Least precise term 2.5 or 3.5 has two significant figures.

Hence, the answer should have two significant figures

$$\frac{2.5 \times 1.25 \times 3.5}{2.01} \approx 5.4415=5.4$$

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.

$$\frac{1}{12} \mathrm{~g} \text { of }{ }^{12} \mathrm{C} \text {-isotope }=1 \mathrm{~mole}$$

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).

$$\begin{aligned} \text { Mass per cent of calcium } & =\frac{3 \times(\text { atomic mass of calcium })}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{120 \mathrm{u}}{310 \mathrm{u}} \times 100=38.71 \% \end{aligned}$$

$$\begin{aligned} \text { Mass per cent of phosphorus } & =\frac{2 \times(\text { atomic mass of phosphorus) }}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{2 \times 31 \mathrm{u}}{310 \mathrm{u}} \times 100=20 \% \end{aligned}$$

$$\begin{aligned} \text { Mass per cent of oxygen } & =\frac{8 \times(\text { atomic mass of oxygen })}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{8 \times 16 \mathrm{u}}{310 \mathrm{u}} \times 100=41.29 \% \end{aligned}$$

For the reaction,

$$\underset{2\mathrm{V}}{2 \mathrm{~N}_2(g)}+\underset{1 \mathrm{V}}{\mathrm{O}_2(g)} \longrightarrow \underset{2 \mathrm{~V}}{2 \mathrm{~N}_2 \mathrm{O}}(\mathrm{g})$$

Hence, the ratio between the volumes of the reactants and the product in the given question is simple i.e., $2: 1: 2$. It proves the Gay-Lussac's law of gaseous volumes.