaOH$ to form $NaHSeO_4$,so the $n$-factor is $1$. Thus,$E_1 = \frac{M}{1} = M$.
In the second reaction,$H_2SeO_4$ reacts with $2$ moles of $NaOH$ to form $Na_2SeO_4$,so the $n$-factor is $2$. Thus,$E_2 = \frac{M}{2} = 0.5M$.
Comparing the two,$E_1 = M$ and $E_2 = 0.5M$,therefore $E_1 > E_2$.

(B) The minimum molecular weight corresponds to the presence of at least one atom of sulphur in the molecule.
$\% \text{ of } S = \frac{\text{Atomic weight of } S \times 1}{\text{Minimum molecular weight}} \times 100$
$5 = \frac{32 \times 100}{M_{min}}$
$M_{min} = \frac{3200}{5} = 640$.

(C) In reaction $(i)$,$C_2H_5OH$ undergoes oxidation to $CH_3CHO$. The change in oxidation state of carbon is from $-2$ to $-1$. Since there are two carbon atoms,the total change in oxidation number is $2 \times ((-1) - (-2)) = 2$. Thus,the $n$-factor is $2$,and the equivalent weight is $E_1 = \frac{M}{2}$.
In reaction $(ii)$,$C_2H_5OH$ acts as an acid reacting with $Na$ to form $C_2H_5ONa$. The number of replaceable $H^+$ ions is $1$. Thus,the $n$-factor is $1$,and the equivalent weight is $E_2 = \frac{M}{1}$.
The ratio of equivalent weights $E_1 : E_2 = \frac{M}{2} : \frac{M}{1} = 1 : 2$.

(D) The atomic weight of the metal is calculated as: $\text{Atomic weight} = \text{Equivalent mass} \times \text{Valency}$.
Given,$\text{Equivalent mass} = 31.82$ and the metal is divalent $(n = 2)$.
$\text{Atomic weight} = 31.82 \times 2 = 63.64 \ g/mol$.
To find the mass of a single atom,we divide the molar mass by Avogadro's number $(N_A \approx 6.023 \times 10^{23} \ mol^{-1})$.
$\text{Mass of single atom} = \frac{63.64}{6.023 \times 10^{23}} \ g$.

(B) The atomic mass of sulphur is $32 \ g/mol$.
In $SCl_2$,the oxidation state of sulphur is $+2$,which corresponds to its valency.
Equivalent mass is calculated as $\frac{\text{Atomic mass}}{\text{Valency}}$.
Equivalent mass of sulphur $= \frac{32}{2} = 16 \ g/eq$.

(C) The equivalent weight of a base is defined as the ratio of its molecular weight to its acidity. Thus,the assertion is correct.
Acidity is defined as the number of replaceable $-OH$ groups present in one molecule of a base,not the number of replaceable hydrogen atoms. Thus,the reason is incorrect.

(C) The normality of a solution is given by the formula: $\text{Normality} = \text{Molarity} \times \text{n-factor}$.
For $H_3PO_3$ (phosphorous acid),the structure contains two $P-OH$ bonds,making it a dibasic acid. Thus,its n-factor is $2$.
Normality $= 0.3 \ M \times 2 = 0.6 \ N$. Therefore,the Assertion is correct.
The equivalent weight of an acid is defined as $\frac{\text{Molecular weight}}{\text{Basicity}}$.
Since the basicity of $H_3PO_3$ is $2$,its equivalent weight is $\frac{\text{Molecular weight}}{2}$.
The Reason states the denominator is $3$,which is incorrect. Therefore,the Reason is incorrect.