It is because most of the elements occur in nature as a miture of isotopes and their atomic masses are the average relative atomic masses of the isotopes depending on their abundance.

The compound is glucose.Its molecular formula is C₆H₁₂O₆ while empirical formula is CH₂O.

1 mole of carbon atoms = 6.023 × 10²³ atoms of carbon
Mass of 1 atom of 12C =Atomic mass of C/Avogadro′s number
\(=  \frac{ 12}{6.022×10^{23}} g\)
= 1.9927 ×10^-23 g

1 mole of O₂ gas at STP = 6.022 ×× 1023 molecules of O₂
( Avogadro number ) = 32 g of O₂
Hence, 1 mole of oxygen gas is equal to molecular weight of oxygen as well as Avogadro number.

The SI unit of mass is kilogram kg.
The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram.

Molality is defined as the number of moles of solute present in 1kg of solvent. It is denoted by m.
Thus Molality (m)
\(=\frac { moles\ of\ solute }{ mass\ of\ solvent } \)
Given that, Mass of solvent (H₂O)
= 500g
= 0.5kg
Weight of HCI
= \(1\times 1+1\times 35.5=36.5g\)
Molar of HCI(solute) 
\(=\frac { 18.25 }{ 36.5 } =0.5\)
\(m=\frac { 0.5 }{ 0.5 } =1m\)

In the given question 0.9g L^-1 means that 1000 mL solution contains 0.9 g of glugose
Number of moles = 0.9g glucose = \(\frac { 0.9 }{ 180 } \)mol glugose
= \(5\times { 10 }^{ -3 }\)mol glugose.
(where molecular mass of glugose(C₂H₁₂O₆) = \(12\times 6+12\times 1+6\times 16=180u)\)
i.e IL solution contains 0.05 mole glugose or the molarity of glucose is 0.005M.

(a) Steps:
108   Ã⋅7.2 = 14.5833
Three S.F. Two S.F.
Here answer should have two significant figures. Therefore, correct answer, after rounding off two significant figures is 15.
(b) Steps: 2 x 1.6 x 10²
2 x 163.2 = 326.4 = 300
(c) Steps: (1.0042 - 0.0034) x 1.23
1.0008 x 1.23 = 1.230984
= 1.23

Molecular mass of \(\mathrm{MgSO}_4\)
=24+32+4 \times 16=120
Mass of \(\mathrm{Mg}=24\)
Mass percentage of \(\mathrm{Mg}=\frac{24}{120} \times 100=20 \%\)
Mass of \(\mathrm{S}=32\)
mass percentage of \(S=\frac{32}{120} \times 100=26.67 \%\)
Mass of \(\mathrm{O}=4 \times 16=64\)
Mass of percentage of \(\mathrm{O}=\frac{64}{120} \times 100=53.33 \%\)

1.54 ×10³pm μ s ^-1.

mass of \({ O }_{ 2\\ }\)molecule = \(=5.32\times { 10 }^{ -23 }g\) ;
mass of \({ SO }_{ 2 }\) molecule \(=1.06\times { 10 }^{ -22 }g\)

\(1 \mathrm{LO}_2 \text { or } \mathrm{SO}_2=\frac{1}{22.4} \mathrm{~mol}=\frac{1}{22.4} \times 6.02 \times 10^{23}\)
molecules \(=2.688 \times 10^{22}\) molecules
Mass of 1 molecule of \(O_2=\frac{1.43}{2.688 \times 10^2}=1.0629 \times 10^{-23} g\)
Gram Molecular mass of
\(\mathrm{O}_2=\text { Mass of } 22.4 \mathrm{~L} \text { at STP }=1.43 \times 22.4=32.032 g\)
Molecular mass of O₂=32.032 u
Similarly, molecular mass of SO₂ can be calculated.
(i) \({ 2.69\times 10 }^{ 22 }\) molecules of each
(ii) mass of \({ O }_{ 2\\ }\)molecule = \(=5.32\times { 10 }^{ -23 }g\) ;
mass of \({ SO }_{ 2 }\) molecule \(=1.06\times { 10 }^{ -22 }g\)
(iii) O₂ = 32.032 u and SO₂ = 63.997 u

Step 1. Conversion of mass per cent to grams : Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present.
Step 2. Convert into number moles of each element : Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound
Moles of hydrogen = \(\frac{4.07 \mathrm{~g}}{1.008 \mathrm{~g}}=4.04\)
Moles of carbon = \(\frac{24.27 \mathrm{~g}}{12.01 \mathrm{~g}}=2.021\)
Moles of chlorine = \(\frac{71.65 \mathrm{~g}}{35.453 \mathrm{~g}}=2.021\)
Step 3. Divide each of the mole values obtained above by the smallest number amongst them : Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl. In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.
Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements : CH₂Cl is, thus, the empirical formula of the above compound.
Step 5. Writing molecular formula : (a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula.
For CH₂Cl, empirical formula mass is
12.01 + (2 x 1.008) + 35.453
= 49.48 g
(b) Divide Molar mass by empirical formula mass
\(\frac{\text { Molar mass }}{\text { Empirical formula mass }}=\frac{98.96 \mathrm{~g}}{49.48 \mathrm{~g}}\)
= 2 = (n)
(c) Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula = CH₂Cl, n = 2. Hence molecular formula is C₂H₄Cl₂.

Weight in pounds = 12.0 in x 6.00 in \(\times 3.00 \text { in } \times \frac{2.54 \mathrm{~cm}}{1 \mathrm{in}} \times \frac{2.54 \mathrm{~cm}}{1 \mathrm{in}} \times \frac{2.54 \mathrm{~cm}}{1 \mathrm{in}} \times \frac{19.3 \mathrm{~g}}{1 \mathrm{~cm}^3} \times \frac{1 \mathrm{lb}}{453.6 \mathrm{~g}}\)
\(=150.60491 \mathrm{lb}=151 \mathrm{lb} \)(after rounding off because least precise term has 3 significant figures)

There are three common scales to measure temperature ^oC (degree celsius), ^oF(degree Fahrenheit) and K (kelvin). The K is the SI unit.
The temperature on two scales are related to each other by the following relationship \(^{ 0 }F=\frac { 9 }{ 5 } t^{ 0 }C+32\) 
Putting the values in above equation.
\(200-32=\frac { 9 }{ 5 } t^{ 0 }C\Rightarrow \frac { 9 }{ 5 } t^{ 0 }C=168\)   
\(\Rightarrow \ t^{ 0 }C=\frac { 168\times 5 }{ 9 } =93.3^{ 0 }C\)

(a) Given, C = 25^o C
\(^{ 0 }F=\frac { 9 }{ 5 } C^{ 0 }+32=\frac { 9 }{ 5 } \times 25+32=45+32=77^{ 0 }F\)
(b) Given, C = 35^o C
\(^{ 0 }F=\frac { 9 }{ 5 } C^{ 0 }+32=\frac { 9 }{ 5 } \times 35+32=63+32=95^{ 0 }F\)