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

Titration calculations are based on the principle of equivalence, where the number of milliequivalents of the analyte must equal the number of milliequivalents of the titrant. The reaction between $\text{Na}_2\text{CO}_3$ and $\text{HCl}$ is defined by the balanced equation $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$, which identifies that the n-factor for $\text{Na}_2\text{CO}_3$ is 2 while the n-factor for $\text{HCl}$ is 1.

The titre readings provide a concordant volume of 5.0 mL of 0.2 M $\text{HCl}$ solution required to neutralize 10.0 mL of $\text{Na}_2\text{CO}_3$ solution. Using the relationship $M_1V_1n_1 = M_2V_2n_2$, the calculation is structured as:
$M_{\text{Na}_2\text{CO}_3} \times 10.0 \times 2 = 0.2 \times 5.0 \times 1$
Solving this equation results in a molarity of 0.05 M, which is equivalent to 50 mM.

Neutralization reactions rely on the principle of equivalents, which states that for any acid-base titration, the number of gram equivalents of the acid must equal the number of gram equivalents of the base. Because phosphinic acid functions as a monobasic acid-possessing only one replaceable hydrogen atom-it reacts with sodium hydroxide in a simple one-to-one stoichiometric ratio. By applying the standard titration relationship $N_1V_1 = N_2V_2$, we can equate the normality and volume of the acid to those of the base. Substituting the known values into the equation $0.1 \times 10 = 0.1 \times V$ allows us to solve for the unknown volume of the base, resulting in a requirement of 10 mL of sodium hydroxide to reach the equivalence point.

Molarity $C_{2}$ is defined as the number of moles of solute per liter of solution, expressed as $C_{2}=\frac{n_{2} \times 1000}{V_{\text{mL}}}$, where $V_{\text{mL}}$ is the volume of the solution in milliliters. The volume is calculated from the mass of the solution and its density $d$ as $V_{\text{mL}}=\frac{n_{1} M_{1}+n_{2} M_{2}}{d}$, which when substituted into the molarity equation yields $C_{2}=\frac{1000 n_{2} d}{n_{1} M_{1}+n_{2} M_{2}}$.

Dividing both the numerator and the denominator by the total moles $n_{1}+n_{2}$ allows for the expression of mole fractions $x_{1}=\frac{n_{1}}{n_{1}+n_{2}}$ and $x_{2}=\frac{n_{2}}{n_{1}+n_{2}}$, transforming the equation to $C_{2}=\frac{1000 x_{2} d}{x_{1} M_{1}+x_{2} M_{2}}$. Replacing $x_{1}$ with $1-x_{2}$ in the denominator gives $C_{2}=\frac{1000 d x_{2}}{(1-x_{2}) M_{1}+x_{2} M_{2}}$, which simplifies to $C_{2}=\frac{1000 d x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}$.

The flocculation of a negatively charged arsenic sulphide sol relies on the Hardy-Schulze rule, which states that the coagulation power depends on the valency of the effective ion. Since H₂SO₄ releases two hydrogen ions per molecule compared to the single ion released by HCl, it requires only half the molar quantity to achieve the same flocculating effect. For 250 mL of solution, the equivalent amount of HCl would be 7.5 m mol, which means the required H₂SO₄ is 3.75 m mol. The final mass is determined by $3.75 \times 10^{-3} \times 98 = 0.3675$ grams.

Expressing water hardness as $CaCO_3$ equivalents requires converting the molar concentration of a dissolved salt into an equivalent mass concentration of $CaCO_3$, where $1$ ppm is defined as $1$ mg/L. Because $1$ mole of $MgSO_4$ provides the same degree of hardness as $1$ mole of $CaCO_3$, a solution with a concentration of $10^{-3}$ M $MgSO_4$ is stoichiometrically equivalent to $10^{-3}$ moles of $CaCO_3$ per litre. Multiplying this molarity by the molar mass of $CaCO_3$, which is approximately $100$ g/mol, yields a mass concentration of $0.1$ g/L. Converting this value into milligrams per litre results in $100$ mg/L, which defines the hardness as $100$ ppm.

Molarity is defined as the number of moles of solute per liter of solution, which links the mass percentage and density of the sample to its concentration. By considering a 100 g sample, the 63% composition indicates that 63 g of $\mathrm{HNO}_3$ are present, which constitutes $n = \frac{63}{63} = 1$ mole given the molecular weight of $\mathrm{HNO}_3$ is 63. The volume of this sample is determined by the formula $V = \frac{\text{mass}}{\text{density}}$, resulting in $\frac{100}{1.4}$ mL. Dividing the 1 mole of solute by this volume converted to liters, specifically $\frac{100}{1.4 \times 1000}$ L, leads to a molarity of 14 M.

Balancing chemical equations, particularly redox reactions, relies on the principle of conservation of mass and charge, where the number of electrons lost during oxidation must exactly equal the number of electrons gained during reduction. In the basic medium reaction between iron(II) and hydrogen peroxide, the oxidation of two ferrous ions to ferric ions releases two electrons, which are then accepted by the peroxide to yield two hydroxide ions according to the balanced equation $2\mathrm{Fe}^{2+}+\mathrm{H}_{2}\mathrm{O}_{2}\rightarrow 2\mathrm{Fe}^{3+}+2\mathrm{OH}^{-}$ establishing the stoichiometric coefficients as $x=2$ and $y=2$.

Moving to the acidic reaction involving permanganate and hydrogen peroxide, the manganese in the permanganate ion undergoes reduction from an oxidation state of $+7$ to $+2$, while the hydrogen peroxide is oxidized to oxygen gas. Ensuring charge neutrality and elemental balance across the reactants and products leads to the balanced equation $2\mathrm{MnO}_{4}^{-}+6\mathrm{H}^{+}+5\mathrm{H}_{2}\mathrm{O}_{2}\rightarrow 2\mathrm{Mn}^{2+}+8\mathrm{H}_{2}\mathrm{O}+5\mathrm{O}_{2}$ where the stoichiometric coefficients are identified as $x'=2$ for the manganese product, $y'=8$ for water, and $z'=5$ for oxygen. Summing these specific values, the expression $2+2+2+8+5=19$ provides the final total.

Hydrogen peroxide solutions are categorized by their volume strength, which represents the volume of oxygen gas liberated at standard temperature and pressure from one unit volume of the peroxide solution. The fundamental conversion factor relating this volume strength to molarity is $V = 11.2 \times M$. Given that the solution has a volume strength of 5.6, we can determine the molarity by dividing the strength by this constant, yielding 0.5 M.

With the molarity established, finding the mass percentage involves determining the mass of the solute per unit mass of the solution. Considering one liter of this solution, which contains 0.5 moles of hydrogen peroxide, the mass of the solute is calculated as $0.5 \times 34 = 17$ grams. Since the density of the solution is 1 g/mL, one liter of the solution possesses a total mass of 1000 grams. Consequently, the mass percentage is computed as $\frac{17}{1000} \times 100 = 1.7\%$.

The stoichiometry of precipitation reactions dictates that silver ions react with chloride ions in a one-to-one molar ratio. Because each unit of $\mathrm{[Co(NH_3)_6]Cl_3}$ dissociates to release three chloride ions, the total number of chloride ions, and thus the required moles of silver nitrate, is three times the number of moles of the cobalt complex.

The number of moles of the complex is found by dividing 0.3 g by its molar mass of 267.46 g/mol. Multiplying this value by three yields the moles of chloride ions, which must be balanced by the silver nitrate solution. Given the molarity of 0.125 M, the volume in milliliters is calculated using the following equality:

$3 \times \frac{0.3}{267.46} = \frac{0.125 \times V}{1000}$

Solving this equation for $V$ yields a result of 26.92 mL.

Dalton's atomic theory centers on the fundamental nature of atoms, proposing that matter is composed of indivisible units that maintain their mass and identity throughout chemical transformations. While Dalton established that these atoms merely rearrange rather than disappear during a reaction, the observation that gases combine in simple ratios by volume when temperature and pressure are held constant is actually defined by Gay-Lussac's Law of Gaseous Volumes. Because this volumetric relationship describes gas behavior rather than the structural properties of atoms themselves, it is distinct from the specific postulates set forth in Dalton's original framework.

Determining the yield of a chemical reaction requires identifying the limiting reagent, which is the reactant that is consumed first and dictates the total amount of product formed. In the synthesis of ammonia, the balanced chemical equation is $\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3$. By converting the given masses into kilomoles using their molar masses, we find that 2.8 kg of nitrogen is equivalent to 0.1 kmol, while 1 kg of hydrogen is equivalent to 0.5 kmol.

According to the stoichiometry of the reaction, 0.1 kmol of nitrogen requires 0.3 kmol of hydrogen to reach completion. Because the available 0.5 kmol of hydrogen exceeds this requirement, nitrogen acts as the limiting reagent. The reaction generates 0.2 kmol of ammonia based on the nitrogen consumption. Multiplying this quantity by the molar mass of ammonia results in a total mass of 3.4 kg, which converts to 3400 grams.

The fundamental relationship between mass and particle count allows us to determine the molar mass of substance x by first calculating the number of moles in the 10 g sample. Using Avogadro's constant, the amount of substance is:

$\text{Moles} = \frac{6.023 \times 10^{22}}{6.023 \times 10^{23}} = 0.1$

Given that 10 g of the substance equals 0.1 moles, the molar mass is 100 g/mol. With this mass established, the molarity of the solution is determined by dividing the moles of solute in 5 g of the substance by the total volume of 2 L. Since 5 g represents 0.05 moles, the molarity calculation is:

$\text{Molarity} = \frac{0.05}{2} = 0.025$

This result corresponds to the required value expressed in scientific notation:

$0.025 = 25 \times 10^{-3}$

The volume strength of a hydrogen peroxide solution indicates the volume of oxygen gas produced in liters at standard temperature and pressure when one liter of the solution undergoes decomposition. Based on the balanced reaction $2H_2O_2 \rightarrow 2H_2O + O_2$, two moles of hydrogen peroxide generate one mole of oxygen. Since one mole of an ideal gas occupies 22.4 L at 273 K and 1 atm, each mole of hydrogen peroxide produces 11.2 L of oxygen gas.

Multiplying the concentration of 8.9 M by the conversion factor of 11.2 provides the volume of oxygen generated per liter. Performing the calculation yields $8.9 \times 11.2 = 99.68$. Rounding this result to the nearest whole number results in 100.

The average atomic mass of an element represents the weighted mean of its naturally occurring isotopes based on their relative abundances. By assuming a molar ratio of $x$ to 1 for Cl-35 and Cl-37, we define the relationship through the following equation:

$35.5 = \frac{35x + 37}{x + 1}$

Expanding this yields $35.5x + 35.5 = 35x + 37$, which simplifies to $0.5x = 1.5$ after rearranging the terms. Solving for $x$ results in 3, confirming that the isotopes exist in a ratio of 3:1.

Chemical stoichiometry relies on the balanced reaction between urea and sodium hydroxide:

$\mathrm{NH_2CONH_2} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2CO_3} + 2\mathrm{NH_3}$

This equation shows that 1 mole of urea releases 2 moles of $\mathrm{NH_3}$. Given the molar mass of urea is 60 g/mol, 0.6 g of the substance represents 0.01 moles, which subsequently produces 0.02 moles, or 20 mmol, of $\mathrm{NH_3}$.

Because ammonia acts as a base, the amount of acid required for neutralization must equal the milliequivalents of the ammonia produced, meaning 20 meq of $\mathrm{HCl}$ is necessary. Multiplying the volume of the acid by its normality for the given options shows that 100 ml of 0.2 N $\mathrm{HCl}$ provides exactly 20 meq, which balances the reaction perfectly.

Concentration in parts per million (ppm) is determined by calculating the ratio of the mass of a solute to the total mass of the solution, then scaling the result by one million. Convert the 10.30 mg of oxygen into grams to obtain 0.0103 g. Since one liter of seawater is equivalent to 1000 mL, calculate its total mass by multiplying the volume by the density of 1.03 g/mL, which yields 1030 g. Substituting these values into the formula $\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$ gives the final calculation:

$\frac{0.0103}{1030} \times 10^6 = 10$

Mole fraction provides a direct ratio of the moles of a specific component to the total number of moles in a solution. In a system where the total molar amount is assumed to be 1 mole, a mole fraction of 0.1 for glucose leaves 0.9 moles for water.

Converting these quantities to mass involves multiplying by the respective molar masses of 180 g/mol for glucose and 18 g/mol for water, resulting in 18 g of glucose and 16.2 g of water. The mass percentage of water is found by comparing the mass of water to the total mass of the mixture:

$\text{Mass } \% \text{ of water} = \left( \frac{0.9 \times 18}{0.1 \times 180 + 0.9 \times 18} \right) \times 100 = 47.36\%$

Rounding this percentage to the nearest whole number gives 47.

Stoichiometry relies on identifying the exact molecular composition, which is derived by converting mass percentages into atomic mole ratios. Once the empirical ratio is established, the specific characteristics of the compound, such as being saturated and acyclic, allow for the determination of the correct molecular formula and the corresponding balanced combustion reaction.

Starting with mass ratios $C:H=4:1$ and $C:O=3:4$, dividing by their respective atomic weights of 12, 1, and 16 provides the mole ratio 1:3:1. This corresponds to an empirical formula of $CH_3O$, and given that the compound is saturated and acyclic, its molecular formula is $C_2H_6O_2$. The stoichiometry of the complete combustion is given by the balanced equation
$C_2H_6O_2 + \frac{5}{2}O_2 \rightarrow 2CO_2 + 3H_2O$
which shows that 2.5 moles of oxygen gas are needed for one mole of the substance. Multiplying this requirement by the two moles of compound specified in the problem confirms that 5 moles of oxygen gas are required.

The stoichiometry of redox reactions is governed by the principle that the number of equivalents of the oxidizing agent exactly matches the number of equivalents of the reducing agent at the point of complete reaction. For this reaction, we determine the equivalents of $\mathrm{KMnO}_4$ by dividing its mass by its molar mass and multiplying by the $n$-factor, which is $5$ in an acidic medium. Using $0.316\ \mathrm{g}$ of $\mathrm{KMnO}_4$ and a molar mass of $158\ \mathrm{g/mol}$, the equivalents are calculated as $\frac{0.316}{158} \times 5 = 0.01$.

These $0.01$ equivalents must react with an equal quantity of $\mathrm{H}_2\mathrm{O}_2$. Given that the $n$-factor of $\mathrm{H}_2\mathrm{O}_2$ is $2$, we find the number of moles by dividing the equivalents by the $n$-factor, giving $\frac{0.01}{2} = 0.005$ moles. Multiplying this molar amount by the molar mass of $34\ \mathrm{g/mol}$ yields $0.17\ \mathrm{g}$ of pure $\mathrm{H}_2\mathrm{O}_2$. Finally, comparing the mass of pure $\mathrm{H}_2\mathrm{O}_2$ to the initial impure sample mass of $0.2\ \mathrm{g}$ results in a purity percentage of $\frac{0.17}{0.2} \times 100 = 85\%$.

The concept of parts per million (ppm) represents a mass-based ratio, indicating the quantity of a specific substance within one million mass units of a total mixture. This principle allows for the calculation of the absolute mass of a nutrient required to attain a specified concentration within a larger sample.

Calculating the amount of ferrous sulphate heptahydrate needed for $100 \text{ kg}$ of wheat at $10 \text{ ppm}$ begins by converting the wheat mass to $10^5 \text{ g}$. At a concentration of $10 \text{ ppm}$, the necessary mass of pure iron is $1 \text{ g}$. Given the molar mass of the salt is $277.85 \text{ g/mol}$ and the atomic weight of iron is $55.85 \text{ g/mol}$, the iron content fraction in the salt is $55.85 / 277.85$. Setting the mass of salt, $W$, to satisfy the iron requirement yields the following equation:

$W = \frac{1 \times 277.85}{55.85} = 4.97 \, \text{g}$