Some Basic Concepts Of Chemistry – JEE Main ChemistryPractice Questions & PYQs

Calculating the total number of atoms in a given mass of a substance relies on first determining the molar amount, multiplying by Avogadro's number ($N_{A}$) to find the number of molecules or formula units, and finally multiplying by the atomicity, which is the total count of atoms within each individual molecule or unit. It is essential to convert all mass values from milligrams to grams by multiplying by $10^{-3}$ before proceeding with the molar calculations.

For $1.8 \, \mathrm{mg}$ of water, the number of moles is $\frac{1.8 \times 10^{-3}}{18} = 10^{-4}$. Given that water consists of 3 atoms per molecule, the total count is $10^{-4} \times 3 \times N_{A}$, which equals $3 \times 10^{-4} N_{A}$. In the case of $9.8 \, \mathrm{mg}$ of sulphuric acid, the moles are calculated as $\frac{9.8 \times 10^{-3}}{98} = 10^{-4}$. Since each molecule of sulphuric acid comprises 7 atoms, the product is $10^{-4} \times 7 \times N_{A}$, resulting in $7 \times 10^{-4} N_{A}$.

Moving to $1.8 \, \mathrm{mg}$ of carbon, the moles are $\frac{1.8 \times 10^{-3}}{12} = 1.5 \times 10^{-4}$. As carbon is a monatomic element, the atom count directly corresponds to $1.5 \times 10^{-4} \times N_{A}$. Finally, for $5.85 \, \mathrm{mg}$ of sodium chloride, the moles are $\frac{5.85 \times 10^{-3}}{58.5} = 10^{-4}$. Because one formula unit of sodium chloride contains 2 atoms, multiplying by 2 yields $2 \times 10^{-4} \times N_{A}$.

The reaction stoichiometry governs the amount of reactant consumed and the excess reactant remaining. The molar mass of $\mathrm{CaCO}_3$ is calculated as $40+12+(3 \times 16) = 100$ g/mol, while the molar mass of $\mathrm{HCl}$ is $1+35.5 = 36.5$ g/mol. The number of moles of $\mathrm{CaCO}_3$ is $90 / 100 = 0.9$ mol. The total mass of the $\mathrm{HCl}$ solution is $300 \times 1.13 = 339$ g, and the mass of pure $\mathrm{HCl}$ present in the solution is $339 \times 0.3855 = 130.6845$ g.

$\mathrm{CaCO}_3 + 2\mathrm{HCl} \longrightarrow \mathrm{CaCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$

According to the balanced equation, one mole of $\mathrm{CaCO}_3$ reacts with two moles of $\mathrm{HCl}$. Therefore, $0.9$ mol of $\mathrm{CaCO}_3$ requires $0.9 \times 2 = 1.8$ mol of $\mathrm{HCl}$ for complete reaction. The mass of $\mathrm{HCl}$ reacted is $1.8 \times 36.5 = 65.7$ g. The amount of $\mathrm{HCl}$ remaining is the initial mass minus the reacted mass, which is $130.6845 - 65.7 = 64.9845$ g, confirming that approximately $64.97$ g remains unreacted.

The neutralization reaction is represented by the balanced equation $2\text{HCl} + \text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2\text{H}_2\text{O}$, which indicates that one mole of $\text{Ba(OH)}_2$ reacts with two moles of $\text{HCl}$.

The number of moles of $\text{Ba(OH)}_2$ present in the titration is calculated as $0.1 \times 0.025 = 0.0025$ mol. According to the stoichiometry of the reaction, the required number of moles of $\text{HCl}$ is $n = 2 \times 0.0025 = 0.005$ mol.

The mass of $\text{HCl}$ is determined by multiplying the number of moles by its molar mass, yielding $0.005 \times 36.5 = 0.1825$ g. Converting this mass to milligrams gives $0.1825 \times 1000 = 182.5$ mg. Since the value $x$ is defined such that $x = 182.5$ and the result is required in the form $x \times 10^{-1}$, then $182.5 = 1825 \times 10^{-1}$, making the required numerical value 1825.

The stoichiometric reaction is represented by $Ca + 2HCl \rightarrow CaCl_2 + H_2$. The number of moles of calcium is determined by $n = 14.0 \text{ g} / 40 \text{ g/mol} = 0.35 \text{ mol}$. Given that hydrochloric acid is in excess, calcium serves as the limiting reagent, and the reaction produces 0.35 mol of both $CaCl_2$ and $H_2$. The mass of $CaCl_2$ formed is calculated as $0.35 \text{ mol} \times (40 + 2 \times 35.5) \text{ g/mol} = 0.35 \times 111 \text{ g/mol} = 38.85 \text{ g}$, which invalidates the claim that 33.3 g of $CaCl_2$ is produced. Furthermore, at 273 K and 1 atm, the volume of $H_2$ evolved is $0.35 \text{ mol} \times 22.4 \text{ L/mol} = 7.84 \text{ L}$, confirming the correctness of the remaining statements.

The mass of carbon is derived from the amount of $\mathrm{CO}_2$ produced using the expression $\frac{12}{44} \times 0.1837 = 0.0501$ g, while the mass of hydrogen is derived from the amount of $\mathrm{H}_2\mathrm{O}$ produced using $\frac{2}{18} \times 0.15 = 0.0167$ g. The mass of oxygen in the organic compound is determined by subtracting the combined mass of carbon and hydrogen from the initial mass, $\frac{0.25 - (0.0501 + 0.0167)}{0.25} \times 100 \approx 73.28\%$. Rounding this result to the nearest integer gives 73.

The thermal decomposition of silver carbonate follows a stoichiometric relationship where the solid reactant converts into solid metallic silver and gaseous byproducts. Upon heating, $\mathrm{Ag}_2\mathrm{CO}_3(\mathrm{s})$ decomposes according to the balanced equation $\mathrm{Ag}_2\mathrm{CO}_3(\mathrm{s}) \rightarrow 2\mathrm{Ag}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g})$. Since carbon dioxide and oxygen escape as gases, the remaining solid residue consists exclusively of silver.

Determining the mass of this residue requires converting the reactant mass to moles using its molar mass, which is calculated as $(2 \times 108) + 12 + (3 \times 16) = 276 \text{ g mol}^{-1}$. Dividing the initial 2.76 g by this molar mass confirms the presence of 0.01 mol of silver carbonate. The stoichiometry of the reaction dictates that 1 mole of $\mathrm{Ag}_2\mathrm{CO}_3$ yields 2 moles of silver, meaning 0.01 mol of the reactant produces 0.02 mol of $\mathrm{Ag}$. Converting this molar quantity back to grams by multiplying by the atomic mass of silver yields $0.02 \times 108$, resulting in a final residue mass of 2.16 g.

Chemical reactions proceed according to fixed stoichiometric ratios, where the actual yield of a product is restricted by whichever reactant is fully consumed first, known as the limiting reagent. For the reaction $\mathrm{Zn}(\mathrm{s}) + \mathrm{H}_2\mathrm{SO}_4(\mathrm{aq}) \rightarrow \mathrm{ZnSO}_4(\mathrm{aq}) + \mathrm{H}_2(\mathrm{g})$, this principle dictates that the number of moles of hydrogen gas formed is determined strictly by the amount of zinc reacting, provided there is an excess of sulfuric acid.

The total mass of the sulfuric acid solution is 65 g, derived from 50 mL of acid with a density of 1.3 g/mL. Given the 50 percent concentration, the mass of pure sulfuric acid is 32.5 g, which corresponds to approximately 0.33 moles when divided by its molar mass of 98 g/mol. Since 20 g of zinc corresponds to $\frac{20}{65}$ or approximately 0.31 moles, zinc is the limiting reagent in this mixture. Consequently, the reaction will consume all 0.31 moles of zinc to produce an equivalent amount of hydrogen gas. Multiplying this stoichiometric quantity by the standard molar volume of 22.4 L/mol at STP leads to the liberation of 6.892 L of hydrogen.

Calculating the total number of atoms requires finding the total moles of molecules and multiplying by the number of atoms in a single molecule. For 2 moles of cyclohexane ($\mathrm{C}_6\mathrm{H}_{12}$), where each molecule contains 18 atoms, the total is $2 \times 18 = 36$ moles of atoms. Regarding 684 g of sucrose ($\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}$), dividing the mass by the molar mass of 342 g/mol yields 2 moles of sucrose, and since one molecule contains 45 atoms, the total is $2 \times 45 = 90$ moles of atoms. For 90.8 L of dihydrogen gas at STP, dividing by the standard molar volume of 22.7 L/mol gives 4 moles of $\mathrm{H}_2$, which results in $4 \times 2 = 8$ moles of atoms. Comparing these values shows that sucrose provides 90 moles of atoms, cyclohexane provides 36, and dihydrogen provides 8, establishing an order where sucrose exceeds cyclohexane, which in turn exceeds dihydrogen.

A primary standard must be chemically stable, highly pure, and non-reactive with atmospheric components like moisture or carbon dioxide, while undergoing rapid and stoichiometric reactions to ensure precise titrimetric analysis. Unlike the requirements stated in D and E, a reliable primary standard must be highly soluble in water and possess a high molar mass to minimize the relative error during mass measurements. Because substances such as oxalic acid dihydrate function effectively as primary standards, statements A, B, and C are valid criteria, while the suggestions that primary standards should have low solubility or low molar mass are incorrect.

Stoichiometry relies on the law of definite proportions, where a balanced chemical equation provides the fixed molar ratios necessary to determine the quantities of reactants involved in a reaction. The interaction between iron and steam is represented by the equation

$3\,\mathrm{Fe}(s) + 4\,\mathrm{H}_2\mathrm{O}(g) \longrightarrow \mathrm{Fe}_3\mathrm{O}_4(s) + 4\,\mathrm{H}_2(g)$

Dividing the $18\,\mathrm{g}$ of steam by its molar mass of $18\,\mathrm{g}\,\mathrm{mol}^{-1}$ reveals that $1\,\mathrm{mol}$ of water is reacting. Because the balanced equation shows that four moles of steam consume three moles of iron, reacting one mole of steam requires $0.75\,\mathrm{mol}$ of iron. Consequently, the total mass of iron consumed is $0.75\,\mathrm{mol} \times 56\,\mathrm{g}\,\mathrm{mol}^{-1}$, which equals $42\,\mathrm{g}$.

Mole fraction represents the proportion of moles of a specific component relative to the total number of moles in a solution. For a solution defined as 10% by weight of urea, every 100 g of the mixture consists of 10 g of urea and 90 g of water, which serves as the solvent.

First, determine the number of moles for each component by dividing the mass by the respective molar mass. The molar mass of urea ($NH_2CONH_2$) is $14(2) + 1(4) + 12 + 16 = 60\,\mathrm{g}\,\mathrm{mol}^{-1}$, resulting in $n_{\mathrm{urea}} = \frac{10}{60} \approx 0.167\,\mathrm{mol}$. The molar mass of water ($H_2O$) is $1(2) + 16 = 18\,\mathrm{g}\,\mathrm{mol}^{-1}$, which gives $n_{\mathrm{water}} = \frac{90}{18} = 5\,\mathrm{mol}$.

The mole fraction of water, $x_{\mathrm{water}}$, is calculated by dividing the moles of water by the total number of moles present in the solution. This is expressed as $x_{\mathrm{water}} = \frac{n_{\mathrm{water}}}{n_{\mathrm{urea}} + n_{\mathrm{water}}} = \frac{5}{0.167 + 5} = \frac{5}{5.167} \approx 0.967$.

To determine the molecular formula of the hydrocarbon, compare the given mass ratio of carbon to hydrogen, which is 12:1, against their respective atomic masses of 12 and 1; this confirms that the molar ratio of carbon to hydrogen is 1:1. Given that the molecule contains two carbon atoms, the chemical formula is identified as $\mathrm{C}_{2}\mathrm{H}_{2}$, resulting in a total molar mass of 26 $\mathrm{g\,mol}^{-1}$. Using the combustion stoichiometry $\mathrm{C}_{2}\mathrm{H}_{2}(\mathrm{g}) + \frac{5}{2}\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(l)$, we see that one mole of the hydrocarbon yields two moles of carbon dioxide. Since 3.38 g of the hydrocarbon corresponds to $\frac{3.38}{26}$ moles, the amount of carbon dioxide produced is $\left( \frac{3.38}{26} \times 2 \right)$ moles. Multiplying this value by the molar mass of $\mathrm{CO}_{2}$, which is 44 $\mathrm{g\,mol}^{-1}$, gives a total mass of 11.44 g.

To determine the number of atoms in a substance, multiply the number of moles by the atomicity of the molecule and Avogadro's constant, $N_{\mathrm{A}}$. For 2 grams of $\mathrm{O}_2$, the number of atoms is calculated as $\frac{2}{32} \times 2 N_{\mathrm{A}} = 0.125 N_{\mathrm{A}}$. For 4 grams of $\mathrm{SO}_2$, the calculation is $\frac{4}{64} \times 3 N_{\mathrm{A}} = 0.1875 N_{\mathrm{A}}$. Considering 1400 mL of $\mathrm{O}_2$ gas at STP, the atoms present are $\frac{1400}{22400} \times 2 N_{\mathrm{A}} = 0.125 N_{\mathrm{A}}$. The 0.05 L of helium at STP contains $\frac{0.05}{22.4} N_{\mathrm{A}} \approx 0.002 N_{\mathrm{A}}$ atoms. Finally, 0.0625 moles of $\mathrm{H}_2$ gas contains $0.0625 \times 2 N_{\mathrm{A}} = 0.125 N_{\mathrm{A}}$ atoms. These results demonstrate that the samples in A, C, and E contain the same number of atoms.

The limiting reagent determines the extent of a chemical reaction, calculated by identifying which reactant produces the smallest amount of product based on the stoichiometry of $A + 2B \rightarrow AB_2$. The initial moles available are $n_A = \frac{36.0}{60} = 0.60\,\text{mol}$ and $n_B = \frac{56.0}{80} = 0.70\,\text{mol}$. Because the stoichiometry requires 2 moles of B for every 1 mole of A, the available 0.70 mol of B is insufficient to react with all 0.60 mol of A, identifying B as the limiting reagent and A as the excess reagent, which renders statement A incorrect.

The amount of product $AB_2$ formed is governed by the limiting reagent B, where 2 moles of B yield 1 mole of $AB_2$, resulting in $n_{AB_2} = \frac{0.70}{2} = 0.35\,\text{mol}$. The molar mass of $AB_2$ is $60 + 2(80) = 220\,\text{g mol}^{-1}$, which makes statement C incorrect, while the mass of $AB_2$ produced is $0.35 \times 220 = 77.0\,\text{g}$, confirming statement B is correct. Finally, the unreacted mass of A is calculated by subtracting the consumed moles of A from the initial moles, yielding $(0.60 - 0.35) \times 60 = 0.25 \times 60 = 15.0\,\text{g}$, which validates statement D.

Determining an empirical formula relies on converting the mass percentage composition of elements into their respective molar ratios based on their atomic masses. Assuming a 100 g sample, the compound consists of 69.9 g of iron and 30.1 g of oxygen, which corresponds to $69.9 / 56 \approx 1.248$ moles of iron and $30.1 / 16 \approx 1.881$ moles of oxygen. Dividing these molar amounts by the smaller value of 1.248 results in a molar ratio of $1 : 1.5$, which simplifies to a whole-number ratio of $2 : 3$ when both values are multiplied by two, leading to the empirical formula $\mathrm{Fe}_{2}\mathrm{O}_{3}$.

The principle of atom conservation implies that the phosphorus content in the magnesium pyrophosphate ($Mg_2P_2O_7$) originates entirely from the compound $(X)$. The molar mass of $Mg_2P_2O_7$ equals $2(24) + 2(31) + 7(16) = 222$ g mol$^{-1}$. The number of moles of phosphorus in the 1.79 g of product is $\frac{1.79 \times 2}{222} = \frac{1.79}{111} \approx 0.0161$ mol. The mass of phosphorus is calculated as $0.0161 \times 31 \approx 0.50$ g. The percentage of phosphorus in the 1.00 g sample is $\frac{0.50}{1.00} \times 100 = 50\%$.

The stoichiometry of the reaction $2 Al(s)+6 HCl(aq) \longrightarrow 2 Al^{3+}(aq)+6 Cl^{-}(aq)+3 H_2(g)$ establishes that 6 moles of $HCl$ react to produce 3 moles of $H_2$ gas. This molar ratio simplifies to a reaction where 1 mole of $HCl$ produces 0.5 moles of $H_2$. Applying the ideal molar volume of a gas at STP, which is 22.4 L/mol, the volume of $H_2$ produced per mole of $HCl$ is $0.5 \text{ mol} × 22.4 L/\text{mol} = 11.2 L$.

The balanced chemical equation for the reaction between manganese dioxide and hydrochloric acid is given by $MnO _2(s) + 4HCl(aq) \rightarrow MnCl _2(aq) + Cl _2(g) + 2H _2O(l)$. First, calculate the molar mass of $MnO _2$ as $55 + 2 \times 16 = 87 \text{ g mol}^{-1}$, which allows for the determination of moles of $MnO _2$ used as $\frac{8.7 \text{ g}}{87 \text{ g mol}^{-1}} = 0.1 \text{ mol}$. According to the stoichiometric coefficients in the balanced equation, $1 \text{ mole}$ of $MnO _2$ produces $1 \text{ mole}$ of $Cl _2$ gas. Therefore, $0.1 \text{ mol}$ of $MnO _2$ results in the production of $0.1 \text{ mol}$ of $Cl _2$. Using the molar mass of $Cl _2$, which is $2 \times 35.5 = 71 \text{ g mol}^{-1}$, the final mass of $Cl _2$ liberated is $0.1 \text{ mol} \times 71 \text{ g mol}^{-1} = 7.1 \text{ g}$.

At Standard Temperature and Pressure, the molar volume of an ideal gas is defined as $22.7 \text{ L/mol}$. To determine the number of moles in a given volume, we use the formula $n = \frac{V}{22.7}$. By substituting the sample volume of $1.4187 \text{ L}$ into this expression, we obtain $0.0633$ moles of $\mathrm{SO}_2$. To find the total number of molecules, we multiply the number of moles by Avogadro's constant, which is $6.023 \times 10^{23} \text{ mol}^{-1}$. Performing the calculation $0.0633 \times 6.023 \times 10^{23}$ yields $3.812 \times 10^{22}$ molecules.

The reaction between benzoic acid and sodium bicarbonate proceeds in a one-to-one molar ratio, meaning every mole of $\mathrm{C_6H_5COOH}$ reacts to release exactly one mole of $\mathrm{CO_2}$ gas. To determine the quantity of acid used, we first convert the given volume of 11.2 L at standard temperature and pressure into moles by dividing by the standard molar volume of 22.4 L per mole, which results in 0.5 moles of gas. Since the stoichiometry dictates an equal molar ratio, the reaction must have involved 0.5 moles of benzoic acid. Calculating the mass by multiplying this molar quantity by the molar mass of benzoic acid, 122 g/mol, yields the final value of 61 g.