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

The determination of a molecular formula from combustion products relies on the principle of conservation of mass, which dictates that the number of atoms for each element must remain consistent between the reactants and the products. Since the combustion of the organic compound produces 6 moles of $\mathrm{CO}_2$, we can deduce that the original compound must contain 6 carbon atoms to account for this carbon output. Similarly, the production of 4 moles of $\mathrm{H}_2\mathrm{O}$ indicates the presence of 8 hydrogen atoms, given that each water molecule contributes two hydrogen atoms. Finally, the release of 1 mole of nitrogen gas, which is diatomic $\mathrm{N}_2$, confirms that 2 nitrogen atoms are present in the structure, leading to the molecular formula $\mathrm{C}_6\mathrm{H}_8\mathrm{N}_2$.

Determining the mole percentage of carbon in methane requires identifying the total ratio of atoms within a single molecule of $CH_4$. Because every molecule contains one carbon atom and four hydrogen atoms, there are five total moles of atoms for every mole of methane.

To find the composition by mole, divide the quantity of carbon by the total number of atoms and multiply the result by one hundred.

$\text{Mole \% of carbon} = \frac{1}{5} \times 100 = 20\%$

The concept of a limiting reagent is determined by the stoichiometric proportions defined in a balanced chemical equation, where the substance that runs out first dictates the amount of product formed. In the reaction $\mathrm{N}_2(g)+3\,\mathrm{H}_2(g) \rightarrow 2\,\mathrm{NH}_3(g)$, one mole of nitrogen requires three moles of dihydrogen to react fully. When considering the mixture of 56g of $\mathrm{N}_2$ and 10g of $\mathrm{H}_2$, we first convert these masses to moles, finding that 56g of $\mathrm{N}_2$ represents 2 moles and 10g of $\mathrm{H}_2$ represents 5 moles. Because the 2 moles of $\mathrm{N}_2$ would stoichiometrically require 6 moles of $\mathrm{H}_2$ for a complete reaction but only 5 moles are actually available, the dihydrogen is the reactant that limits the extent of the process.

Molality is defined as the number of moles of solute dissolved per kilogram of solvent, providing a temperature-independent way to express concentration where $\text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$. A 20% mass by mass aqueous solution indicates that every 100g of the mixture contains 20g of potassium iodide and 80g of water. Dividing the mass of potassium iodide by its molar mass of 166 g mol^-1 gives 0.1204 mol of solute. Converting the solvent mass of 80g to 0.08 kg and dividing the moles of solute by this value results in a final molality of 1.51 mol kg^-1.

The principle of conservation of mass dictates that when a hydrocarbon burns, every atom of carbon and hydrogen initially present in the sample must be accounted for in the resulting products, $CO_2$ and $H_2O$. Since these products serve as the only receptacles for the fuel's elements, we can determine the original composition by identifying the mass contribution of each element within the molecules produced. Analyzing the burning of 25 g of the unknown compound reveals that the 88 g of $CO_2$ contains exactly 24 g of carbon, while the 9 g of $H_2O$ formed accounts for the remaining 1 g of hydrogen. Together, these elements equal the total mass of the 25 g sample, confirming that the hydrocarbon consists of 24 g of carbon and 1 g of hydrogen.

The stoichiometry of the reaction between sodium bicarbonate ($NaHCO_3$) and oxalic acid ($H_2C_2O_4$) follows the reaction $NaHCO_3 + H^+ \rightarrow Na^+ + H_2O + CO_2$, which dictates that one mole of sodium bicarbonate produces one mole of carbon dioxide ($CO_2$). The number of moles of $CO_2$ released from the tablet is determined by dividing the volume of the gas by the given molar volume:

$n_{CO_2} = \frac{V}{V_m} = \frac{0.25 \times 10^{-3} \text{ L}}{25.0 \text{ L mol}^{-1}} = 1.0 \times 10^{-5} \text{ mol}$

Since the molar ratio of sodium bicarbonate to carbon dioxide is 1:1, the number of moles of sodium bicarbonate in the tablet is equal to the moles of $CO_2$ produced. Using the given molar mass of $84 \text{ g mol}^{-1}$, the mass of sodium bicarbonate in the tablet is calculated as:

$m_{NaHCO_3} = n \times M = 1.0 \times 10^{-5} \text{ mol} \times 84 \text{ g mol}^{-1} = 8.4 \times 10^{-4} \text{ g} = 0.84 \text{ mg}$

The percentage of sodium bicarbonate in the 10 mg tablet is calculated by dividing the mass of the bicarbonate by the total mass of the tablet and multiplying by 100:

$\text{Percentage} = \frac{0.84 \text{ mg}}{10 \text{ mg}} \times 100 = 8.4\%$

Molality represents the moles of solute per kilogram of solvent, providing a direct link between the molar ratio of components and the physical mass of the solvent. Given that the sum of mole fractions for all components in a solution must equal one, the mole fraction of the solute is calculated as $1 - 0.8 = 0.2$. By assuming a total of one mole for the mixture, we identify 0.2 moles of solute and 0.8 moles of water. To determine the mass of the solvent, we multiply 0.8 moles of water by its molar mass of 18 g/mol, which gives 14.4 g or 0.0144 kg. Dividing the solute moles by the solvent mass provides the final result: $m = \frac{0.2}{0.0144} = 13.88 \text{ mol kg}^{-1}$.

Determining the molar mass of individual elements within a compound relies on the relationship between the total mass of a sample and the number of moles present, where molar mass is calculated by dividing total mass by the amount in moles. For $AB_{2}$, dividing 125 x $10^{-3}$ kg by 5 mol gives a molar mass of 25 x $10^{-3}$ kg mol$^{-1}$, which establishes the algebraic relationship
$M_{A} + 2 M_{B} = 25 \times 10^{-3}$
Similarly, for $A_{2} B_{2}$, dividing 300 x $10^{-3}$ kg by 10 mol yields a molar mass of 30 x $10^{-3}$ kg mol$^{-1}$, resulting in the second equation
$2 M_{A} + 2 M_{B} = 30 \times 10^{-3}$
Subtracting the first equation from the second eliminates $M_{B}$ to isolate $M_{A}$, providing
$M_{A} = 30 \times 10^{-3} - 25 \times 10^{-3} = 5 \times 10^{-3}$
Substituting this calculated molar mass into the initial equation yields 5 x $10^{-3}$ + 2 $M_{B}$ = 25 x $10^{-3}$, which simplifies to 2 $M_{B}$ = 20 x $10^{-3}$ and identifies $M_{B}$ as 10 x $10^{-3}$ kg mol$^{-1}$.

The law of equivalence is the fundamental principle for titration, stating that at the equivalence point, the equivalents of acid must equal the equivalents of base. An equivalent is the product of the solution volume, molarity, and the n-factor, which represents the stoichiometry of the proton or hydroxide ion exchange during the reaction.

In the first reaction, hydrochloric acid reacts with sodium carbonate, which has an n-factor of two. The equivalence relationship is:
$25 \times M \times 1 = 30 \times 0.1 \times 2$
Solving this confirms that the molarity, $M$, of the hydrochloric acid is 0.24 M.

In the subsequent reaction with sodium hydroxide, both reactants have an n-factor of one. The equivalence relationship is defined as:
$0.24 \times V \times 1 = 30 \times 0.2 \times 1$
Determining the value of $V$ indicates that 25 mL of the hydrochloric acid solution is required.

Molarity defines the concentration of a solution by relating the number of moles of solute to the volume of the solution in liters. To determine the mass of sugar needed, use the relationship:

$M = \frac{\text{mass}}{\text{Molar Mass} \times V}$

The molar mass of sucrose ($\mathrm{C_{12}H_{22}O_{11}}$) is calculated by adding the atomic masses of its constituent atoms, resulting in 342 g/mol. Substituting the given values into the molarity equation provides $0.1 = \frac{\text{mass}}{342 \times 2}$. Multiplying 342 by 2 and then by 0.1 yields a total mass of 68.4 g.

Stoichiometry enables us to predict product yields by applying the molar ratios established by the balanced chemical equation. With a molar mass of 890 g/mol, the 445 g sample of $\mathrm{C}_{57}\mathrm{H}_{110}\mathrm{O}_6$ corresponds to 0.5 moles of reactant. Because the reaction relates 2 moles of $\mathrm{C}_{57}\mathrm{H}_{110}\mathrm{O}_6$ to 110 moles of $\mathrm{H}_2\mathrm{O}$, our 0.5 moles of reactant yields 27.5 moles of water. Multiplying these 27.5 moles by the molar mass of 18 g/mol results in a final mass of 495 g.

Thermal decomposition of potassium permanganate ($\mathrm{KMnO}_4$) serves as a reliable laboratory method for oxygen preparation. When heated to 513 K, it undergoes a transformation described by the equation $2\,\mathrm{KMnO}_{4} \rightarrow \mathrm{K}_2\mathrm{MnO}_{4} + \mathrm{MnO}_2 + \mathrm{O}_2$, identifying $\mathrm{KMnO}_4$ as the compound $X$ and potassium manganate as the product $Y$.

In the subsequent step, the manganese dioxide generated in the decomposition acts as an oxidizing agent. When subjected to a mixture of sodium chloride and concentrated sulfuric acid, it facilitates the oxidation of chloride ions according to the reaction $4\,\mathrm{NaCl} + \mathrm{MnO}_2 + 4\,\mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{MnCl}_2 + 4\,\mathrm{NaHSO}_4 + 2\,\mathrm{H}_2\mathrm{O} + \mathrm{Cl}_{2}$. This process yields chlorine gas, which is the pungent gas $Z$.

According to Avogadro's Law, at constant temperature and pressure, the volume ratio of gases in a reaction is equivalent to the ratio of their stoichiometric coefficients. The balanced chemical equation for the combustion of a hydrocarbon is $C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$. Given the reaction volumes of 10 mL hydrocarbon, 55 mL $O_2$, and 40 mL $CO_2$, the molar ratio is simplified to $1 : 5.5 : 4$. Since 1 mole of hydrocarbon produces $x$ moles of $CO_2$, it follows that $x = 4$. Substituting this into the stoichiometric expression for oxygen, $x + \frac{y}{4} = 5.5$, yields $4 + \frac{y}{4} = 5.5$, which simplifies to $\frac{y}{4} = 1.5$ and gives $y = 6$. Consequently, the molecular formula is $C_4H_6$.

To determine which reactant consumes the least oxygen, one must calculate the mass of $\mathrm{O}_2$ required per single gram of reactant for each reaction. This value is obtained by dividing the total mass of oxygen consumed in the balanced chemical equation by the total molar mass of the reactant participating in that specific reaction.

For the reaction $2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{MgO}(\mathrm{s})$, the combustion of 48 g of magnesium requires 32 g of oxygen, resulting in $\frac{32}{48}$ or $\frac{2}{3}$ g of $\mathrm{O}_2$ per gram of reactant. In the reaction $4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{Fe}_2\mathrm{O}_3(\mathrm{s})$, the reaction of 224 g of iron consumes 96 g of oxygen, yielding $\frac{96}{224}$ or $\frac{3}{7}$ g of $\mathrm{O}_2$ per gram of reactant. The combustion of $\mathrm{P}_4(\mathrm{s})+5 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{P}_4\mathrm{O}_{10}(\mathrm{s})$ involves 160 g of oxygen for 124 g of reactant, which gives $\frac{160}{124}$ g of $\mathrm{O}_2$ per gram. Finally, the propane reaction $\mathrm{C}_3\mathrm{H}_8(\mathrm{g})+5 \mathrm{O}_2(\mathrm{g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{g})+4 \mathrm{H}_2\mathrm{O}(\mathrm{l})$ uses 160 g of oxygen for 44 g of propane, resulting in $\frac{160}{44}$ g of $\mathrm{O}_2$ per gram.

Comparing these ratios, the fraction $\frac{3}{7}$ is the smallest value among the group, indicating that the iron-based reaction requires the minimum amount of oxygen per gram of reactant.

Molality measures the number of moles of solute present per kilogram of solvent. To determine this concentration, first calculate the number of moles of sodium ions by dividing their mass by the molar mass of sodium, which is 23 g/mol. Using the relationship $n = \frac{\text{mass}}{\text{molar mass}}$, the calculation becomes $\frac{92}{23} = 4$ mol. Since this amount is dissolved in 1 kg of water, the molality is 4 mol kg^-1.

Concentration metrics like mole fraction and molality are based on comparing the amount of solute in moles to either the total moles or the mass of the solvent. By calculating the molar mass of NaOH as 40 grams per mole and water as 18 grams per mole, it becomes clear that 8 grams of NaOH equals 0.2 moles, while 18 grams of water equals 1 mole.

The mole fraction is the proportion of solute moles relative to the total number of moles in the solution, determined by the expression $\frac{0.2}{0.2 + 1}$, which yields 0.167. Molality represents the moles of solute per kilogram of solvent, so by dividing 0.2 moles of NaOH by the 0.018 kilograms of water, the final concentration is 11.11 mol per kg.

The moles of $N_2$ produced per gram of reactant is determined by the ratio of the stoichiometric coefficient of $N_2$ in the balanced chemical equation to the total molar mass of the reactants.

For reaction A, $2 NH_4NO_3 \rightarrow 2 N_2 + 4 H_2O + O_2$, the molar mass is 160 g/mol, which yields $2/160 = 0.0125$ mol/g of $N_2$. For reaction B, $Ba(N_3)_2 \rightarrow Ba + 3 N_2$, the molar mass is 221 g/mol, which yields $3/221 \approx 0.0136$ mol/g of $N_2$. For reaction C, $(NH_4)_2Cr_2O_7 \rightarrow N_2 + 4 H_2O + Cr_2O_3$, the molar mass is 252 g/mol, which yields $1/252 \approx 0.0040$ mol/g of $N_2$. For reaction D, $2 NH_3 \rightarrow N_2 + 3 H_2$, the molar mass is 34 g/mol, which yields $1/34 \approx 0.0294$ mol/g of $N_2$. Comparing these values, reaction D results in the maximum quantity of $N_2$ per gram of reactant.

The number of atoms in a substance is determined by calculating the moles of the constituent element present and multiplying by the Avogadro constant ($N_A$). With 3.55% of chlorine by mass in 1 g of the chlorohydrocarbon, the mass of chlorine is $\frac{3.55}{100} \times 1 \text{ g} = 0.0355 \text{ g}$. Dividing this mass by the atomic weight of chlorine ($35.5 \text{ g/mol}$) yields the number of moles of chlorine: $\frac{0.0355}{35.5} = 10^{-3} \text{ mol}$. Multiplying these moles by the given Avogadro constant results in the total number of chlorine atoms: $10^{-3} \times 6.023 \times 10^{23} = 6.023 \times 10^{20}$.

The thermal decomposition of sodium chlorate is governed by the balanced stoichiometric equation $2NaClO_3 \rightarrow 2NaCl + 3O_2$. This relationship dictates that 3 moles of oxygen gas are released for every 2 moles of sodium chloride residue produced.

Given that the mass of oxygen lost is 0.16 g, the moles of oxygen released are $n(O_2) = 0.16 / 32 = 0.005$ mol. According to the stoichiometry, the moles of $NaCl$ generated are $n(NaCl) = 2/3 \times 0.005 = 0.01 / 3$ mol. Since the reaction with silver nitrate proceeds with a 1:1 molar ratio between $NaCl$ and $AgCl$, the moles of $AgCl$ formed are also $0.01 / 3$ mol. The final mass of $AgCl$ is calculated as $m(AgCl) = (0.01 / 3) \times 143.5 = 0.4783$ g, which rounds to 0.48 g.

The relationship between the carbon and hydrogen atoms in the compound follows from the given mass ratio of $6:1$. Since carbon has an atomic mass of $12$ and hydrogen has an atomic mass of $1$, the proportion $\frac{12X}{Y} = \frac{6}{1}$ indicates that there are twice as many hydrogen atoms as carbon atoms, expressed as $Y = 2X$.

To find the number of oxygen atoms, evaluate the oxygen requirement for the combustion of $\mathrm{C}_X\mathrm{H}_Y$, described by the equation $\mathrm{C}_X\mathrm{H}_Y + \left(X + \frac{Y}{4}\right) \mathrm{O}_2 \rightarrow X \mathrm{CO}_2 + \frac{Y}{2} \mathrm{H}_2\mathrm{O}$. The total number of oxygen atoms consumed is $2 \times \left(X + \frac{Y}{4}\right) = 2X + \frac{Y}{2}$, and since the molecule $\mathrm{C}_X\mathrm{H}_Y\mathrm{O}_Z$ contains half this amount, we establish $Z = \frac{1}{2} \left(2X + \frac{Y}{2}\right)$, which simplifies to $Z = X + \frac{Y}{4}$.

Substituting $Y = 2X$ into the expression for $Z$ results in $Z = X + \frac{2X}{4} = 1.5X$. Assigning the smallest integer value $X = 2$ satisfies the condition for whole numbers, yielding $Y = 4$ and $Z = 3$, which identifies the empirical formula as $\mathrm{C}_2\mathrm{H}_4\mathrm{O}_3$.