Atomic, Molecular and Equivalent masses Questions in English

(C) According to the Law of Reciprocal Proportions,when two different elements combine separately with a fixed mass of a third element,the ratio of the masses in which they do so is either the same or a simple multiple of the ratio of the masses in which they combine with each other. This ratio is related to their $Equivalent \ weight$.

(A) According to the law of equivalence,when one metal displaces another from its salt solution,the number of equivalents of the two metals must be equal.
Number of equivalents of metal $A$ = Number of equivalents of metal $B$.
Since,$\text{Number of equivalents} = \frac{\text{mass}}{\text{equivalent weight}}$,we have:
$\frac{m_1}{E_1} = \frac{m_2}{E_2}$.
Rearranging the equation to find $E_1$ (equivalent weight of $A$):
$E_1 = \frac{m_1}{m_2} \times E_2$.

(A) The average atomic mass is calculated as the weighted average of the isotopes:
Average Atomic Mass $= \frac{(10 \times 19) + (11 \times 81)}{100}$
$= \frac{190 + 891}{100}$
$= \frac{1081}{100}$
$= 10.81 \approx 10.8$

(A) Given that oxygen is $20\%$ by weight in the oxide,the weight of the element is $100\% - 20\% = 80\%$.
Let the total weight of the oxide be $100 \ g$.
Weight of element $(W_M)$ = $80 \ g$.
Weight of oxygen $(W_O)$ = $20 \ g$.
The equivalent weight $(E)$ of an element is given by the formula: $E = \frac{W_M}{W_O} \times 8$.
Substituting the values: $E = \frac{80}{20} \times 8 = 4 \times 8 = 32$.
Thus,the equivalent weight of the element is $32$.

(C) The mass of $6.022 \times 10^{23}$ atoms (Avogadro's number,$N_A$) of oxygen is equal to its atomic mass,which is $16 \ g/mol$.
To find the mass of one atom,we use the formula:
$\text{Mass of one atom} = \frac{\text{Atomic mass}}{N_A}$
$\text{Mass of one atom} = \frac{16 \ g}{6.022 \times 10^{23}}$
$\text{Mass of one atom} = 2.656 \times 10^{-23} \ g$

(C) The mass of the element is $A_1 \ g$.
The mass of the oxide is $A_2 \ g$.
Therefore,the mass of oxygen in the oxide is $(A_2 - A_1) \ g$.
According to the law of equivalent proportions,the equivalent mass of the element is given by the formula:
$\text{Equivalent mass of element} = \frac{\text{Mass of element}}{\text{Mass of oxygen}} \times \text{Equivalent mass of oxygen}$.
Since the equivalent mass of oxygen is $8$,we have:
$\text{Equivalent mass} = \frac{A_1}{A_2 - A_1} \times 8$.

(C) The percentage of oxygen in the metal oxide is $32\%$.
This means in $100 \ g$ of metal oxide,there is $32 \ g$ of oxygen and $(100 - 32) = 68 \ g$ of metal.
The equivalent mass of an element is given by the formula: $E = \frac{\text{Mass of metal}}{\text{Mass of oxygen}} \times 8$.
Substituting the values: $E = \frac{68}{32} \times 8$.
$E = \frac{68}{4} = 17$.
Therefore,the equivalent mass of the metal is $17$.

(C) Let the molecular mass of $X$ be $m \ g/mol$.
The number of moles of $X$ is $n_X = \frac{1.65 \times 10^{21}}{6.022 \times 10^{23}} \ mol$.
The mass of $X$ is $m_X = n_X \times m = \frac{1.65 \times 10^{21}}{6.022 \times 10^{23}} \times m \ g$.
The number of moles of $Y$ is $n_Y = \frac{1.85 \times 10^{21}}{6.022 \times 10^{23}} \ mol$.
The mass of $Y$ is $m_Y = n_Y \times 187 = \frac{1.85 \times 10^{21}}{6.022 \times 10^{23}} \times 187 \ g$.
Given that the total mass $m_X + m_Y = 0.688 \ g$.
Substituting the values: $\frac{1.65 \times 10^{21} \times m}{6.022 \times 10^{23}} + \frac{1.85 \times 10^{21} \times 187}{6.022 \times 10^{23}} = 0.688$.
Multiplying by $N_A = 6.022 \times 10^{23}$:
$(1.65 \times 10^{21}) \times m + (1.85 \times 10^{21}) \times 187 = 0.688 \times 6.022 \times 10^{23}$.
$1.65 \times m + 1.85 \times 187 = 0.688 \times 602.2$.
$1.65 \times m + 345.95 = 414.3136$.
$1.65 \times m = 68.3636$.
$m = \frac{68.3636}{1.65} \approx 41.43 \approx 42 \ g/mol$.

(C) Let the total mass of the metal oxide be $100 \, g$.
Since the oxide contains $32\%$ oxygen,the mass of oxygen is $32 \, g$.
The mass of the metal is $100 \, g - 32 \, g = 68 \, g$.
The equivalent mass of a metal is calculated using the formula: $\text{Equivalent mass} = \frac{\text{Mass of metal}}{\text{Mass of oxygen}} \times 8$.
Substituting the values: $\text{Equivalent mass} = \frac{68}{32} \times 8 = \frac{68}{4} = 17$.

(B) The mass of $6.022 \times 10^{23}$ atoms of $H$ is $1.008 \ g$.
The mass of $1$ atom of $H$ is calculated as:
$\text{Mass} = \frac{1.008 \ g}{6.022 \times 10^{23}} \approx 1.67 \times 10^{-24} \ g$.

(D) Let the equivalent weight of the metal be $x$.
The equivalent weight of $OH^-$ is $17$ and the equivalent weight of $O^{2-}$ is $8$.
According to the law of equivalence:
$\frac{\text{Mass of metal hydroxide}}{\text{Mass of metal oxide}} = \frac{\text{Equivalent weight of metal} + \text{Equivalent weight of } OH^-}{\text{Equivalent weight of metal} + \text{Equivalent weight of } O^{2-}}$
$\frac{1.520}{0.995} = \frac{x + 17}{x + 8}$
$1.520(x + 8) = 0.995(x + 17)$
$1.520x + 12.160 = 0.995x + 16.915$
$1.520x - 0.995x = 16.915 - 12.160$
$0.525x = 4.755$
$x = \frac{4.755}{0.525} \approx 9.057 \approx 9$
Thus,the equivalent weight of the metal is $9$.

(B) According to the law of equivalence,the mass of one element displaced is proportional to its equivalent mass.
$\frac{W_{Fe}}{W_{Cu}} = \frac{E_{Fe}}{E_{Cu}}$
Given: $W_{Fe} = 2.8 \ g$,$W_{Cu} = 3.2 \ g$,$E_{Fe} = 28$.
Substituting the values:
$\frac{2.8}{3.2} = \frac{28}{E_{Cu}}$
$E_{Cu} = \frac{28 \times 3.2}{2.8}$
$E_{Cu} = 32$.

(C) For the oxide $M_mO_n$,let the atomic weight of $M$ be $A$.
The mass of $M$ in the compound is $(m \times A) \text{ g}$ and the mass of oxygen is $(n \times 16) \text{ g}$.
According to the law of equivalence:
$\frac{\text{Mass of } M}{\text{Equivalent weight of } M} = \frac{\text{Mass of } O}{\text{Equivalent weight of } O}$
Given that the equivalent weight of $M$ is $x$ and the equivalent weight of oxygen is $8$:
$\frac{m \times A}{x} = \frac{n \times 16}{8}$
$\frac{m \times A}{x} = 2n$
$A = \frac{2nx}{m}$

(C) Mass of metal hydride = $0.84 \, g$
Mass of hydrogen = $0.04 \, g$
Mass of metal $(W_M)$ = $0.84 \, g - 0.04 \, g = 0.80 \, g$
Equivalent mass of metal $(E)$ = $\frac{\text{Mass of metal}}{\text{Mass of hydrogen}} \times 1.008$
$E = \frac{0.80}{0.04} \times 1.008 = 20 \times 1.008 \approx 20.16$
Rounding to the nearest integer,the equivalent mass is $20$.

(C) The formula of the oxide is $M_2O_3$. In $M_2O_3$,the valency of metal $M$ is $3$ (since oxygen has a valency of $2$).
Atomic mass of $M = \text{Equivalent mass} \times \text{Valency} = 9 \times 3 = 27 \ g/mol$.
The molecular mass of $M_2O_3 = (2 \times \text{Atomic mass of } M) + (3 \times \text{Atomic mass of } O)$.
Molecular mass $= (2 \times 27) + (3 \times 16) = 54 + 48 = 102 \ g/mol$.

(B) The $Victor \ Meyer$ method is a standard laboratory technique used to determine the molar mass of a volatile liquid or solid substance.
In this method,a known mass of the volatile substance is vaporized,and the volume of air displaced by the vapor is measured at a specific temperature and pressure.
Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$,the molar mass $M$ can be calculated as $M = \frac{mRT}{PV}$.

(A) According to Dulong-Petit law,approximate atomic weight $\approx \frac{6.4}{\text{specific heat}} = \frac{6.4}{0.064} = 100$.
In the metal chloride,mass of chlorine = $49.5 \ g$,so mass of metal = $100 - 49.5 = 50.5 \ g$.
Equivalent weight of metal = $\frac{\text{mass of metal}}{\text{mass of chlorine}} \times 35.5 = \frac{50.5}{49.5} \times 35.5 \approx 36.21$.
Valency = $\frac{\text{approximate atomic weight}}{\text{equivalent weight}} = \frac{100}{36.21} \approx 2.76 \approx 3$.
Exact atomic weight = $\text{equivalent weight} \times \text{valency} = 36.21 \times 3 = 108.63$.

(C) Given mass of chlorine $(W_{Cl_2})$ = $71 \ g$.
Mass of metal chloride = $111 \ g$.
Mass of metal $(W_M)$ = $111 - 71 = 40 \ g$.
Equivalent weight of metal $(E_M)$ = $\frac{W_M \times 35.5}{W_{Cl}} = \frac{40 \times 35.5}{71} = 20$.
Since the chloride is isomorphous with $MgCl_2 \cdot 6H_2O$,the valency of the metal is $2$.
Atomic weight of metal = $E_M \times \text{valency} = 20 \times 2 = 40$.

(C) The molar mass $(M_W)$ is calculated as $M_W = 2 \times \text{Vapour Density} = 2 \times 70 = 140 \text{ g/mol}$.
The formula mass is given by $x \times (12 + 16) = 28x$.
Equating the two: $28x = 140$.
Solving for $x$: $x = \frac{140}{28} = 5$.

(C) According to the $Dulong$ and $Petit's$ law,the approximate atomic weight is given by:
Approximate atomic weight $\approx \frac{6.4}{\text{Specific heat}} = \frac{6.4}{0.031} \approx 206.45$.
Now,calculate the valency:
Valency = $\frac{\text{Approximate atomic weight}}{\text{Equivalent weight}} = \frac{206.45}{103.6} \approx 1.99 \approx 2$.
Therefore,the exact atomic weight = $\text{Equivalent weight} \times \text{Valency} = 103.6 \times 2 = 207.2$.

(A) The equivalent mass of an element is given by the formula: $\text{Equivalent mass} = \frac{\text{Atomic mass}}{\text{Valency}}$.
In $SO_2$,the oxidation state of $S$ is $+4$. The equivalent mass is $\frac{32}{4} = 8$.
In $SO_3$,the oxidation state of $S$ is $+6$. The equivalent mass is $\frac{32}{6}$.
Since $8 = \frac{32}{4}$,we can write $32 = 8 \times 4$.
Substituting this in the expression for $SO_3$: $\text{Equivalent mass} = \frac{8 \times 4}{6} = \frac{8 \times 2}{3}$.

(C) Given mass of metallic chloride = $74.5 \ g$
Mass of chlorine = $35.5 \ g$
Mass of metal = $74.5 \ g - 35.5 \ g = 39 \ g$
Equivalent mass of metal $(E)$ is calculated using the formula:
$E = \frac{\text{Mass of metal}}{\text{Mass of chlorine}} \times 35.5$
$E = \frac{39}{35.5} \times 35.5 = 39$

(B) Calculation of equivalent weight:
Weight of oxygen $= 67.67 \text{ g}$
Weight of element $= 100 - 67.67 = 32.33 \text{ g}$
$67.67 \text{ g}$ of oxygen combines with $32.33 \text{ g}$ of element.
$8 \text{ g}$ of oxygen combines with $\frac{32.33 \times 8}{67.67} = 3.82 \text{ g}$ of element.
Equivalent weight of the element $= 3.82 \text{ g}$.
Let the valency of the element be $n$. The chloride is $MCl_n$.
Equivalent weight of chloride $= \text{Equivalent weight of element} + \text{Equivalent weight of chlorine} = 3.82 + 35.5 = 39.32$.
Molecular weight of chloride $= 2 \times \text{Vapour density} = 2 \times 79 = 158$.
Valency $n = \frac{\text{Molecular weight}}{\text{Equivalent weight of chloride}} = \frac{158}{39.32} \approx 4$.
Atomic weight $= \text{Equivalent weight} \times \text{Valency} = 3.82 \times 4 = 15.28$.

(B) The number of atoms in $1 \, mole$ of an element is equal to Avogadro's number,$N_A = 6.022 \times 10^{23} \, atoms/mol$.
Given that $4.6 \times 10^{22}$ atoms weigh $13.8 \, g$.
Therefore,the mass of $6.022 \times 10^{23}$ atoms (atomic mass) is calculated as:
$\text{Atomic Mass} = \frac{13.8 \, g \times 6.022 \times 10^{23} \, atoms/mol}{4.6 \times 10^{22} \, atoms} = 180 \, g/mol$.

(A) The relationship between atomic mass $(A)$,equivalent mass $(E)$,and valency $(n)$ is given by the formula: $E = \frac{A}{n}$.
Rearranging this formula,we get $A = E \times n$.
Therefore,the correct relation is $A = E \times n$.

(C) Mass of metallic chloride $= 74.5 \ g$.
Mass of chlorine $= 35.5 \ g$.
Mass of metal $= 74.5 \ g - 35.5 \ g = 39 \ g$.
The equivalent mass of a metal is given by the formula: $\text{Equivalent mass of metal} = \frac{\text{Mass of metal}}{\text{Mass of chlorine}} \times \text{Equivalent mass of chlorine}$.
Since the equivalent mass of chlorine is $35.5$,we have:
$\text{Equivalent mass of metal} = \frac{39}{35.5} \times 35.5 = 39$.

(A) The formula of the chloride is $MCl_3$ where $M$ is the element with atomic mass $27$.
The molar mass of $MCl_3$ is calculated as: $M_W = 27 + (3 \times 35.5) = 27 + 106.5 = 133.5 \ g/mol$.
Vapor density is defined as: $\text{Vapor Density} = \frac{\text{Molar Mass}}{2}$.
Therefore,$\text{Vapor Density} = \frac{133.5}{2} = 66.75$.

(C) For a divalent metal $M$,the valency is $2$.
The relationship between atomic weight $(A)$ and equivalent weight $(W)$ is $A = W \times \text{valency}$.
So,$A = W \times 2 = 2W$.
The formula of the metal chloride is $MCl_2$.
The molecular weight of $MCl_2 = \text{Atomic weight of } M + 2 \times \text{Atomic weight of } Cl$.
Molecular weight $= 2W + 2 \times 35.5 = 2W + 71$.

(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is calculated as: $(2 \times 1) + (2 \times 12) + (4 \times 16) + 2 \times (2 \times 1 + 16) = 2 + 24 + 64 + 36 = 126 \ g/mol$.
Since oxalic acid is a dibasic acid,its n-factor is $2$.
Equivalent mass = $\frac{\text{Molar mass}}{\text{n-factor}} = \frac{126}{2} = 63 \ g \ eq^{-1}$.

(C) Given mass of the compound $w = 116 \, mg = 0.116 \, g$.
Volume of air displaced at $S.T.P.$ $V = 44.8 \, mL = 0.0448 \, L$.
At $S.T.P.$,$22.4 \, L$ of any gas corresponds to $1 \, mole$.
Number of moles $n = \frac{V}{22400 \, mL} = \frac{44.8}{22400} = 0.002 \, mol$.
Molar mass $M = \frac{w}{n} = \frac{0.116 \, g}{0.002 \, mol} = 58 \, g/mol$.

(B) The equivalent weight of an element is the mass that combines with $8 \, g$ of oxygen or $1 \, g$ of hydrogen or $35.5 \, g$ of chlorine or $80 \, g$ of bromine.
Given that $1 \, g$ of calcium combines with $4 \, g$ of bromine.
Therefore,the amount of calcium that combines with $80 \, g$ of bromine is calculated as:
$\text{Equivalent weight of Ca} = \frac{1 \, g \, Ca}{4 \, g \, Br} \times 80 \, g \, Br = 20 \, g$.
Thus,the equivalent weight of calcium is $20$.

(B) The equivalent weight of the metal is $31.82$.
Since the metal is divalent,its valency is $2$.
Atomic weight $= \text{Equivalent weight} \times \text{Valency} = 31.82 \times 2 = 63.64$.
The weight of one atom is given by $\frac{\text{Atomic weight}}{N_A}$,where $N_A = 6.02 \times 10^{23}$.
Therefore,the weight of one atom $= \frac{63.64}{6.02 \times 10^{23}}$.

(B) At $N.T.P.$,the volume of $1 \ mole$ of any ideal gas is $22.4 \ L$.
Given that $1 \ L$ of gas weighs $1.25 \ g$.
Therefore,the weight of $22.4 \ L$ (molar mass) of the gas is $1.25 \ g \times 22.4 \ L = 28 \ g/mol$.
The molar mass of $N_2$ is $2 \times 14 = 28 \ g/mol$.
Thus,the gas is $N_2$.

(A) The average atomic mass is calculated using the formula: $\text{Average Atomic Mass} = \sum (\text{Isotopic Mass} \times \text{Fractional Abundance})$.
Given:
Isotope $1$: Mass = $85$,Abundance = $75\% = 0.75$.
Isotope $2$: Mass = $87$,Abundance = $25\% = 0.25$.
Calculation:
$\text{Average Atomic Mass} = (85 \times 0.75) + (87 \times 0.25)$
$= 63.75 + 21.75$
$= 85.5 \text{ u}$.
Therefore,the correct option is $A$.

(A) Let the percentage abundance of isotope $(I)$ be $x_1$ and isotope $(II)$ be $x_2$. Since $x_1 + x_2 = 100$,we can write $x_2 = 100 - x_1$.
The average atomic mass is given by the formula: $\text{Average Mass} = \frac{M_1 x_1 + M_2 x_2}{100}$.
Substituting the given values: $10.81 = \frac{10.01 x_1 + 11.01 (100 - x_1)}{100}$.
$1081 = 10.01 x_1 + 1101 - 11.01 x_1$.
$1081 - 1101 = -1.00 x_1$.
$-20 = -1.00 x_1$.
$x_1 = 20$.
Therefore,$x_2 = 100 - 20 = 80$.
The percentage abundances are $20\%$ and $80\%$ respectively.

(D) Let the abundance of $Ne^{20}$ be $x_1$ and $Ne^{22}$ be $x_2$.
Given the average atomic mass is $20.2$.
The formula is $20.2 = \frac{20x_1 + 22x_2}{x_1 + x_2}$.
Assuming $x_1 + x_2 = 100$,we have $20.2 = 0.20x_1 + 0.22x_2$.
Since $x_1 = 100 - x_2$,we substitute: $20.2 = 0.20(100 - x_2) + 0.22x_2$.
$20.2 = 20 - 0.20x_2 + 0.22x_2$.
$0.2 = 0.02x_2$.
$x_2 = \frac{0.2}{0.02} = 10$.
Thus,the abundance of the heavier isotope $Ne^{22}$ is $10\%$.

(B) The formula for normality is $N = \frac{\text{Weight in gram}}{\text{Equivalent weight} \times \text{Volume in Liters}}$.
Equivalent weight of $H_2SO_4 = \frac{\text{Molar mass}}{\text{Valency factor}} = \frac{98}{2} = 49$.
Weight in gram = $N \times \text{Equivalent weight} \times \text{Volume in Liters}$.
Weight = $\frac{1}{7} \times 49 \times \frac{150}{1000}$.
Weight = $7 \times 0.15 = 1.05 \, g$.

(A) The equivalent weight of an acid is calculated as: $\text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Basicity (n-factor)}}$.
The molar mass of $H_3PO_4$ is $98 \ g/mol$.
$1$. In the first reaction,$H_3PO_4 + OH^{-} \rightarrow H_2PO_4^- + H_2O$,the $n$-factor is $1$. So,$\text{Eq. Wt.} = \frac{98}{1} = 98$.
$2$. In the second reaction,$H_3PO_4 + 2OH^{-} \rightarrow HPO_4^{2-} + 2H_2O$,the $n$-factor is $2$. So,$\text{Eq. Wt.} = \frac{98}{2} = 49$.
$3$. In the third reaction,$H_3PO_4 + 3OH^{-} \rightarrow PO_4^{3-} + 3H_2O$,the $n$-factor is $3$. So,$\text{Eq. Wt.} = \frac{98}{3} = 32.67$.
Thus,the values are $98, 49, 32.67$.

(C) The number of gram equivalents is equal to the number of Faradays,which is equal to the number of moles of electrons.
Number of moles of electrons = $\frac{6 \times 10^{20}}{6.022 \times 10^{23}} \approx \frac{6 \times 10^{20}}{6 \times 10^{23}} = 10^{-3}$.
Therefore,the number of gram equivalents is $0.001$.

(C) $1$. Identify the oxidizing agent: In the reaction $SO_2 + 2H_2S \rightarrow 3S + 2H_2O$,the oxidation state of $S$ in $SO_2$ changes from $+4$ to $0$ (reduction),so $SO_2$ is the oxidizing agent.
$2$. Calculate the change in oxidation state: The change in oxidation state per molecule of $SO_2$ is $|4 - 0| = 4$.
$3$. Calculate the equivalent mass: The equivalent mass of an oxidizing agent is given by $\frac{\text{Molar mass}}{\text{Change in oxidation state}}$.
$4$. Molar mass of $SO_2 = 32 + (2 \times 16) = 64 \ g/mol$.
$5$. Equivalent mass $= \frac{64}{4} = 16$.

(A) The equivalent weight of an acid is given by the formula: $\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}$.
The molar mass of $H_3PO_4$ is $98 \ g/mol$.
In the first reaction,$H_3PO_4 + OH^{-} \rightarrow H_2PO_4^{-} + H_2O$,the $n$-factor is $1$ (one $H^+$ is replaced).
Equivalent weight $= 98 / 1 = 98$.
In the second reaction,$H_3PO_4 + 2OH^{-} \rightarrow HPO_4^{2-} + 2H_2O$,the $n$-factor is $2$ (two $H^+$ are replaced).
Equivalent weight $= 98 / 2 = 49$.
In the third reaction,$H_3PO_4 + 3OH^{-} \rightarrow PO_4^{3-} + 3H_2O$,the $n$-factor is $3$ (three $H^+$ are replaced).
Equivalent weight $= 98 / 3 = 32.67$.
Thus,the equivalent weights are $98, 49, 32.67$.

(A) The oxidation state of $Fe$ changes from $+2$ to $+3$ (change of $1$ electron).
The oxidation state of $C$ in $C_2O_4^{2-}$ changes from $+3$ to $+4$ (change of $1$ electron per $C$ atom).
Since there are $2$ carbon atoms,the total change for carbon is $2 \times 1 = 2$ electrons.
Total change in oxidation number $= 1 (Fe) + 2 (C) = 3$.
Therefore,the equivalent weight $E = \frac{M}{3}$.

(B) The accurate determination of atomic masses is performed using a $Mass \ spectrometer$. This instrument measures the mass-to-charge ratio of ions,allowing for the precise identification and quantification of isotopes and their relative abundances.

(C) The normality $(N)$ of a solution is defined as the number of gram equivalents of solute per litre of solution.
Formula: $\text{Mass of solute (in g)} = N \times \text{Equivalent mass} \times \text{Volume (in L)}$.
For $H_2SO_4$,the molar mass is $98\, g/mol$ and the n-factor (basicity) is $2$.
Equivalent mass of $H_2SO_4 = \frac{98}{2} = 49\, g/eq$.
Given $N = 2\, N$ and volume $= 1\, L$.
Mass of $H_2SO_4 = 2 \times 49 \times 1 = 98\, g$.
Therefore,the solution contains $98\, g$ of $H_2SO_4$ per litre of solution.

(C) The atomic weight is calculated as the weighted average of the isotopic masses based on their natural abundance.
Atomic weight = $(0.75 \times 85) + (0.25 \times 87)$
Atomic weight = $63.75 + 21.75 = 85.5$.

(A) For $XY_2$:
$0.1 \ mol$ of $XY_2 = 10 \ g$
$1 \ mol$ of $XY_2 = 100 \ g$
So,$X + 2Y = 100$ (Equation $1$)
For $X_3Y_2$:
$0.05 \ mol$ of $X_3Y_2 = 9 \ g$
$1 \ mol$ of $X_3Y_2 = 180 \ g$
So,$3X + 2Y = 180$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$:
$(3X + 2Y) - (X + 2Y) = 180 - 100$
$2X = 80 \implies X = 40$
Substituting $X = 40$ in Equation $1$:
$40 + 2Y = 100$
$2Y = 60 \implies Y = 30$
Therefore,the atomic weights of $X$ and $Y$ are $40$ and $30$ respectively.

(D) The weighted average atomic mass is calculated as:
$\text{Average atomic mass} = (\% \text{ abundance of isotope } 1 \times \text{mass of isotope } 1) + (\% \text{ abundance of isotope } 2 \times \text{mass of isotope } 2) + (\% \text{ abundance of isotope } 3 \times \text{mass of isotope } 3)$
$\text{Average atomic mass} = (0.90 \times 200) + (0.08 \times 199) + (0.02 \times 202)$
$\text{Average atomic mass} = 180.00 + 15.92 + 4.04 = 199.96 \ amu$
Rounding to the nearest whole number,the value is $200 \ amu$.

(C) The average atomic mass is calculated using the formula:
$A_{avg} = \frac{\sum (A_i \times X_i)}{\sum X_i}$
Given:
$A_1 = 10, X_1 = 20\% = 0.2$
$A_2 = 11, X_2 = 80\% = 0.8$
Substituting the values:
$A_{avg} = \frac{(10 \times 0.2) + (11 \times 0.8)}{0.2 + 0.8}$
$A_{avg} = \frac{2.0 + 8.8}{1.0}$
$A_{avg} = 10.8$
Thus,the atomic weight of boron is $10.80$.