Atomic, Molecular and Equivalent masses Questions in English

(A) Given,equivalent weight of metal $= 4.0$ and vapour density of metal chloride $= 59.25$.
Molecular weight of metal chloride $= 2 \times \text{V.D.} = 2 \times 59.25 = 118.5$.
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 $\text{Atomic weight} = \text{Equivalent weight} \times n$,we have $118.5 = 4n + 35.5n = 39.5n$.
$n = \frac{118.5}{39.5} = 3$.
Atomic weight $= 4 \times 3 = 12$.

(B) The equivalent weight of a base is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Acidity}}$.
In the given reaction,$Zn(OH)_2 + HNO_3 \to Zn(OH)(NO_3) + H_2O$,only one $OH^-$ group is replaced by one $NO_3^-$ group.
Therefore,the acidity (number of replaceable $OH^-$ ions) is $1$.
Thus,the equivalent weight is $\frac{M}{1}$.

(C) At the equivalence point,the number of equivalents of acid equals the number of equivalents of base.
$Equivalents \text{ of acid} = \text{Equivalents of } NaOH$
$\frac{\text{Mass of acid}}{\text{Equivalent weight}} = \text{Normality} \times \text{Volume (in L)}$
$\frac{0.1914}{E} = 0.12 \times \frac{25}{1000}$
$E = \frac{0.1914 \times 1000}{0.12 \times 25}$
$E = \frac{191.4}{3} = 63.8$
Therefore,the equivalent weight of the acid is $63.8$.

(B) The equivalent weight of an acid is calculated by the formula: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Basicity}}$.
For a tribasic acid,the basicity is $3$.
Therefore,the equivalent weight $= \frac{W}{3}$.

(D) The relationship between atomic weight,equivalent weight,and valency is given by: $\text{Atomic weight} = \text{Equivalent weight} \times \text{Valency}$.
First,calculate the approximate valency: $\text{Valency} = \frac{\text{Approximate atomic weight}}{\text{Equivalent weight}} = \frac{26.89}{8.9} \approx 3.02$.
Since valency must be a whole number,we take $\text{Valency} = 3$.
Now,calculate the exact atomic weight: $\text{Exact atomic weight} = 8.9 \times 3 = 26.7$.

(B) The equivalent weight of an element is defined as the mass of the element 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 $4 \, g$ of bromine combines with $1 \, g$ of calcium.
Therefore,$80 \, g$ of bromine will combine with $\frac{1}{4} \times 80 = 20 \, g$ of calcium.
Thus,the equivalent weight of calcium is $20$.

(C) Chloroform $(CHCl_3)$ is a volatile liquid with a boiling point of $61.2 \ ^\circ C$.
Victor Meyer's method is the most appropriate and standard laboratory technique for determining the molecular weight of volatile liquids like chloroform by measuring the volume of air displaced by the vaporized sample.

(A) The relationship between molecular weight,equivalent weight,and basicity for an acid is given by the formula:
$\text{Molecular weight} = \text{Equivalent weight} \times \text{Basicity}$.
Therefore,option $A$ is correct.

(B) The molecular mass of the methyl ester is calculated as: $M_{\text{ester}} = 2 \times \text{Vapour Density} = 2 \times 37 = 74 \ g/mol$.
Let the monocarboxylic acid be $RCOOH$.
The methyl ester is $RCOOCH_3$.
The molecular mass of the ester is $M_R + 44 + 15 = 74$,where $M_R$ is the mass of the alkyl group $R$.
$M_R + 59 = 74 \implies M_R = 15$.
An alkyl group with mass $15$ is the methyl group $(-CH_3)$.
Thus,the acid is $CH_3COOH$ (acetic acid).
The molecular weight of the acid $CH_3COOH$ is $12 + 3 + 12 + 16 + 16 + 1 = 60 \ g/mol$.

(C) The molecular weight of a volatile substance using Victor Mayer's method is calculated as:
$Molecular \ weight = \frac{\text{Mass of substance (g)}}{\text{Volume of air at STP (mL)}} \times 22400 \ mL$
Given:
Mass of substance = $0.2 \ g$
Volume of air at $STP$ = $56 \ mL$
Calculation:
$Molecular \ weight = \frac{0.2}{56} \times 22400$
$Molecular \ weight = 0.2 \times 400 = 80$
Therefore,the correct option is $(C)$.

(C) Given mass of the compound $= 116 \ mg = 116 \times 10^{-3} \ g$.
Volume of air displaced at $S.T.P. = 44.8 \ mL = 44.8 \times 10^{-3} \ L$.
At $S.T.P.$,$1 \ mole$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.
Number of moles of the compound $= \frac{\text{Volume at S.T.P. (mL)}}{22400 \ mL/mol} = \frac{44.8}{22400} = 0.002 \ mol$.
Molecular weight $= \frac{\text{Mass}}{\text{Moles}} = \frac{116 \times 10^{-3} \ g}{0.002 \ mol} = 58 \ g/mol$.

(C) Step $1$: Calculate the equivalent mass of the silver salt $(E_{salt})$.
Using the relation: $\frac{\text{Mass of salt}}{\text{Mass of Ag}} = \frac{\text{Eq. mass of salt}}{\text{Eq. mass of Ag}}$
$\frac{0.228}{0.162} = \frac{E_{salt}}{108}$
$E_{salt} = \frac{0.228}{0.162} \times 108 = 152 \, g/eq$.
Step $2$: Calculate the equivalent mass of the acid $(E_{acid})$.
The silver salt of a dibasic acid $(H_2A)$ is $Ag_2A$. The equivalent mass of the salt is $E_{acid} + E_{Ag} - 1$ (since $2$ $H^+$ are replaced by $2$ $Ag^+$).
$E_{salt} = E_{acid} + 108 - 1 = E_{acid} + 107$.
$152 = E_{acid} + 107 \implies E_{acid} = 45 \, g/eq$.
Step $3$: Calculate the molecular mass of the acid.
$\text{Molecular mass} = \text{Equivalent mass} \times \text{Basicity} = 45 \times 2 = 90$.

(D) The formula to calculate the mass $(W)$ of a solute in grams is given by:
$W = \frac{N \times \text{equivalent weight} \times V(mL)}{1000}$
Given:
Normality $(N)$ = $0.05 \ N$
Equivalent weight = $49.04$
Volume $(V)$ = $100 \ mL$
Substituting the values:
$W = \frac{0.05 \times 49.04 \times 100}{1000} = \frac{245.2}{1000} = 0.2452 \ g$
Therefore,the correct option is $(D)$.

(D) According to this law,the product of atomic mass and specific heat of a solid element is approximately equal to $6.4 \, cal \cdot mol^{-1} \cdot K^{-1}$. This law is valid for solid elements only.

(B) The mass-energy equivalence is given by Einstein's equation,$E = mc^2$.
For $1 \ amu$ (atomic mass unit),the equivalent energy is calculated as:
$E = (1.6605 \times 10^{-27} \ kg) \times (2.9979 \times 10^8 \ m/s)^2 \approx 1.4924 \times 10^{-10} \ J$.
Converting this to $MeV$ $(1 \ eV = 1.602 \times 10^{-19} \ J)$:
$E = \frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-13}} \approx 931.5 \ MeV$.
Thus,$1 \ amu$ is equivalent to $931.5 \ MeV$.

(D) . The modern standard for atomic weight is based on the carbon isotope $C^{12}$,which is assigned a mass of exactly $12.000 \ amu$.

(B) The average atomic mass of chlorine is $35.5 \ u$.
Let the percentage abundance of $Cl^{37}$ be $x$ and $Cl^{35}$ be $(100 - x)$.
Using the formula: $\text{Average Atomic Mass} = \frac{x \times 37 + (100 - x) \times 35}{100}$.
$35.5 = \frac{37x + 3500 - 35x}{100}$.
$3550 = 2x + 3500$.
$2x = 50$,so $x = 25$.
Thus,the abundance of $Cl^{37}$ is $25\%$ and $Cl^{35}$ is $75\%$.
The ratio of $Cl^{35} : Cl^{37}$ is $75 : 25$,which simplifies to $3 : 1$.

(B) The average atomic mass is calculated by taking the weighted average of the isotopic masses.
Atomic mass $= \left( \frac{90}{100} \times 16 \right) + \left( \frac{10}{100} \times 18 \right)$
$= 14.4 + 1.8 = 16.2$.

(B) The average atomic weight is calculated as the weighted average of the isotopes:
Average atomic weight $= (\text{Atomic weight}_1 \times \text{Abundance}_1 + \text{Atomic weight}_2 \times \text{Abundance}_2) / 100$
Average atomic weight $= (85 \times 75 + 87 \times 25) / 100$
Average atomic weight $= (6375 + 2175) / 100$
Average atomic weight $= 8550 / 100 = 85.5$

(B) Given that the metal oxide contains $20\%$ oxygen by weight.
Therefore,the percentage of metal by weight is $100\% - 20\% = 80\%$.
The equivalent weight of an element is calculated using the formula: $\text{Equivalent weight of metal} = \frac{\text{Weight of metal}}{\text{Weight of oxygen}} \times \text{Equivalent weight of oxygen}$.
The equivalent weight of oxygen is $8$.
Substituting the values: $\text{Equivalent weight of metal} = \frac{80}{20} \times 8 = 4 \times 8 = 32$.

(C) Let the trivalent element be $M$. Since it is trivalent,its valency is $3$.
The formula of its oxide is $M_2O_3$.
The equivalent weight of $M$ is $20$.
Atomic weight of $M = \text{Equivalent weight} \times \text{Valency} = 20 \times 3 = 60$.
The molecular weight of the oxide $M_2O_3$ is calculated as:
$2 \times (\text{Atomic weight of } M) + 3 \times (\text{Atomic weight of } O) = 2 \times 60 + 3 \times 16 = 120 + 48 = 168$.

(B) The mass of copper $(Cu)$ is $4 \ g$. The mass of the oxide formed is $5 \ g$.
The mass of oxygen $(O)$ in the oxide is $5 \ g - 4 \ g = 1 \ g$.
According to the law of equivalent proportions,the mass of an element that combines with $8 \ g$ of oxygen is its equivalent weight.
Equivalent weight of $Cu = \frac{\text{Mass of } Cu}{\text{Mass of } O} \times 8 \ g$.
Equivalent weight of $Cu = \frac{4 \ g}{1 \ g} \times 8 \ g = 32 \ g$.
Thus,the equivalent weight of copper is $32$.

(C) The equivalent weight is calculated using the formula: $\text{Equivalent weight} = \frac{\text{Atomic weight}}{\text{n-factor}}$.
The reaction is: $Fe^{2+} \to Fe^{3+} + e^-$.
Here,the number of electrons lost ($n$-factor) is $1$.
Therefore,$\text{Equivalent weight} = \frac{\text{Atomic weight}}{1} = \text{Atomic weight}$.

(A) The atomic masses of the given elements are:
$1$. Uranium $(U)$: $238.03 \ u$
$2$. Radium $(Ra)$: $226.03 \ u$
$3$. Lead $(Pb)$: $207.2 \ u$
$4$. Mercury $(Hg)$: $200.59 \ u$
Comparing these values,Uranium has the highest atomic mass.
Therefore,the correct option is $A$.

(B) The equivalent weight of an oxidizing agent is calculated as: $\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}$.
For $KMnO_4$ in an acidic medium,the reduction reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
The change in oxidation state of $Mn$ is from $+7$ to $+2$,so the $n$-factor is $5$.
The molar mass of $KMnO_4$ is $158 \ g/mol$.
Therefore,$\text{Equivalent weight} = \frac{158}{5} = 31.6$.

(C) In an acidic medium,the reduction half-reaction for $K_2Cr_2O_7$ is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$
The change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Thus,the $n$-factor for $K_2Cr_2O_7$ is $6$.
The molar mass of $K_2Cr_2O_7$ is $294 \ g/mol$.
Equivalent weight $= \frac{\text{Molar mass}}{n\text{-factor}} = \frac{294}{6} = 49$.

(A) The equivalent weight of an oxidizing agent is given by $\frac{\text{Molecular Weight}}{n\text{-factor}}$.
In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and undergoes the following change: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$.
The change in oxidation state of $Mn$ is from $+7$ to $+4$,so the $n$-factor is $3$.
Therefore,the equivalent weight in an alkaline medium is $\frac{M}{3}$.
Thus,the correct statement is that it is one third of its molecular weight in an alkaline medium.

(C) The relationship between equivalent mass,atomic mass,and valency is given by: $\text{Equivalent mass} = \frac{\text{Atomic mass}}{\text{Valency}}$.
Given that the metal is divalent,its valency is $2$. Therefore,the atomic mass of the metal is $\text{Atomic mass} = 32 \times 2 = 64 \ g/mol$.
The formula for the nitrate salt of a divalent metal $M$ is $M(NO_3)_2$.
The molar mass of $M(NO_3)_2$ is calculated as: $\text{Molar mass} = \text{Atomic mass of } M + 2 \times (\text{Atomic mass of } N + 3 \times \text{Atomic mass of } O)$.
Substituting the values: $\text{Molar mass} = 64 + 2 \times (14 + 3 \times 16) = 64 + 2 \times (14 + 48) = 64 + 2 \times (62) = 64 + 124 = 188 \ g/mol$.

(B) The equivalent mass of a metal is given by the formula: $\text{Equivalent Mass} = \frac{\text{Atomic Mass}}{\text{Valency}}$.
Therefore,the atomic mass of the metal $= 12 \times 2 = 24 \ \text{g mol}^{-1}$.
The metal $M$ forms an oxide with valency $+2$,which is $MO$ (since oxygen has a valency of $-2$).
The molar mass of the oxide $MO = \text{Atomic mass of } M + \text{Atomic mass of } O$.
Molar mass of $MO = 24 + 16 = 40 \ \text{g mol}^{-1}$.

(D) The molecular weight of the metal chloride is $M = 2 \times \text{V.D.} = 2 \times 77 = 154$.
Let the valency of the metal be $x$. The formula of the metal chloride is $MCl_x$.
The molecular weight is given by $M = \text{Equivalent weight of metal} + \text{Equivalent weight of chlorine} \times x$.
$154 = 3 + 35.5x$.
$151 = 35.5x$.
$x = \frac{151}{35.5} \approx 4.25$.
Since valency must be an integer,we check the calculation. If the equivalent weight is $3$,and we assume the formula is $MCl_3$,then $M = 3 + 35.5 \times 3 = 3 + 106.5 = 109.5$. If $MCl_4$,$M = 3 + 35.5 \times 4 = 145$. If $MCl_4$ is the closest,the atomic weight $A_w = E \times x = 3 \times 4 = 12$.

(C) In $SCl_2$,the oxidation state of sulfur is $+2$. The equivalent mass is calculated as $\frac{\text{Atomic mass of S}}{\text{Valency of S}} = \frac{32}{2} = 16$.
In $S_2Cl_2$,the total mass of $2$ sulfur atoms is $64$,and they are combined with $2$ chlorine atoms $(71 \ g)$.
Since the valency of chlorine is $1$,the total valency of $2$ sulfur atoms is $2$,meaning the valency per sulfur atom is $1$.
Therefore,the equivalent mass of sulfur in $S_2Cl_2$ is $\frac{\text{Atomic mass of S}}{\text{Valency of S}} = \frac{32}{1} = 32$.

(B) The equivalent weight of an element is defined as the mass of the element 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 molecular mass of the chloride is given by: $\text{Molecular mass} = 2 \times \text{Vapor density} = 2 \times 59.25 = 118.5$.
Let the valency of the element be $n$ and its equivalent mass be $E = 4$.
The molecular mass of the chloride $(MCl_n)$ is given by: $M + n \times 35.5 = 118.5$.
Since $\text{Equivalent mass} (E) = \frac{\text{Atomic mass} (M)}{n}$,we have $M = E \times n = 4n$.
Substituting $M$ in the molecular mass equation:
$4n + 35.5n = 118.5$
$39.5n = 118.5$
$n = \frac{118.5}{39.5} = 3$.
Therefore,the valency of the element is $3$.

(B) The atomic mass of sulfur is $32 \ g/mol$.
In $SCl_2$,the oxidation state of sulfur is $+2$ (since chlorine is $-1$).
The valency factor (n-factor) for sulfur in $SCl_2$ is $2$.
Equivalent mass $= \frac{\text{Atomic mass}}{\text{Valency factor}} = \frac{32}{2} = 16 \ g/mol$.

(C) Mass of metal oxide $(W_{oxide})$ = $3.15 \, g$
Mass of metal $(W_M)$ = $1.05 \, g$
Mass of oxygen $(W_O)$ = $W_{oxide} - W_M = 3.15 - 1.05 = 2.10 \, g$
The equivalent mass of metal $(E_M)$ is calculated using the formula:
$E_M = \frac{W_M \times 8}{W_O}$
$E_M = \frac{1.05 \times 8}{2.10} = \frac{8.4}{2.10} = 4$
Thus,the equivalent mass of the metal is $4$.

(D) Let the equivalent weight of the metal be $E$.
The equivalent weight of the nitrate ion $(NO_3^-)$ is $62/1 = 62$.
The equivalent weight of the sulfate ion $(SO_4^{2-})$ is $96/2 = 48$.
According to the law of equivalence,the number of equivalents of metal nitrate equals the number of equivalents of metal sulfate.
$\frac{\text{Mass of metal nitrate}}{\text{Equivalent weight of metal nitrate}} = \frac{\text{Mass of metal sulfate}}{\text{Equivalent weight of metal sulfate}}$
$\frac{1.0}{E + 62} = \frac{0.86}{E + 48}$
$1.0(E + 48) = 0.86(E + 62)$
$E + 48 = 0.86E + 53.32$
$E - 0.86E = 53.32 - 48$
$0.14E = 5.32$
$E = \frac{5.32}{0.14} = 38$
Thus,the equivalent weight of the metal is $38$.

(A) The hardness is removed by the reaction: $Mg^{2+} + Na_2CO_3 \rightarrow MgCO_3 + 2Na^+$.
Number of milliequivalents of $Mg^{2+} = \frac{\text{mass in mg}}{\text{equivalent weight of } Mg^{2+}}$.
Equivalent weight of $Mg^{2+} = \frac{\text{Atomic mass}}{\text{Valency}} = \frac{24}{2} = 12$.
Milliequivalents of $Mg^{2+} = \frac{12.00 \, mg}{12} = 1 \, meq$.
Since $1 \, meq$ of $Mg^{2+}$ reacts with $1 \, meq$ of washing soda $(Na_2CO_3)$,the required amount is $1 \, meq$.

(C) At $STP$,the volume of $1 \ mole$ of any gas is $22.4 \ L$.
Given that $1 \ L$ of gas weighs $2 \ g$,therefore $22.4 \ L$ of gas weighs $2 \times 22.4 = 44.8 \ g$.
Thus,the molecular weight $(M)$ of the gas is $44.8 \ g/mol$.
The relationship between molecular weight and vapor density $(VD)$ is given by: $M = 2 \times VD$.
Therefore,$VD = M / 2 = 44.8 / 2 = 22.4$.
The vapor density is $22.4$ and the molecular weight is $44.8$.

(B) The equivalent weight of a metal $(E_M)$ is calculated using the formula:
$E_M = \frac{\text{Weight of Metal}}{\text{Weight of Bromine}} \times \text{Equivalent weight of Bromine}$
Given:
Weight of metal $(W_M)$ = $1.0 \ g$
Weight of bromine $(W_{Br_2})$ = $8.89 \ g$
Equivalent weight of bromine $(E_{Br_2})$ = $80 \ g/eq$
Substituting the values:
$E_M = \frac{1.0}{8.89} \times 80 \approx 8.998 \approx 9$
Therefore,the approximate equivalent weight of the metal is $9$.

(B) The modern atomic mass scale is based on the carbon-$12$ isotope $(^{12}C)$.
By international agreement,one atomic mass unit $(amu)$ or unified mass $(u)$ is defined as exactly one-twelfth $(1/12)$ of the mass of one carbon-$12$ atom.

(B) The equivalent weight $E$ of a metal is given by the formula: $E = \frac{\text{Mass of metal}}{\text{Volume of } H_2 \text{ displaced at } NTP} \times 11.2$.
Given: Mass of metal $= 1.2 \ g$,Volume of $H_2 = 1.12 \ L$.
Substituting the values: $E = \frac{1.2 \times 11.2}{1.12}$.
$E = \frac{1.2 \times 11.2}{1.12} = 12$.
Thus,the equivalent weight of the metal is $12$.

(B) According to Dulong-Petit law,$\text{Atomic weight} \approx \frac{6.4}{\text{Specific heat}}$.
$\text{Approximate atomic weight} = \frac{6.4}{0.1} = 64$.
Valency $(x)$ = $\frac{\text{Approximate atomic weight}}{\text{Equivalent weight}} = \frac{64}{31.8} \approx 2$.
$\text{Actual atomic weight} = \text{Valency} \times \text{Equivalent weight} = 2 \times 31.8 = 63.6$.

(B) Let the percentage abundance of $C^{12}$ be $x\%$ and that of $C^{13}$ be $(100-x)\%$.
The average atomic mass is given by the formula:
$\text{Average Atomic Mass} = \frac{(x \times 12) + ((100 - x) \times 13)}{100}$
Substituting the given values:
$12.011 = \frac{12x + 1300 - 13x}{100}$
$12.011 = \frac{1300 - x}{100}$
$1201.1 = 1300 - x$
$x = 1300 - 1201.1$
$x = 98.9\%$
Therefore,the abundance of $C^{12}$ is $98.9\%$.

(A) At $NTP$ (Normal Temperature and Pressure),$1 \ mole$ of any ideal gas occupies a volume of $22.4 \ L$. The mass of $1 \ mole$ of a substance is defined as its molar mass. Therefore,the weight of a substance that occupies $22.4 \ L$ at $NTP$ is its molar mass.

(D) Given: Mass of liquid $(W)$ = $510 \ mg = 0.51 \ g$.
Volume of air displaced at $NTP$ $(V)$ = $67.2 \ cm^3 = 0.0672 \ L$.
At $NTP$,$22400 \ cm^3$ of any gas corresponds to $1 \ mole$.
Number of moles $(n)$ = $\frac{67.2}{22400} = 0.003 \ mol$.
Molecular weight $(M_w)$ = $\frac{W}{n} = \frac{0.51 \ g}{0.003 \ mol} = 170 \ g/mol$.

(B) The atomic mass of an element is a constant property defined by the number of protons and neutrons in its nucleus.
Valency can vary for elements showing variable oxidation states (e.g.,$Fe^{2+}, Fe^{3+}$).
Equivalent mass is defined as $\frac{\text{Atomic mass}}{\text{Valency}}$,and since valency can change,the equivalent mass can also vary.
Therefore,atomic mass is the only property among the given options that remains constant.

(B) The equivalent weight of an element is defined as the mass of the element that combines with $8 \, g$ of oxygen.
Given that $4 \, g$ of $Cu$ produces $5 \, g$ of copper oxide.
Therefore,$4 \, g$ of $Cu$ combines with $(5 - 4) = 1 \, g$ of oxygen.
Using the unitary method:
$1 \, g$ of oxygen combines with $4 \, g$ of $Cu$.
So,$8 \, g$ of oxygen will combine with $\frac{4 \times 8}{1} = 32 \, g$ of $Cu$.
The equivalent weight of copper is $32 \, g \, eq^{-1}$.

(C) For a divalent metal $M$,the valency factor $(V.F.)$ is $2$.
Equivalent weight $(E_w)$ = $\frac{\text{Atomic weight } (A_w)}{V.F.}$
$32.7 = \frac{A_w}{2} \implies A_w = 32.7 \times 2 = 65.4 \ g/mol$.
The formula of the chloride of a divalent metal is $MCl_2$.
Molecular weight $(M_w)$ = $A_w(M) + 2 \times A_w(Cl)$
$M_w = 65.4 + 2 \times 35.5 = 65.4 + 71.0 = 136.4 \ g/mol$.

(C) According to Dulong-Petit's law,the approximate atomic weight is calculated as:
Atomic weight $\approx \frac{6.4}{\text{Specific heat}} = \frac{6.4}{0.25} = 25.6$.
The valency $(n)$ of the metal is calculated as:
$n = \frac{\text{Approximate atomic weight}}{\text{Equivalent weight}} = \frac{25.6}{12} \approx 2.13$.
Since valency must be a whole number,we take $n = 2$.
Therefore,the exact atomic weight is:
Atomic weight = $\text{Valency} \times \text{Equivalent weight} = 2 \times 12 = 24$.