Assertion (A) Significant figures for 0.200 is 3 where as for 200 it is 1. Reason (R) Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.
Assertion (A) Combustion of 16 g of methane gives 18 g of water. Reason (R) In the combustion of methane, water is one of the products.
A vessel contains 1.6 g of dioxygen at STP ( $273.15 \mathrm{~K}, 1 \mathrm{~atm}$ pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate
(a) volume of the new vessel.
(b) number of molecules of dioxygen.
$$\text{(a)} \quad p_1=1 \mathrm{~atm}, p_2=\frac{1}{2}=0.5 \mathrm{~atm}, T_1=273.15, V_2=?, V_1=?$$
$$\begin{aligned} & 32 \mathrm{~g} \text { dioxygen occupies }=22.4 \mathrm{~L} \text { volume at STP } \\ & \therefore 1.6 \mathrm{~g} \text { dioxygen will occupy }=\frac{22.4 \mathrm{~L} \times 1.6 \mathrm{~g}}{32 \mathrm{~g}}=1.12 \mathrm{~L} \end{aligned}$$
$$V_1=1.12 \mathrm{~L}$$
$$\begin{aligned} &\text { From Boyle's law (as temperature is constant), }\\ &\begin{aligned} p_1 V_1 & =p_2 V_2 \\ V_2 & =\frac{p_1 V_1}{p_2} \end{aligned} \end{aligned}$$
$$=\frac{1 \mathrm{~atm} \times 1.12 \mathrm{~L}}{0.5 \mathrm{~atm}}=2.24 \mathrm{~L}$$
$$\begin{aligned} \text { (b) } \text { Number of moles of dioxygen } & =\frac{\text { Mass of dioxygen }}{\text { Molar mass of dioxygen }} \\ n_{\mathrm{O}_2} & =\frac{1.6}{32}=0.05 \mathrm{~mol} \end{aligned}$$
$$\begin{aligned} 1 \mathrm{~mol} \text { of dioxygen contains } & =6.022 \times 10^{23} \text { molecules of dioxygen } \\ \therefore \quad 0.05 \mathrm{~mol} \text { of dioxygen } & =6.022 \times 10^{23} \times 0.05 \text { molecule of } \mathrm{O}_2 \\ & =0.3011 \times 10^{23} \text { molecules } \\ & =3.011 \times 10^{22} \text { molecules } \end{aligned}$$
Calcium carbonate reacts with aqueous HCl to give $\mathrm{CaCl}_2$ and $\mathrm{CO}_2$ according to the reaction given below
$$\mathrm{CaCO}_3(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{CaCl}_2(a q)+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)$$
What mass of $\mathrm{CaCl}_2$ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of $\mathrm{CaCO}_3$ ? Name the limiting reagent. Calculate the number of moles of $\mathrm{CaCl}_2$ formed in the reaction.
Molar mass of $\mathrm{CaCO}_3=40+12+3 \times 16=100 \mathrm{~g} \mathrm{~mol}^{-1}$
Moles of $\mathrm{CaCO}_3$ in $1000 \mathrm{~g}, n_{\mathrm{CaCO}_3}=\frac{\text { Mass }(\mathrm{g})}{\text { Molar mass }}$
$$\begin{aligned} &\begin{aligned} n_{\mathrm{CaCO}_3} & =\frac{1000 \mathrm{~g}}{100 \mathrm{~g} \mathrm{~mol}^{-1}}=10 \mathrm{~mol} \\ \text { Molarity } & =\frac{\text { Moles of solute }(\mathrm{HCl}) \times 1000}{\text { Volume of solution }} \end{aligned}\\ &\text { (It is given that moles of } \mathrm{HCl} \text { in } 250 \mathrm{~mL} \text { of } 0.76 \mathrm{M} \mathrm{HCl}=n_{\mathrm{HCI}} \text { ) }\\ &\begin{aligned} 0.76 & =\frac{n_{\mathrm{HCl}} \times 1000}{250} \\ n_{\mathrm{HCl}} & =\frac{0.76 \times 250}{1000}=0.19 \mathrm{~mol} . \end{aligned} \end{aligned}$$
$$\underset{1 \mathrm{~mol}}{\mathrm{CaCO}_3(\mathrm{~s})}+\underset{2 \mathrm{~mol}}{2 \mathrm{HCl}(\mathrm{aq})} \longrightarrow \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)$$
According to the equation,
1 mole of $\mathrm{CaCO}_3$ reacts with 2 moles HCl
$\therefore \quad 10$ moles of $\mathrm{CaCO}_3$ will react with $\frac{10 \times 2}{1}=20$ moles HCl .
But we have only 0.19 moles HCl , so HCl is limiting reagent and it limits the yield of $\mathrm{CaCl}_2$.
Since, $\quad 2$ moles of HCl produces 1 mole of $\mathrm{CaCl}_2$
$$\begin{aligned} 0.19 \text { mole of } \mathrm{HCl} \text { will produce } \frac{1 \times 0.19}{2} & =0.095 \mathrm{~mol} \mathrm{CaCl}_2 \\ \text { Molar mass of } \mathrm{CaCl}_2 & =40+(2 \times 35.5)=111 \mathrm{~g} \mathrm{~mol}^{-1} \\ \therefore \quad 0.095 \text { mole of } \mathrm{CaCl}_2 & =0.095 \times 111=10.54 \mathrm{~g} \end{aligned}$$
Define the law of multiple proportions. Explain it with two examples. How does this law point to the existence of atoms?
'Law of multiple proportions' was first studied by Dalton in 1803 which may be defined as follows
When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other, bear a simple ratio to one another.
e.g., hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.
$$\begin{aligned} &\begin{aligned} & \underset{2 \mathrm{~g}}{\text { Hydrogen }}+\underset{32 \mathrm{~g}}{\text { Oxygen }} \longrightarrow \underset{34 \mathrm{~g}}{\text { Hydrogen peroxide }} \end{aligned}\\ &\underset{2 \mathrm{~g}}{\text { Hydrogen }}+\underset{32 \mathrm{~g}}{\text { Oxygen }} \longrightarrow \underset{34 \mathrm{~g}}{\text { Hydrogen peroxide }} \end{aligned}$$
Here, the masses of oxygen (i.e., 16 g and 32 g ) which combine with a fixed mass of hydrogen $(2 \mathrm{~g})$ bear a simple ratio, i.e., $16: 32$ or $1: 2$.
As we know that, when compounds mixed in different proportionation, Then they form different compounds. In the above examples, when hydrogen is mixed with different proportion of oxygen, then they form water or hydrogen peroxide.
It shows that there are constituents which combine in a definite proportion. These constituents may be atoms. Thus, the law of multiple proportions shows the existence of atoms which combine into molecules.