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

The law of equivalents forms the foundation of titration, stating that at the equivalence point, the number of milliequivalents of acid exactly equals the number of milliequivalents of base. The reaction between hydrochloric acid and calcium hydroxide serves to define the concentration of the latter, where the volume of calcium hydroxide used is found by the change in the burette reading, $35.5 \ \mathrm{mL} - 25.5 \ \mathrm{mL} = 10 \ \mathrm{mL}$. Equating the milliequivalents using the product of volume, molarity, and n-factor gives $20 \ \mathrm{mL} \times 0.5 \ \mathrm{M} \times 1 = 10 \ \mathrm{mL} \times M_{\mathrm{Ca(OH)}_2} \times 2$, which identifies the molarity of calcium hydroxide as $0.5 \ \mathrm{M}$.

Applying this established concentration to the reaction involving sulfuric acid yields the unknown value. Because $20 \ \mathrm{mL}$ of the $0.5 \ \mathrm{M}$ calcium hydroxide solution is consumed by $10 \ \mathrm{mL}$ of the sulfuric acid solution, the equality $20 \ \mathrm{mL} \times 0.5 \ \mathrm{M} \times 2 = 10 \ \mathrm{mL} \times M_{\mathrm{H}_2\mathrm{SO}_4} \times 2$ holds true. Solving this equation reveals the concentration of the sulfuric acid to be $1 \ \mathrm{M}$.

At Standard Temperature and Pressure, one mole of an ideal gas occupies a volume of 22.7 L. By dividing the given volume of 2.8375 L by this molar volume, we calculate the amount of oxygen using $n = \frac{2.8375}{22.7}$, which results in 0.125 mol.

To find the total number of molecules, we multiply the molar amount by Avogadro's constant. Using the expression $0.125 \times 6.022 \times 10^{23}$, we determine there are $7.527 \times 10^{22}$ molecules in the gas sample.

The balanced chemical equation $\mathrm{Zn} + 2\ \mathrm{HCl} \rightarrow \mathrm{ZnCl}_2 + \mathrm{H}_2$ shows that the reaction proceeds in a one-to-one molar ratio between zinc and hydrogen. To find the moles of zinc, divide the mass of 11.5 grams by the molar mass of 65.4 grams per mole using the expression $\frac{11.5}{65.4} \approx 0.1758$ moles.

Since the molar ratio of zinc to hydrogen is one-to-one, this same amount of hydrogen gas is formed in the reaction. Calculating the volume using the provided molar volume at STP involves the expression $\frac{11.5}{65.4} \times 22.7 \approx 3.99$ liters. Rounding this value to the nearest integer gives 4 liters.

Molality is defined as the amount of solute in moles per kilogram of solvent, which necessitates calculating the mass of both components from their volume and density. A 10% (v/v) solution indicates that 10 mL of the di-bromine solute is mixed with 90 mL of the CCl4 solvent. Using the given densities, the mass of the solute is 10 mL × 3.2 g/cm^3, which equals 32 g, and the mass of the solvent is 90 mL × 1.6 g/cm^3, which equals 144 g.

Determining the molar content involves dividing the solute mass by its molar mass, 160 g/mol, resulting in 0.2 mol of solute. The molality is calculated as the ratio of these moles to the mass of the solvent in kilograms:
$m = \frac{0.2 \text{ mol}}{0.144 \text{ kg}} \approx 1.388 \text{ mol/kg}$
Representing this as $x \times 10^{-2}$ leads to $138.8 \times 10^{-2}$, which rounds to the nearest integer of 139.

The principle of atom conservation ensures that all the carbon within an organic compound is converted into carbon in the carbon dioxide produced during complete combustion. Since the compound weighs 0.5 grams and contains 60 percent carbon, the mass of carbon present is $0.5 \times 0.6 = 0.3$ grams. Dividing this mass by the atomic weight of carbon, 12 grams per mole, results in $0.3 / 12 = 0.025$ moles of carbon atoms. Because every mole of carbon atoms produces one mole of carbon dioxide, the number of moles of carbon dioxide formed is also 0.025 moles. Multiplying this quantity by the molar mass of carbon dioxide, 44 grams per mole, yields the final mass:

$0.025 \times 44 = 1.1\ \text{grams}$

This final value is 11 times $10^{-1}$ grams.

This problem requires stoichiometric calculation based on the balanced reaction between sodium and water and the subsequent neutralization of the resulting base. First, determine the moles of metallic sodium by dividing the mass of 0.69 g by the molar mass of 23 g mol^-1, which equals 0.03 mol. Given the reaction $\mathrm{Na} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{NaOH} + \frac{1}{2}\mathrm{H}_2$, one mole of sodium yields one mole of $\mathrm{NaOH}$, meaning we have 0.03 mol of $\mathrm{NaOH}$.

Next, determine the concentration of the acid by dividing 73 g L^-1 by the molar mass of $\mathrm{HCl}$, which is 36.5 g mol^-1, resulting in a molarity of 2 M. Since $\mathrm{NaOH}$ and $\mathrm{HCl}$ react in a 1:1 molar ratio, we use the equation $n = M \times V$ to equate the moles of base to the moles of acid. By substituting the values 0.03 mol = 2 M $\times$ V, we solve for V to find 0.015 L, which equals 15 mL.

Molarity is defined as the concentration of a solution expressed as the number of moles of solute per liter of solution. To determine the molarity of the hydrated oxalic acid solution, one calculates the moles by dividing the mass of $3.15\ \mathrm{g}$ by the provided molar mass of $126\ \mathrm{g\ mol^{-1}}$, resulting in $0.025\ \mathrm{mol}$ of solute. Dividing this amount by the total volume of $0.250\ \mathrm{L}$ yields a solution concentration of $0.1\ \mathrm{M}$. Because the molar mass specified in the reason is the essential factor used to derive the moles and verify the final molarity, the reason correctly supports and explains the assertion.

Neutralization relies on the principle that the number of equivalents of acid must equal the number of equivalents of base, a relationship defined by $N_1 V_1 = N_2 V_2$. HBr acts as a monoprotic acid with an n-factor of 1, so its normality is equal to its molarity of 0.02 N, whereas Ba(OH)2 is a dibasic compound with an n-factor of 2, making its normality 0.02 N. By substituting these values into the neutralization formula $0.02 \times V_1 = 0.02 \times 10.0$ we can determine the volume, which results in 10.0 mL.

The molecular formula of a compound is determined by relating its empirical formula-the simplest ratio of atoms-to its total molar mass. Since the hydrocarbon consists only of carbon and hydrogen, subtracting the carbon percentage of 85.8 from 100 reveals that hydrogen comprises 14.2 percent of the compound. In a 100 g sample, this mass translates to 7.15 moles of carbon and 14.2 moles of hydrogen, resulting in a molar ratio of 1 to 2. This corresponds to an empirical formula of $\text{CH}_2$, which has a molar mass of 14 g/mol. Comparing this empirical mass to the provided molar mass of 84 g/mol establishes a scaling factor of 6. Consequently, the molecular formula is $\text{C}_6\text{H}_{12}$, which indicates that there are 12 hydrogen atoms in each molecule.

Stoichiometric calculations rely on balanced chemical equations to establish the molar relationships between reactants and products. The reaction is represented by $\mathrm{Mg} + 2\,\mathrm{HCl} \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2 \uparrow$ which shows that one mole of magnesium produces one mole of hydrogen gas. To determine the number of moles reacting, divide the mass of 2.4 g by the molar mass of 24 g/mol, which results \in 0.1 mol of magnesium. Given the 1:1 molar ratio, this amount of magnesium yields 0.1 mol of hydrogen gas. Since one mole of gas occupies 22.4 L at standard temperature and pressure, the volume liberated is calculated as $0.1 × 22.4 = 2.24\,\mathrm{L}$. Converting this volume into the required units leads to $224 × 10^{-2}\,\mathrm{L}$.

The ionization of the coordination complex $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_5\mathrm{Cl}]\mathrm{Cl}_2$ dictates that only the two chloride ions situated outside the coordination sphere are available to react with silver nitrate, establishing a molar ratio of $1:2$ between the complex and the precipitating agent. Given the solution contains $0.2$ millimoles of the complex, calculated as $0.01 \mathrm{M} \times 20 \mathrm{mL}$, it will release $0.4$ millimoles of chloride ions upon dissociation. Consequently, exactly $0.4$ millimoles of $\mathrm{AgNO}_3$ are necessary to complete the reaction. Applying the volume formula $V = \frac{n}{M}$ with $0.4$ millimoles and $0.1 \mathrm{M}$ results in $V = 4 \mathrm{mL}$.

Coordination compounds dissociate in aqueous solution to release ions located outside the coordination sphere, while ligands bonded directly to the central metal ion remain attached. When the two initial solutions are combined, the resulting volume is $2\text{ L}$, so dividing this mixture into two equal parts ensures that each $1\text{ L}$ sample of $X$ contains exactly $0.01\text{ mol}$ of $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4]\mathrm{Br}$ and $0.01\text{ mol}$ of $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}]\mathrm{SO}_4$.

In the portion treated with excess $\mathrm{AgNO}_3$, only the bromide ion in $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4]\mathrm{Br}$ acts as a counter-ion and is available to precipitate. This reaction yields $0.01\text{ mol}$ of $\mathrm{AgBr}$, whereas the sulfate group in the other complex is coordinated to the metal and remains in solution. Conversely, when the second portion is treated with excess $\mathrm{BaCl}_2$, the sulfate ion in $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}]\mathrm{SO}_4$ functions as the counter-ion and precipitates. This produces $0.01\text{ mol}$ of $\mathrm{BaSO}_4$, while the bromide ion in the first complex remains bound to the cobalt metal.

The fundamental principle of combustion analysis relies on the conservation of mass, ensuring that every carbon atom from the organic compound is accounted for in the resulting carbon dioxide. Because each molecule of carbon dioxide contains exactly one carbon atom, the total molar amount of carbon in the original sample is identical to the molar amount of carbon dioxide produced.

Dividing the produced mass of carbon dioxide, $4.4\,\mathrm{g}$, by its molar mass of $44\,\mathrm{g}\,\mathrm{mol}^{-1}$ results in $0.1\,\mathrm{mol}$ of carbon. Given that the organic compound contains $60\%$ carbon by mass, the mass of carbon within $0.01\,\mathrm{mol}$ of the compound is $0.01 \times M \times 0.60$, where $M$ is the molar mass. Dividing this carbon mass by the atomic weight of carbon, $12\,\mathrm{g}\,\mathrm{mol}^{-1}$, provides the moles of carbon, which must equal $0.1\,\mathrm{mol}$. Setting the equation $\frac{0.01 \times M \times 0.60}{12} = 0.1$ results in a molar mass $M$ of $200\,\mathrm{g}\,\mathrm{mol}^{-1}$.

The principle of charge neutrality requires that the total positive charge from metal ions perfectly balances the negative charge from oxygen, which has a value of $-2$ in the formula $M_{0.83} O_{1.00}$. By defining $x$ as the number of $M^{2+}$ ions, the remaining $0.83 - x$ ions must exist in the $+3$ state, leading to the charge balance equation $2x + 3(0.83 - x) = 2$. Solving this equation gives $2x + 2.49 - 3x = 2$, which simplifies to $x = 0.49$. Calculating the percentage of $M^{2+}$ ions by dividing this amount by the total metal content of $0.83$ and multiplying by $100$ results in $59\%$.

Significant figures represent the reliability and precision of a numerical value, dictated by rules regarding non-zero digits and the placement of zeros. Leading zeros in a number like 0.00253 serve only as placeholders, meaning only the digits 2, 5, and 3 are significant, resulting in a count of three. In the case of 1.0003, all digits are significant because the zeros are trapped between non-zero integers, creating a total of five significant figures. For 15.0, the trailing zero after the decimal point is considered significant alongside the non-zero digits, totaling three, while 163 contains three non-zero digits that are all inherently significant. Comparing these counts reveals that 0.00253, 15.0, and 163 all contain exactly three significant figures.

The combustion of any hydrocarbon with the general formula $\mathrm{C}_{x}\mathrm{H}_{y}$ follows the stoichiometric relationship
$\mathrm{C}_{x}\mathrm{H}_{y} + \left(x + \frac{y}{4}\right)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O}$
Since the reaction produces 4 equivalents of water, setting the term $\frac{y}{2} = 4$ confirms that the hydrocarbon contains $y = 8$ hydrogen atoms. Applying the oxygen consumption requirement where $x + \frac{y}{4} = 11$, substituting the value of $y$ results in $x + \frac{8}{4} = 11$. Simplifying this gives $x + 2 = 11$, which determines that $x = 9$, yielding the molecular formula $\mathrm{C}_9\mathrm{H}_8$.

The volume strength of hydrogen peroxide signifies the total volume of oxygen gas liberated at standard temperature and pressure when one unit volume of the solution decomposes. This relationship is quantified by the standard conversion formula $Volume \text{ } Strength = 11.35 \times M$, where $M$ represents the molarity of the solution.

To determine the mass concentration for a 25 volume solution, calculate the molarity as $M = \frac{25}{11.35}$. Multiplying this value by the molar mass of hydrogen peroxide, which is 34 g/mol, provides the strength in grams per liter. Performing the calculation $\frac{25 \times 34}{11.35}$ results in approximately 74.889 g/L, confirming that 1 L of this specific marketed solution contains 75 g of hydrogen peroxide.

The molar mass of $\text{CH}_4$ is 16 g/mol, so 16 g represents 1 mole of the gas. Since each molecule of $\text{CH}_4$ contains 10 electrons, 1 mole corresponds to the total number of electrons calculated by $10 \times 6.02 \times 10^{23} = 60.2 \times 10^{23}$ electrons. For $\text{H}_2$, 1 g is equal to 0.5 mole because the molar mass is 2 g/mol, and this volume of gas occupies 11.4 L at STP.

A 1 mole quantity of $\text{N}_2$ is equivalent to 28 g, which is the sum of the atomic masses of the two nitrogen atoms in the molecule. For $\text{SO}_2$, the weight is found using the equation $0.5 \times 64 = 32$ g, given that the molar mass of $\text{SO}_2$ is 64 g/mol.

Molality defines the concentration of a solute relative to the mass of the solvent, whereas molarity relates the amount of solute to the total volume of the solution. To determine the mass of ethylene glycol required in both instances, we apply the appropriate concentration formulas while assuming that the mass of the solvent is approximately equal to the mass of the solution in these dilute aqueous mixtures.

A 0.25 molal solution prepared with 500 g of solvent requires a mass $W_1$ defined by the following expression:
$0.25 = \frac{W_1}{62} \times \frac{1000}{500}$
This equation reveals that $W_1$ is equal to 0.25 multiplied by 62 multiplied by 0.5.

For 250 mL of a 0.25 molar solution, the mass $W_2$ is determined through the molarity definition:
$0.25 = \frac{W_2}{62} \times \frac{1000}{250}$
This relationship indicates that $W_2$ equals 0.25 multiplied by 62 multiplied by 0.25.

Comparing the two resulting masses, $W_1$ and $W_2$, demonstrates that $W_1$ is twice the value of $W_2$ because 0.5 divided by 0.25 is 2. The mass ratio of ethylene glycol for these two solutions is 2:1.

Mass percentage defines the concentration of a solution by expressing the mass of the solute as a fraction of the total mass of the solution, represented as $\frac{\text{mass of solute}}{\text{total mass of solution}} \times 100$. Because the solvent is 1 mole of water, its mass is 18 g, and combining this with the 2 g of solute results in a total solution mass of 20 g. Evaluating the ratio $\frac{2}{20} \times 100$ confirms that the concentration is 10%.