Atomic, Molecular and Equivalent masses Questions in English

(A) The atomic weight of a metal is calculated as: $\text{Atomic weight} = \text{Equivalent weight} \times \text{Valency}$.
Given that the metal is bivalent,its valency is $2$.
$\text{Atomic weight} = 20 \times 2 = 40 \ g \ mol^{-1}$.
The molecular formula of the anhydrous chloride of a bivalent metal $M$ is $MCl_2$.
The molecular mass of $MCl_2 = \text{Atomic mass of } M + 2 \times \text{Atomic mass of } Cl$.
$\text{Molecular mass} = 40 + 2 \times 35.5 = 40 + 71 = 111 \ g \ mol^{-1}$.

(D) The atomic mass of a metal is given by the formula: $\text{Atomic mass} = \text{Equivalent mass} \times \text{Valency}$.
Given,$\text{Equivalent mass} = 32$ and the metal is bivalent (valency $= 2$).
Therefore,$\text{Atomic mass} = 32 \times 2 = 64$.
The formula of the metal nitrate for a bivalent metal $M$ is $M(NO_3)_2$.
The molecular mass of $M(NO_3)_2 = \text{Atomic mass of } M + 2 \times (\text{Atomic mass of } N + 3 \times \text{Atomic mass of } O)$.
Substituting the values: $64 + 2 \times (14 + 3 \times 16) = 64 + 2 \times (14 + 48) = 64 + 2 \times 62 = 64 + 124 = 188$.

(C) Let the total weight of the oxide be $100 \ g$.
Since the oxide contains $20 \%$ oxygen by weight,the weight of oxygen is $20 \ g$ and the weight of the element is $100 - 20 = 80 \ g$.
The equivalent weight of oxygen is $8 \ g$.
According to the law of equivalence,the number of gram equivalents of the element must be equal to the number of gram equivalents of oxygen.
$\frac{\text{Weight of element}}{\text{Equivalent weight of element}} = \frac{\text{Weight of oxygen}}{\text{Equivalent weight of oxygen}}$
$\frac{80}{E} = \frac{20}{8}$
$E = \frac{80 \times 8}{20} = 32$
Therefore,the equivalent weight of the element is $32$.

(B) The equivalent weight of an element is given by the formula: $\text{Equivalent weight} = \frac{\text{Atomic mass}}{\text{Valency factor}}$.
In $Fe_2O_3$,the oxidation state of $Fe$ is $+3$,so the valency factor is $3$.
Given the atomic mass of $Fe = 56 \ g \ mol^{-1}$.
Therefore,$\text{Equivalent weight} = \frac{56}{3} = 18.66 \ g \ \approx 18.6 \ g$.

(C) The average atomic mass is calculated by the weighted average of the isotopic masses:
Average atomic mass of $Fe = (54 \times 0.10) + (56 \times 0.85) + (57 \times 0.05)$
$= 5.4 + 47.6 + 2.85$
$= 55.85 \ u$

(B) One mole of gas at $STP$ occupies $22.4 \ L$.
Thus,$5.6 \ L$ of gas at $STP$ contains $n = \frac{5.6}{22.4} = 0.25 \ \text{moles}$.
Number of moles = $\frac{\text{weight in gram}}{\text{molecular weight (M)}}$.
$0.25 = \frac{7.5}{M}$.
$M = \frac{7.5}{0.25} = 30 \ \text{g/mol}$.
The molar mass of $NO$ is $14 + 16 = 30 \ \text{g/mol}$.
Thus,the gas is $NO$.
Hence,the correct option is $B$.

(B) Let the element be $M$ and its valency be $n$. The chloride formed is $MCl_n$.
Mass of $Cl$ in $315.5 \ g$ of $MCl_n = 315.5 \ g - 209 \ g = 106.5 \ g$.
Equivalent mass of $Cl = \frac{35.5}{1} = 35.5 \ g$.
Number of equivalents of $Cl = \frac{106.5}{35.5} = 3$.
Since equivalents of $M$ = equivalents of $Cl$,equivalents of $M = 3$.
Equivalent mass of $M = \frac{\text{Mass of } M}{\text{Equivalents}} = \frac{209}{3} \ g$.
Now,for the oxide $M_xO_y$,equivalents of $M$ = equivalents of $O$.
For $418 \ g$ of $M$,equivalents of $M = \frac{418}{209/3} = 418 \times \frac{3}{209} = 2 \times 3 = 6$.
Equivalent mass of $O = \frac{16}{2} = 8 \ g$.
Mass of $O = \text{Equivalents} \times \text{Equivalent mass} = 6 \times 8 = 48 \ g$.

(D) Mass of metal $= 10 \text{ g}$.
Mass of oxide $= 11.6 \text{ g}$.
Mass of oxygen $= 11.6 \text{ g} - 10 \text{ g} = 1.6 \text{ g}$.
Equivalent weight of metal $= \frac{\text{Mass of metal}}{\text{Mass of oxygen}} \times 8$.
Equivalent weight of metal $= \frac{10}{1.6} \times 8 = 50$.

(B) The molecular weight of the metal chloride is $2 \times \text{vapour density} = 2 \times 83 = 166$.
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 $\text{Atomic weight} = n \times \text{Equivalent weight} = n \times 6$,we have:
$166 = 6n + 35.5n$.
$166 = 41.5n$.
$n = \frac{166}{41.5} = 4$.
Atomic weight of metal $= n \times \text{Equivalent weight} = 4 \times 6 = 24$.
Hence,the correct option is $B$.

(A) Mass of metal $X = 12 \ g \times 0.20 = 2.4 \ g$.
Mass of metal $Y = 12 \ g - 2.4 \ g = 9.6 \ g$.
Moles of $X = \frac{2.4 \ g}{40 \ g/mol} = 0.06 \ mol$.
Since the ratio of atoms $X:Y$ is $2:5$,the ratio of moles $n_X:n_Y$ is also $2:5$.
$n_Y = n_X \times \frac{5}{2} = 0.06 \times 2.5 = 0.15 \ mol$.
Atomic mass of $Y = \frac{\text{Mass of } Y}{n_Y} = \frac{9.6 \ g}{0.15 \ mol} = 64 \ g/mol$.
Thus,the atomic mass of $Y$ is $64 \ amu$.

(D) $1$. Calculate the mass of oxygen reacted: $Mass_{oxide} - Mass_{metal} = 9 \ g - 8 \ g = 1 \ g$ of oxygen.
$2$. Equivalent weight of metal $(E_M)$ is given by: $E_M = \frac{Mass_{metal}}{Mass_{oxygen}} \times 8 = \frac{8 \ g}{1 \ g} \times 8 = 64$.
$3$. The equivalent weight of hydrogen is $1$.
$4$. To react with $8 \ g$ of hydrogen,the mass of metal required is: $Mass_{metal} = E_M \times Mass_{hydrogen} = 64 \times 8 = 512 \ g$.

(D) Given that $3.011 \times 10^{22}$ atoms of an element weigh $1.15 \ g$.
We know that $1 \ mol$ of an element contains $N_A$ atoms,where $N_A \approx 6.022 \times 10^{23} \ atoms/mol$.
Mass of $6.022 \times 10^{23}$ atoms (molar mass) $= \frac{1.15 \ g}{3.011 \times 10^{22} \ atoms} \times 6.022 \times 10^{23} \ atoms/mol$.
$= 1.15 \times 20 = 23 \ g/mol$.
Therefore,the atomic mass of the element is $23 \ amu$.

(B) Sodium chloride $(NaCl)$ is an ionic compound that exists as a continuous crystal lattice,not as discrete molecules.
Therefore,the term 'molecular mass' is not applicable to it.
Instead,we use the term 'formula mass'.
The formula mass of $NaCl$ is calculated as the sum of the atomic masses of $Na$ $(23 \ amu)$ and $Cl$ $(35.5 \ amu)$.
Formula mass $= 23 + 35.5 = 58.5 \ amu$.

(C) The mass of one mole of a substance in grams is defined as its molar mass.
It is expressed in units of $g \ mol^{-1}$.
Therefore,option $(C)$ is the correct answer.

(A) The equivalent weight is calculated as: $E = \frac{\text{Molar mass}}{n\text{-factor}}$.
$1$. For $Na_2C_2O_4 \cdot 2H_2O$: Molar mass $= 2(23) + 2(12) + 4(16) + 2(18) = 46 + 24 + 64 + 36 = 170 \ g/mol$. Here,$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$,so $n$-factor $= 2$. $E = \frac{170}{2} = 85$.
$2$. For $KMnO_4$ in acidic medium: Molar mass $= 39 + 55 + 4(16) = 158 \ g/mol$. $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$,so $n$-factor $= 5$. $E = \frac{158}{5} = 31.6$.
$3$. For $KMnO_4$ in neutral medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$,so $n$-factor $= 3$. $E = \frac{158}{3} \approx 52.67$.
$4$. For $K_2Cr_2O_7$ in acidic medium: Molar mass $= 2(39) + 2(52) + 7(16) = 78 + 104 + 112 = 294 \ g/mol$. $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,so $n$-factor $= 6$. $E = \frac{294}{6} = 49$.
Comparing the values $(85, 31.6, 52.67, 49)$,the maximum equivalent weight is $85$.

(D) The equivalent weight of an element is defined as the mass of the element that reacts with $8 \ g$ of oxygen.
Given that $12 \ g$ of the element reacts with $32 \ g$ of oxygen.
Therefore,the mass of the element that reacts with $1 \ g$ of oxygen is $\frac{12}{32} \ g$.
Thus,the mass of the element that reacts with $8 \ g$ of oxygen is $\frac{12 \times 8}{32} = 3 \ g$.
Hence,the equivalent weight of the element is $3$.

(A) The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular mass}}{n-\text{factor}}$.
$A. Na_2CO_3$: It dissociates as $Na_2CO_3 \rightarrow 2Na^+ + CO_3^{2-}$. The total positive charge is $2$,so $n-\text{factor} = 2$. Equivalent weight $= M/2$ $(III)$.
$B. KMnO_4 / H^+$: In acidic medium,$Mn^{+7}$ is reduced to $Mn^{+2}$. Change in oxidation state $= 7 - 2 = 5$. $n-\text{factor} = 5$. Equivalent weight $= M/5$ $(I)$.
$C. K_2Cr_2O_7 / H^+$: In acidic medium,$Cr_2^{+6}$ is reduced to $2Cr^{+3}$. Change in oxidation state $= 2 \times (6 - 3) = 6$. $n-\text{factor} = 6$. Equivalent weight $= M/6$ $(IV)$.
$D. KMnO_4 / H_2O$: In neutral medium,$Mn^{+7}$ is reduced to $Mn^{+4}$ in $MnO_2$. Change in oxidation state $= 7 - 4 = 3$. $n-\text{factor} = 3$. Equivalent weight $= M/3$ $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(C) The molecular mass of sucrose $(C_{12}H_{22}O_{11})$ is calculated by summing the atomic masses of all atoms present in the molecule.
Atomic masses: $C = 12 \ u$,$H = 1 \ u$,$O = 16 \ u$.
Molecular mass $= (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \ g/mol$.

(A) The balanced chemical equation for the combustion of $CH_3OH$ to $CO$ and $H_2O$ is:
$CH_3OH + O_2 \rightarrow CO + 2H_2O$
In this reaction,the oxidation state of carbon in $CH_3OH$ is $-2$ and in $CO$ is $+2$.
The change in oxidation state of carbon is $|2 - (-2)| = 4$.
Thus,the valency factor ($n$-factor) for methanol is $4$.
The molecular weight of $CH_3OH$ is $12 + 4(1) + 16 = 32 \ g/mol$.
Equivalent weight = $\frac{\text{Molecular weight}}{n\text{-factor}} = \frac{32}{4} = 8$.

(A) Given that the oxide contains $40 \%$ oxygen,the metal content is $100 \% - 40 \% = 60 \%$.
Since $40 \ g$ of oxygen combines with $60 \ g$ of metal,
$8 \ g$ of oxygen (equivalent mass of oxygen) will combine with:
$\text{Equivalent weight of metal} = \frac{60 \times 8}{40} = 12 \ g$.
Atomic weight is calculated as:
$\text{Atomic weight} = \text{Equivalent weight} \times \text{Valency}$.
$\text{Atomic weight} = 12 \times 2 = 24$.

(C) Mass of $Cd = 0.9 \ g$.
Mass of $Cl_2 = 1.5 \ g - 0.9 \ g = 0.6 \ g$.
Atomic weight of $Cl = 35.5 \ g/mol$.
Mass of $Cl_2$ in $CdCl_2 = 2 \times 35.5 = 71 \ g/mol$.
According to the law of chemical equivalence:
$\frac{\text{Mass of } Cd}{\text{Atomic weight of } Cd} = \frac{\text{Mass of } Cl_2}{\text{Equivalent weight of } Cl_2}$.
$\frac{0.9}{x} = \frac{0.6}{71}$.
$x = \frac{0.9 \times 71}{0.6} = 1.5 \times 71 = 106.5 \ g/mol$.

(D) The average molecular weight is calculated by taking the weighted average of the molar masses of the components based on their volume percentages.
Average molecular weight $= (\% \text{ of } O_2 \times \text{Molar mass of } O_2) + (\% \text{ of } N_2 \times \text{Molar mass of } N_2)$
Average molecular weight $= (\frac{21}{100} \times 32) + (\frac{79}{100} \times 28)$
Average molecular weight $= 6.72 + 22.12 = 28.8 \text{ g/mol}$

(C) Atomic mass is defined as the weighted average of the masses of all naturally occurring isotopes of an element.
This is calculated by multiplying the mass of each isotope by its fractional natural abundance and summing these values.
Option $(a)$ is true as atomic mass is relative to the ${}^{12}C$ standard.
Option $(b)$ is true as the atomic mass of $Na$ is $23 \ u$.
Option $(d)$ is true as it describes the definition of average atomic mass.
Option $(c)$ is incorrect because it suggests a simple sum of masses rather than a weighted average based on natural abundance.

(D) The number of milli-equivalents of $NaOH$ is given by $N \times V = 0.1 \times 20 = 2 \ mEq$.
Since $1 \ mEq = 10^{-3} \ Eq$,the number of equivalents is $2 \times 10^{-3} = 0.002 \ Eq$.
At the point of neutralisation,the number of equivalents of acid equals the number of equivalents of base.
Therefore,$0.002 = \frac{\text{mass of acid}}{\text{equivalent weight}} = \frac{0.126}{E}$.
Solving for $E$,we get $E = \frac{0.126}{0.002} = 63$.

(D) The number of gram equivalents of $HCl$ used is calculated as: $\text{Equivalents} = \text{Normality} \times \text{Volume (in L)} = 0.1 \ N \times 0.1 \ L = 0.01 \ \text{eq}$.
According to the law of equivalence,the number of gram equivalents of the metal carbonate must be equal to the number of gram equivalents of $HCl$ used for neutralization.
Therefore,the number of gram equivalents of metal carbonate $= 0.01 \ \text{eq}$.
The equivalent weight is defined as the mass of the substance divided by the number of gram equivalents.
$\text{Equivalent weight} = \frac{\text{Mass}}{\text{Number of gram equivalents}} = \frac{2 \ g}{0.01 \ \text{eq}} = 200 \ g/\text{eq}$.
Thus,the equivalent weight of the metal carbonate is $200$.

(B) Let the two oxides be $M_x O_y$ and $M_x O_z$.
According to the problem,the ratio of mass of $M$ to mass of $O$ is given as $\frac{M_{mass}}{O_{mass}} = \frac{25}{4}$ and $\frac{25}{6}$.
Let $A$ be the atomic mass of metal $M$.
For the first oxide: $\frac{x \times A}{y \times 16} = \frac{25}{4} \implies \frac{x \times A}{y} = 100$.
For the second oxide: $\frac{x \times A}{z \times 16} = \frac{25}{6} \implies \frac{x \times A}{z} = \frac{200}{3}$.
From the first equation,$x \times A = 100y$.
Substituting this into the second equation: $\frac{100y}{z} = \frac{200}{3} \implies \frac{y}{z} = \frac{2}{3}$.
Thus,$y=2$ and $z=3$ are the smallest integers.
Substituting $y=2$ into $x \times A = 100 \times 2 = 200$.
For minimum atomic mass,we take $x=2$,so $2 \times A = 200 \implies A = 100 \ u$.

(B) The mass of one atom of ${}^{12}C$ is defined as exactly $12 \ \text{a.m.u.}$
Since $1 \ \text{a.m.u.} = 1.66056 \times 10^{-24} \ \text{g}$,
The mass of one atom of ${}^{12}C = 12 \times 1.66056 \times 10^{-24} \ \text{g}$
$= 1.99267 \times 10^{-23} \ \text{g} \approx 1.9923 \times 10^{-23} \ \text{g}$.

(B) According to the Dulong-Petit law,the approximate atomic mass of the metal is $\approx \frac{6.4}{\text{Specific heat}} = \frac{6.4}{0.16} = 40 \ g \ mol^{-1}$.
In the metal chloride,the percentage of chlorine is $65\%$,so the percentage of metal $M$ is $35\%$.
The equivalent weight of the metal is calculated as $\frac{\text{Mass of metal}}{\text{Mass of chlorine}} \times 35.5 = \frac{35}{65} \times 35.5 \approx 19.11$.
The valency $n$ is given by $\frac{\text{Approximate atomic mass}}{\text{Equivalent weight}} = \frac{40}{19.11} \approx 2.09 \approx 2$.
Thus,the formula of the metal chloride is $MCl_2$.