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 $1: 2$.
Calculate the average atomic mass of hydrogen using the following data
| Isotope | % Natural abundance | Molar mass |
|---|---|---|
| $$^1$$H | 99.985 | 1 |
| $$^2$$H | 0.015 | 2 |
Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence), the average atomic mass of the element can be calculated as
$$\text { Average atomic mass }=\frac{\left.\text { (Natural abundance of }{ }^1 \mathrm{H} \times \text { molar mass) }+\left(\text { Natural abundance of }{ }^2 \mathrm{H} \times \text { molar mass of }{ }^2 \mathrm{H}\right)\right\}}{100}$$
$$\begin{aligned} & =\frac{99.985 \times 1+0.015 \times 2}{100} \\ & =\frac{99.985+0.030}{100}=\frac{100.015}{100}=1.00015 \mathrm{u} \end{aligned}$$
Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place
$$\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_2+\mathrm{H}_2$$
Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl .1 mol of a gas occupies 22.7 L volume at STP; atomic mass of $\mathrm{Zn}=65.3 \mathrm{u}$
Given that, Mass of $\mathrm{Zn}=32.65 \mathrm{~g}$
1 mole of gas occupies $=22.7 \mathrm{~L}$ volume at STP
Atomic mass of $\mathrm{Zn}=65.3 \mathrm{u}$
The given equation is
$$\underset{65.3 \mathrm{~g}}{\mathrm{Zn}}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_2+\underset{1 \mathrm{~mol}=22.7 \mathrm{~L} \text { at STP }}{\mathrm{H}_2}$$
From the above equation, it is clear that
65.3 g Zn , when reacts with HCl , produces $=22.7 \mathrm{~L}$ of $\mathrm{H}_2$ at STP
$\therefore 32.65 \mathrm{~g} \mathrm{Zn}$, when reacts with HCl , will produce $=\frac{22.7 \times 32.65}{65.3}=11.35 \mathrm{~L}$ of $\mathrm{H}_2$ at STP.