Atomic, Molecular and Equivalent masses Questions in English

(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.

(A) The molecular mass of glucose $(C_{6}H_{12}O_{6})$ is calculated by summing the atomic masses of all atoms present in the molecule.
Atomic masses: $C = 12.011 \, u$,$H = 1.008 \, u$,$O = 16.00 \, u$.
Molecular mass $= 6 \times (12.011 \, u) + 12 \times (1.008 \, u) + 6 \times (16.00 \, u)$.
$= 72.066 \, u + 12.096 \, u + 96.00 \, u$.
$= 180.162 \, u$.

$(i)$ $H_{2}O$
The molecular mass of water,$H_{2}O$ is calculated as:
$= (2 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of O})$
$= [2(1.008 \, u) + 1(16.00 \, u)]$
$= 2.016 \, u + 16.00 \, u = 18.016 \, u \approx 18.02 \, u$
$(ii)$ $CO_{2}$
The molecular mass of carbon dioxide,$CO_{2}$ is calculated as:
$= (1 \times \text{Atomic mass of C}) + (2 \times \text{Atomic mass of O})$
$= [1(12.011 \, u) + 2(16.00 \, u)]$
$= 12.011 \, u + 32.00 \, u = 44.011 \, u \approx 44.01 \, u$
$(iii)$ $CH_{4}$
The molecular mass of methane,$CH_{4}$ is calculated as:
$= (1 \times \text{Atomic mass of C}) + (4 \times \text{Atomic mass of H})$
$= [1(12.011 \, u) + 4(1.008 \, u)]$
$= 12.011 \, u + 4.032 \, u = 16.043 \, u$

(D) The average atomic mass is calculated by the weighted average of the isotopes:
$\text{Average atomic mass} = \frac{(\text{Abundance of } ^{35}Cl \times \text{Mass of } ^{35}Cl) + (\text{Abundance of } ^{37}Cl \times \text{Mass of } ^{37}Cl)}{100}$
$\text{Average atomic mass} = \frac{(75.77 \times 34.9689) + (24.23 \times 36.9659)}{100}$
$\text{Average atomic mass} = \frac{2649.59 + 895.68}{100} = \frac{3545.27}{100} = 35.4527 \ u$
Rounding to two decimal places,we get $35.45 \ u$.

(N/A) $1$ mole of $^{12}C$ atoms $= 6.022 \times 10^{23}$ atoms.
The molar mass of $^{12}C$ is $12 \, g/mol$.
Mass of one $^{12}C$ atom $= \frac{\text{Molar mass}}{\text{Avogadro's number}}$
$= \frac{12 \, g}{6.022 \times 10^{23}}$
$= 1.993 \times 10^{-23} \, g$.

(N/A) The molar mass of an element with multiple isotopes is calculated by the weighted average of the isotopic masses:
$\text{Molar mass} = \sum (\text{Isotopic mass} \times \text{Fractional abundance})$
$\text{Molar mass} = \left[ \left( 35.96755 \times \frac{0.337}{100} \right) + \left( 37.96272 \times \frac{0.063}{100} \right) + \left( 39.9624 \times \frac{99.600}{100} \right) \right] \, g \, mol^{-1}$
$= [0.12121 + 0.02392 + 39.80255] \, g \, mol^{-1}$
$= 39.94768 \, g \, mol^{-1}$
Rounding to appropriate significant figures,the molar mass is $39.948 \, g \, mol^{-1}$.

(N/A) Mercury $(II)$ chloride: $Hg^{2+}$ and $Cl^-$ ions combine to form $HgCl_2$.
$(b)$ Nickel $(II)$ sulphate: $Ni^{2+}$ and $SO_4^{2-}$ ions combine to form $NiSO_4$.
$(c)$ Tin $(IV)$ oxide: $Sn^{4+}$ and $O^{2-}$ ions combine to form $SnO_2$.
$(d)$ Thallium $(I)$ sulphate: $Tl^+$ and $SO_4^{2-}$ ions combine to form $Tl_2SO_4$.
$(e)$ Iron $(III)$ sulphate: $Fe^{3+}$ and $SO_4^{2-}$ ions combine to form $Fe_2(SO_4)_3$.
$(f)$ Chromium $(III)$ oxide: $Cr^{3+}$ and $O^{2-}$ ions combine to form $Cr_2O_3$.

(N/A) The atomic mass of an atom is extremely small because atoms are very tiny.
Today,sophisticated techniques like mass spectrometry are used to determine atomic masses accurately.
Initially,hydrogen was assigned a mass of $1$ (without units) as it is the lightest atom,and other elements were assigned masses relative to it.
However,the current system of atomic masses is based on carbon-$12$ as the standard,which was agreed upon in $1961$.
Carbon-$12$ is an isotope of carbon represented as ${}^{12}C$.
In this system,${}^{12}C$ is assigned a mass of exactly $12$ atomic mass units $(amu)$,and the masses of all other atoms are determined relative to this standard.
One atomic mass unit $(amu)$ is defined as a mass exactly equal to $\frac{1}{12}$ of the mass of one carbon-$12$ atom.
$1 \ amu = 1.66056 \times 10^{-24} \ g$
The mass of a hydrogen atom is $1.6736 \times 10^{-24} \ g$.
Thus,in terms of $amu$,the mass of a hydrogen atom is $\frac{1.6736 \times 10^{-24}}{1.66056 \times 10^{-24}} \approx 1.0080 \ amu$.
Similarly,the mass of an oxygen-$16$ $({}^{16}O)$ atom is $15.995 \ amu$.
Currently,'$amu$' has been replaced by '$u$',which is known as unified mass.

Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance,the average atomic mass of that element can be computed.
For example,carbon has the following three isotopes with relative abundances and masses as shown against each of them:

Isotope	Relative Abundance $(\%)$	Atomic mass $(\text{amu})$
${}^{12}C$	$98.892$	$12$
${}^{13}C$	$1.108$	$13.00335$
${}^{14}C$	$2 \times 10^{-10}$	$14.00317$

From the above data,the average atomic mass of carbon is calculated as:
$(0.98892)(12 \ u) + (0.01108)(13.00335 \ u) + (2 \times 10^{-12})(14.00317 \ u) = 12.011 \ u$.
Similarly,average atomic masses for other elements can be calculated.

(N/A) Molecular mass is defined as the sum of the atomic masses of all the elements present in a molecule.
It is calculated by multiplying the atomic mass of each constituent element by the number of its atoms in the molecule and then adding these values together.
For example,the molecular mass of methane $(CH_{4})$,which contains one carbon atom and four hydrogen atoms,is calculated as follows:
Molecular mass of $(CH_{4}) = (12.011 \ u) + 4 \times (1.008 \ u) = 16.043 \ u$.
Similarly,the molecular mass of water $(H_{2}O)$ is calculated as:
Molecular mass of $(H_{2}O) = (2 \times \text{atomic mass of hydrogen}) + (1 \times \text{atomic mass of oxygen}) = (2 \times 1.008 \ u) + 16.00 \ u = 18.02 \ u$.

The molecular mass of glucose $(C_{6}H_{12}O_{6})$ is calculated by summing the atomic masses of all atoms present in the molecule.
$= 6 \times (12.011 \ u) + 12 \times (1.008 \ u) + 6 \times (16.00 \ u)$
$= 72.066 \ u + 12.096 \ u + 96.00 \ u$
$= 180.162 \ u$

(N/A) Some substances such as sodium chloride $(NaCl)$ do not contain discrete molecules as their constituent units.
In such compounds,positive (sodium) and negative (chloride) ions are arranged in a three-dimensional structure.
It may be noted that in sodium chloride,one $Na^+$ ion is surrounded by $six$ $Cl^-$ ions and vice-versa.
The formula such as $NaCl$ is used to calculate the formula mass instead of molecular mass,as in the solid state,sodium chloride does not exist as a single discrete entity.
Thus,
Formula mass of sodium chloride = Atomic mass of sodium + Atomic mass of chlorine
$= 23.0 \ u + 35.5 \ u = 58.5 \ u$

(A) The average atomic mass is calculated using the weighted average of the isotopes:
$\text{Average atomic mass} = \sum (\text{abundance}_i \times \text{mass}_i)$
$\text{Average atomic mass} = (0.7577 \times 34.9689) + (0.2423 \times 36.9659)$
$\text{Average atomic mass} = 26.4972 + 8.9567 = 35.4539 \ u$
Rounding to appropriate significant figures,the average atomic mass of chlorine is approximately $35.45 \ u$.

The average molar mass is calculated using the formula: $\overline{A} = \sum (f_i \times A_i)$
$\overline{A} = (0.00337 \times 35.96755) + (0.00063 \times 37.96272) + (0.99600 \times 39.9624)$
$\overline{A} = 0.12121 + 0.02392 + 39.80255$
$\overline{A} = 39.94768 \approx 39.948 \ g \ mol^{-1}$

(N/A) The mass of one mole of a substance in grams is defined as its molar mass.
The molar mass in grams is numerically equal to the atomic,molecular,or formula mass expressed in atomic mass units $(u)$.
For example:
Molar mass of water $(H_2O)$ $= 18.02 \ g \ mol^{-1}$.
Molar mass of sodium chloride $(NaCl)$ $= 58.5 \ g \ mol^{-1}$.

$(i)$ Molar mass of water $(H_2O)$:
Atomic mass of $H = 1 \ u$,Atomic mass of $O = 16 \ u$
Molar mass of $H_2O = (2 \times 1) + (1 \times 16) = 18 \ g/mol$
$(ii)$ Molar mass of carbon dioxide $(CO_2)$:
Atomic mass of $C = 12 \ u$,Atomic mass of $O = 16 \ u$
Molar mass of $CO_2 = (1 \times 12) + (2 \times 16) = 44 \ g/mol$
$(iii)$ Molar mass of methane $(CH_4)$:
Atomic mass of $C = 12 \ u$,Atomic mass of $H = 1 \ u$
Molar mass of $CH_4 = (1 \times 12) + (4 \times 1) = 16 \ g/mol$

(A) The molar mass of $^{12}C$ is $12 \ g/mol$.
One mole of $^{12}C$ contains $6.022 \times 10^{23}$ atoms (Avogadro's number,$N_A$).
The mass of one atom is calculated as:
$\text{Mass of one atom} = \frac{\text{Molar mass}}{N_A} = \frac{12 \ g/mol}{6.022 \times 10^{23} \ atoms/mol}$.
$\text{Mass of one atom} \approx 1.9926 \times 10^{-23} \ g$.

(A) The average atomic mass is calculated by the formula:
$\text{Average Atomic Mass} = \frac{(\text{Abundance of } ^1H \times \text{Mass of } ^1H) + (\text{Abundance of } ^2H \times \text{Mass of } ^2H)}{100}$
Substituting the given values:
$= \frac{(99.985 \times 1) + (0.015 \times 2)}{100}$
$= \frac{99.985 + 0.030}{100}$
$= \frac{100.015}{100} = 1.00015 \ u$

(N/A) Molar weight,commonly referred to as molar mass,is defined as the mass of $1 \ mol$ of a substance in grams.
It is expressed in units of $g \ mol^{-1}$.
For an element,it is numerically equal to the atomic mass in $u$,and for a molecule,it is equal to the molecular mass in $u$.

(A) One $amu$ (atomic mass unit) or $1 \ u$ (unified mass) is defined as exactly $\frac{1}{12}$ of the mass of one carbon-$12$ $(^{12}C)$ atom.

(N/A) For Iron $(III)$ sulphate:
Iron has an oxidation state of $+3$ and the sulphate ion is $SO_{4}^{2-}$. By criss-crossing the valencies,we get $Fe_{2}(SO_{4})_{3}$.
For Chromium $(III)$ oxide:
Chromium has an oxidation state of $+3$ and the oxide ion is $O^{2-}$. By criss-crossing the valencies,we get $Cr_{2}O_{3}$.

(C) In the given reaction,$KIO_{3}$ is converted into $KI$.
The oxidation state of $I$ in $KIO_{3}$ is $+5$.
The oxidation state of $I$ in $KI$ is $-1$.
The change in oxidation number per molecule of $KIO_{3}$ is $|(+5) - (-1)| = 6$.
Therefore,the $n$-factor for $KIO_{3}$ is $6$.
Equivalent mass $= \frac{\text{Molecular mass}}{n\text{-factor}} = \frac{M}{6}$.

(A) The molar mass of hydrogen is $1.008 \ g/mol$.
The number of atoms in one mole is $6.022 \times 10^{23} \ atoms/mol$.
The mass of a single atom is calculated by dividing the molar mass by the Avogadro constant:
$\text{Mass} = \frac{1.008 \ g/mol}{6.022 \times 10^{23} \ atoms/mol} \approx 1.6736 \times 10^{-24} \ g$.

(A) The atomic mass of an atom can be measured accurately using a $Mass \ spectrometer$.

(C) Molecular mass $= 2 \times \text{Vapor density} = 2 \times 66.75 = 133.50$.
The formula of the chloride salt is $XCl_n$,where $n$ is the valency.
Molecular mass $= \text{Atomic mass of } X + n \times \text{Atomic mass of } Cl$.
Atomic mass of $X = \text{Equivalent mass} \times \text{Valency} = 9 \times n$.
Substituting the values: $133.50 = 9n + n \times 35.5$.
$133.50 = 44.5n$.
$n = \frac{133.50}{44.5} = 3$.
Therefore,the valency of element $X$ is $3$.

(A) The molar mass of $H_3PO_4$ is given as $98 \ g/mol$.
The formula for molar mass is: $3 \times \text{Atomic mass of } H + 1 \times \text{Atomic mass of } P + 4 \times \text{Atomic mass of } O = 98 \ g/mol$.
Given atomic masses: $H = 1 \ g/mol$,$O = 16 \ g/mol$.
Substituting the values: $3(1) + P + 4(16) = 98$.
$3 + P + 64 = 98$.
$67 + P = 98$.
$P = 98 - 67 = 31 \ g/mol$.

(A) The atomic mass can be determined accurately using the $Mass \ Spectrometry$ method.

(B) The ratio of isotopes $^{35}Cl$ and $^{37}Cl$ is $3:1$.
This means that out of every $4$ atoms of chlorine,$3$ atoms have a mass of $35 \ amu$ and $1$ atom has a mass of $37 \ amu$.
Average atomic mass $= \frac{(3 \times 35) + (1 \times 37)}{3 + 1} \ amu$.
Average atomic mass $= \frac{105 + 37}{4} \ amu$.
Average atomic mass $= \frac{142}{4} \ amu = 35.5 \ amu$.

(A) The molar masses are calculated as follows:
$(i)$ Barium sulphate $(BaSO_4)$: $137 + 32 + (4 \times 16) = 233 \ g/mol$ $(f)$
$(ii)$ Magnesium pyrophosphate $(Mg_2P_2O_7)$: $(2 \times 24) + (2 \times 31) + (7 \times 16) = 48 + 62 + 112 = 222 \ g/mol$ $(d)$
$(iii)$ Ammonium phosphomolybdate $((NH_4)_3PO_4 \cdot 12MoO_3)$: $\approx 1877 \ g/mol$ $(a)$
$(iv)$ Silver chloride $(AgCl)$: $108 + 35.5 = 143.5 \ g/mol$ $(e)$
$(v)$ Silver bromide $(AgBr)$: $108 + 80 = 188 \ g/mol$ $(c)$
$(vi)$ Silver iodide $(AgI)$: $108 + 127 = 235 \ g/mol$ $(b)$
Therefore,the correct matching is: $(i-f, ii-d, iii-a, iv-e, v-c, vi-b)$.

(D) Let the mole ratio of $^{35}Cl$ to $^{37}Cl$ be $x:1$.
The average molar mass is given by the formula: $\text{Average molar mass} = \frac{n_{1} M_{1} + n_{2} M_{2}}{n_{1} + n_{2}}$
Substituting the given values: $35.5 = \frac{x \times 35 + 1 \times 37}{x + 1}$
$35.5(x + 1) = 35x + 37$
$35.5x + 35.5 = 35x + 37$
$35.5x - 35x = 37 - 35.5$
$0.5x = 1.5$
$x = \frac{1.5}{0.5} = 3$
Therefore,the ratio of $^{35}Cl$ to $^{37}Cl$ is $3:1$.

(B) Let the natural abundance of ${}^{35}Cl$ be $x$ and that of ${}^{37}Cl$ be $y$.
The average atomic mass is given by the formula: $M_{av} = \frac{M_{1}x + M_{2}y}{x + y}$.
Substituting the given values: $35.45 = \frac{35x + 37y}{x + y}$.
$35.45(x + y) = 35x + 37y$.
$35.45x + 35.45y = 35x + 37y$.
$0.45x = 1.55y$.
$\frac{x}{y} = \frac{1.55}{0.45} = \frac{155}{45} \approx 3.44$.
The ratio is closest to $3: 1$.

(D) .
Let the equivalent weight of metal $M = x$.
According to the law of equivalence,the number of equivalents of metal reacted is equal to the number of equivalents of $H_2$ gas produced.
$(eq)_{Ca} = (eq)_{H_2} \text{ (from Ca)} = \frac{2.0}{20} = 0.1$
$(eq)_{M} = (eq)_{H_2} \text{ (from M)} = \frac{2.0}{x}$
Since the volume of gas produced is directly proportional to the number of equivalents under identical conditions:
$\frac{(eq)_{Ca}}{(eq)_{M}} = \frac{V_{H_2} \text{ (from Ca)}}{V_{H_2} \text{ (from M)}}$
$\frac{0.1}{2.0/x} = \frac{1.125}{1.85}$
$\frac{0.1x}{2.0} = \frac{1.125}{1.85}$
$x = \frac{1.125 \times 2.0}{1.85 \times 0.1} = \frac{2.25}{0.185} \approx 12.16$
Thus,the equivalent weight of $M$ is approximately $12$.

(B) The principle of equivalence states that the number of equivalents of the metal equals the number of equivalents of the metal sulphate formed.
Let the equivalent weight of the metal be $x$.
The equivalent weight of the sulphate ion $(SO_4^{2-})$ is $\frac{96}{2} = 48$.
Thus,the equivalent weight of the metal sulphate is $(x + 48)$.
Using the formula: $\frac{\text{Mass of metal}}{\text{Eq. wt. of metal}} = \frac{\text{Mass of metal sulphate}}{\text{Eq. wt. of metal sulphate}}$
$\frac{2.0}{x} = \frac{6.8}{x + 48}$
$2.0(x + 48) = 6.8x$
$2x + 96 = 6.8x$
$4.8x = 96$
$x = \frac{96}{4.8} = 20.0 \ g$.

(B) The average atomic mass of an element is calculated by the weighted average of the masses of its isotopes.
Average atomic mass of $Fe = \frac{(5 \times 54) + (90 \times 56) + (5 \times 57)}{100}$
$= \frac{270 + 5040 + 285}{100}$
$= \frac{5595}{100} = 55.95 \ u$

(C) The molecular formula of the compound $(P)$ is $C_9H_{11}FIN_2O_5$.
Calculating the molar mass:
$M = (9 \times 12) + (11 \times 1) + 127 + 19 + (2 \times 14) + (5 \times 16)$
$M = 108 + 11 + 127 + 19 + 28 + 80 = 373 \ g \ mol^{-1}$.
Weight of $0.1$ mol $= 0.1 \times 373 = 37.3 \ g = 373 \times 10^{-1} \ g$.
Thus,the value is $373$.

(B) The number of atoms is calculated using the formula: $\text{Number of atoms} = \frac{\text{Mass}}{\text{Molar Mass}} \times N_A$.
Since the mass is constant $(10^{-9} \ g)$ for all elements,the number of atoms is inversely proportional to the molar mass.
Therefore,the element with the lowest molar mass will have the highest number of atoms.
The molar masses are:
$\bullet M_{Po} = 209 \ g/mol$
$\bullet M_{Pr} = 141 \ g/mol$
$\bullet M_{Pb} = 207 \ g/mol$
$\bullet M_{Pt} = 195 \ g/mol$
Comparing these values,$Pr$ has the lowest molar mass $(141 \ g/mol)$.
Thus,$Pr$ will have the highest number of atoms.

(A) Let the percentage abundance of $Ne^{20}$ be $x$ and that of $Ne^{22}$ be $(100 - x)$.
Using the formula for average atomic mass:
$Average \ Atomic \ Mass = \frac{(Mass_1 \times Abundance_1) + (Mass_2 \times Abundance_2)}{100}$
$20.2 = \frac{(20 \times x) + (22 \times (100 - x))}{100}$
$2020 = 20x + 2200 - 22x$
$2020 = 2200 - 2x$
$2x = 2200 - 2020$
$2x = 180$
$x = 90$
Thus,the percentage abundance of the lighter isotope $Ne^{20}$ is $90\%$.

(A) At $NTP$,$1 \ mole$ of any gas occupies $22400 \ mL$ volume and its mass is equal to its molecular weight $(M)$.
Given that $V \ mL$ of the vapours weigh $W \ g$.
Therefore,$1 \ mL$ of the vapours weighs $\frac{W}{V} \ g$.
Thus,$22400 \ mL$ of the vapours will weigh $\frac{W}{V} \times 22400 \ g$.
Hence,the molecular weight $M = \frac{W}{V} \times 22400$.

(C) Let the atomic weights of $A$ and $B$ be $a$ and $b$ respectively.
For $B_2A_3$: $0.05 \text{ moles} = 9.0 \ g$,so $1 \text{ mole} = \frac{9.0}{0.05} = 180 \ g/mol$.
Thus,$2b + 3a = 180 \dots (i)$
For $B_2A$: $0.10 \text{ mole} = 10 \ g$,so $1 \text{ mole} = \frac{10}{0.10} = 100 \ g/mol$.
Thus,$2b + a = 100 \dots (ii)$
Subtracting equation $(ii)$ from $(i)$:
$(2b + 3a) - (2b + a) = 180 - 100$
$2a = 80 \implies a = 40$.
Substituting $a = 40$ in equation $(ii)$:
$2b + 40 = 100$
$2b = 60 \implies b = 30$.
Therefore,the atomic weights of $A$ and $B$ are $40$ and $30$ respectively.

(C) The average atomic mass is calculated using the formula:
$Avg. \text{ atomic mass} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{x_1 + x_2 + x_3}$
Substituting the given values:
$Avg. \text{ atomic mass} = \frac{11 \times 80 + 12 \times 15 + 13 \times 5}{80 + 15 + 5}$
$= \frac{880 + 180 + 65}{100} = \frac{1125}{100} = 11.25$

(B) The given structural formula represents ethanol $(CH_3CH_2OH)$.
The molecular formula of ethanol is $C_2H_6O$.
The molar mass is calculated as follows:
$M = (2 \times 12.01) + (6 \times 1.008) + (1 \times 16.00) \approx 24 + 6 + 16 = 46 \ g \ mol^{-1}$.
Therefore,the correct option is $B$.

(B) The chemical formula for aluminium phosphate is $AlPO_4$.
Atomicity of a molecule is the total number of atoms present in one molecule.
In $AlPO_4$,there is $1$ atom of $Al$,$1$ atom of $P$,and $4$ atoms of $O$.
$\therefore$ Atomicity $= 1 + 1 + 4 = 6$.

(D) Let the relative abundance of $^{35}Cl$ be $x\%$ and $^{37}Cl$ be $(100-x)\%$.
Average atomic mass = $\frac{(35 \times x) + (37 \times (100-x))}{100} = 35.5$
$35x + 3700 - 37x = 3550$
$-2x = -150$
$x = 75$
Abundance of $^{35}Cl = 75\%$ and $^{37}Cl = 25\%$.
Ratio = $75:25 = 3:1$.

(B) Let the percentage abundance of $^{35}Cl$ be $x$ and that of $^{37}Cl$ be $(100 - x)$.
Average atomic mass is given by the formula:
$\text{Average atomic mass} = \frac{(\text{mass of } ^{35}Cl \times x) + (\text{mass of } ^{37}Cl \times (100 - x))}{100}$
Substituting the values:
$35.5 = \frac{35x + 37(100 - x)}{100}$
$3550 = 35x + 3700 - 37x$
$3550 = 3700 - 2x$
$2x = 3700 - 3550$
$2x = 150$
$x = 75$
Therefore,the abundance of $^{35}Cl$ is $75 \%$ and the abundance of $^{37}Cl$ is $100 - 75 = 25 \%$.

(B) Atomic mass is the mass of one atom of the element in atomic mass units $(u)$.
Given,mass of $1$ atom $= 10 \ u$.
We know that $1 \ u = 1.66056 \times 10^{-24} \ g$.
Therefore,mass of $1$ atom in grams $= 10 \times 1.66056 \times 10^{-24} \ g = 1.66056 \times 10^{-23} \ g$.

(A) The molar mass of an oxygen molecule $(O_2)$ is $32 \ g/mol$.
The mass of one molecule in $amu$ is equal to its molecular mass,which is $32 \ u$.
To convert this to grams,we use the conversion factor $1 \ u = 1.66056 \times 10^{-24} \ g$.
Mass in grams $= 32 \times 1.66056 \times 10^{-24} \ g \approx 53.14 \times 10^{-24} \ g$.

(A) The unit of atomic mass,$amu$ (atomic mass unit),has been replaced by $u$ in the modern $IUPAC$ system.
Here,$u$ stands for $unified \ mass$.

(A) The symbol $u$ (unified mass) has replaced the older unit of atomic mass,$amu$ (atomic mass unit).

(B) The atomic mass unit $(amu)$,also known as the unified atomic mass unit $(u)$,is defined as $1/12$ of the mass of a carbon-$12$ atom.
$1 \ amu = 1.66057 \times 10^{-27} \ kg$.
Since $1 \ kg = 10^3 \ g$,we convert the value to grams:
$1 \ amu = 1.66057 \times 10^{-27} \ kg \times 10^3 \ g/kg = 1.66057 \times 10^{-24} \ g$.
Therefore,the correct value is $1.66 \times 10^{-24} \ g$.

(A) $1 \ u = 1.66056 \times 10^{-27} \ kg$.
Using Einstein's mass-energy equivalence relation,$E = mc^2$.
$E = (1.66056 \times 10^{-27} \ kg) \times (2.9979 \times 10^8 \ m/s)^2$.
$E \approx 1.4924 \times 10^{-10} \ J$.