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 $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$.
$$\begin{aligned} \text { Mass per cent of calcium } & =\frac{3 \times(\text { atomic mass of calcium })}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{120 \mathrm{u}}{310 \mathrm{u}} \times 100=38.71 \% \end{aligned}$$
$$\begin{aligned} \text { Mass per cent of phosphorus } & =\frac{2 \times(\text { atomic mass of phosphorus) }}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{2 \times 31 \mathrm{u}}{310 \mathrm{u}} \times 100=20 \% \end{aligned}$$
$$\begin{aligned} \text { Mass per cent of oxygen } & =\frac{8 \times(\text { atomic mass of oxygen })}{\text { molecular mass of } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2} \times 100 \\ & =\frac{8 \times 16 \mathrm{u}}{310 \mathrm{u}} \times 100=41.29 \% \end{aligned}$$
45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below
$$2 \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g})$$
Which law is being obeyed in this experiment? Write the statement of the law?
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.
If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio.
(a) Is this statement true?
(b) If yes, according to which law?
(c) Give one example related to this law.
(a) Yes, the given statement is true.
(b) According to the law of multiple proportions
$$\text { (c) } \underset{2 \mathrm{~g}}{\mathrm{H}_2}+\underset{16 \mathrm{~g}}{\mathrm{O}_2} \longrightarrow \underset{18 \mathrm{~g}}{\mathrm{H}_2 \mathrm{O}}$$
$$\qquad \underset{2 g}{\mathrm{H}_2}+\underset{32 \mathrm{~g}}{\mathrm{O}_2} \longrightarrow \underset{34 \mathrm{~g}}{\mathrm{H}_2 \mathrm{O}_2}$$
Here, masses of oxygen, (i.e., 16 g in $\mathrm{H}_2 \mathrm{O}$ and 32 g in $\mathrm{H}_2 \mathrm{O}_2$ ) which combine with fixed mass of hydrogen $(2 \mathrm{~g})$ are in the simple ratio i.e., $16: 32$ or $1: 2$.