23 What will be the mass of one atom of $\mathrm{C}-12$ in grams?
The mass of a carbon-12 atom was determined by a mass spectrometer and found to be equal to $1.992648 \times 10^{-23} \mathrm{~g}$. It is known that 1 mole of $\mathrm{C}-12$ atom weighing 12 g contains $\mathrm{N}_{\mathrm{A}}$ number of atoms. Thus,
1 mole of $\mathrm{C}-12$ atoms $=12 \mathrm{~g}=6.022 \times 10^{23}$ atoms
$\Rightarrow 6.022 \times 10^{23}$ atoms of C - 12 have mass $=12 \mathrm{~g}$
$\therefore \quad 1$ atom of $\mathrm{C}-12$ will have mass $=\frac{12}{6.022 \times 10^{23}} \mathrm{~g}$
$=1.992648 \times 10^{-23} \mathrm{~g} \approx 1.99 \times 10^{-23} \mathrm{~g}$
How many significant figures should be present in the answer of the following calculations?
$$\frac{2.5 \times 1.25 \times 3.5}{2.01}$$
Least precise term 2.5 or 3.5 has two significant figures.
Hence, the answer should have two significant figures
$$\frac{2.5 \times 1.25 \times 3.5}{2.01} \approx 5.4415=5.4$$
25 What is the symbol for SI unit of mole? How is the mole defined?
Symbol for SI unit of mole is mol.
One mole is defined as the amount of a substance that contains as many particles and there are atoms in exactly $12 \mathrm{~g}(0.012 \mathrm{~kg})$ of the ${ }^{12} \mathrm{C}$ - isotope.
$$\frac{1}{12} \mathrm{~g} \text { of }{ }^{12} \mathrm{C} \text {-isotope }=1 \mathrm{~mole}$$
What is the difference between molality and molarity?
Molality It is defined as the number of moles of solute dissolved in 1 kg of solvent. It is independent of temperature.
Molarity It is defined as the number of moles of solute dissolved in 1 L of solution. It depends upon temperature (because, volume of solution $\propto$ temperature).
Calculate the mass per cent of calcium, phosphorus and oxygen in calcium phosphate $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$.
$$\begin{aligned} \text { Mass per cent of calcium } & =\frac{3 \times(\text { atomic mass of calcium })}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{120 \mathrm{u}}{310 \mathrm{u}} \times 100=38.71 \% \end{aligned}$$
$$\begin{aligned} \text { Mass per cent of phosphorus } & =\frac{2 \times(\text { atomic mass of phosphorus) }}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{2 \times 31 \mathrm{u}}{310 \mathrm{u}} \times 100=20 \% \end{aligned}$$
$$\begin{aligned} \text { Mass per cent of oxygen } & =\frac{8 \times(\text { atomic mass of oxygen })}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{8 \times 16 \mathrm{u}}{310 \mathrm{u}} \times 100=41.29 \% \end{aligned}$$