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.
The density of 3 molal solution of NaOH is $1.110 \mathrm{~g} \mathrm{~mL}^{-1}$. Calculate the molarity of the solution.
3 molal solution of NaOH means 3 moles of NaOH are dissolved in 1 kg solvent. So, the mass of solution $=1000 \mathrm{~g}$ solvent $+120 \mathrm{~g} \mathrm{NaOH}=1120 \mathrm{~g}$ solution
(Molar mass of $\mathrm{NaOH}=23+16+1=40 \mathrm{~g}$ and 3 moles of $\mathrm{NaOH}=3 \times 40=120 \mathrm{~g}$ )
$$\text { Volume of solution }=\frac{\text { Mass of solution }}{\text { Density of solution }}\quad \left(\mathrm{Q} d=\frac{m}{V}\right)$$
$$V=\frac{1120 \mathrm{~g}}{1.110 \mathrm{~g} \mathrm{~mL}^{-1}}=1009 \mathrm{~mL}$$
$$\begin{aligned} \text { Molarity } & =\frac{\text { Moles of solute } \times 1000}{\text { Volume of solution }(\mathrm{mL})} \\ & =\frac{3 \times 1000}{1009}=2.973 \mathrm{M} \approx 3 \mathrm{M} \end{aligned}$$
Volume of a solution changes with change in temperature, then what will the molality of the solution be affected by temperature? Give reason for your answer.
No, molality of solution does not change with temperature since mass remains unaffected with temperature.
$$\text { Molality, } m=\frac{\text { moles of solute }}{\text { weight of solvent (in } \mathrm{g})} \times 1000$$