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

The chemical reaction occurring here is represented by the following balanced equation:

$M_2CO_3 + 2HCl \rightarrow 2MCl + H_2O + CO_2$

This equation dictates that every 1 mole of carbonate reacts to release exactly 1 mole of carbon dioxide. Since the problem states that 0.01186 moles of gas are produced, this implies that 0.01186 moles of $M_2CO_3$ were consumed from the initial 1 gram sample. Dividing the mass by the number of moles provides the molar mass of the compound. Consequently, calculating the quotient of 1 divided by 0.01186 results in a molar mass of 84.3 g/mol.

The principle of conservation of mass states that the sum of the quantities of pure solute in the individual solutions must equal the quantity of solute in the final mixture, represented by the equation $V_1 C_1 + V_2 C_2 = (V_1 + V_2)C_{\text{mix}}$ Substituting the given values where $V_1$ is the volume of the 45% solution and $800 - V_1$ is the volume of the 20% solution gives $0.45V_1 + 0.20(800 - V_1) = 800 \times 0.29875$ This simplifies to $0.25V_1 + 160 = 239$, leading to $0.25V_1 = 79$, which yields $V_1 = 316$ mL.

The fundamental principle underlying this problem is isotopic mass. Ordinary hydrogen, denoted as $^1\text{H}$, has an atomic mass of 1, whereas deuterium, denoted as $^2\text{H}$, has an atomic mass of 2 because it contains an additional neutron. Replacing every atom of the lighter isotope with the heavier version effectively doubles the mass contributed by that specific element within the body.

In a 75 kg human adult, the total mass is composed of elements including hydrogen at 10.0%. The initial mass of hydrogen is calculated using the equation $75 \times \frac{10}{100} = 7.5\,\text{kg}$. When these atoms are completely substituted with deuterium, the mass of the hydrogen component effectively doubles from 7.5 kg to 15 kg. Therefore, the net weight gain for the person corresponds exactly to the initial mass of the hydrogen atoms, resulting in an increase of 7.5 kg.

This problem relies on stoichiometric principles, where the balanced chemical equation allows us to relate the amount of precipitate formed to the initial concentration of the reactant. The reaction follows the balanced equation:
$\mathrm{FeCl}_3 + 3\mathrm{NaOH} \rightarrow \mathrm{Fe(OH)}_3 + 3\mathrm{NaCl}$
One mole of $\mathrm{FeCl}_3$ reacts to yield one mole of $\mathrm{Fe(OH)}_3$. The molar mass of $\mathrm{Fe(OH)}_3$ is 107 g/mol. Converting the precipitate mass into moles gives 2.14 g divided by 107 g/mol, which equals 0.02 mol. Because of the direct one-to-one stoichiometric ratio, the initial amount of $\mathrm{FeCl}_3$ must also be 0.02 mol. Dividing this quantity by the solution volume of 0.1 L results in a concentration of 0.2 M.

The Principle of Atom Conservation provides a streamlined approach for solving stoichiometry problems by focusing on the preservation of a specific element during a reaction. Since every arsenic atom present in the reactant, arsenic acid, must necessarily transition into the product, arsenic pentasulphide, one can equate the total molar content of arsenic atoms across both sides of the transformation.

Calculating the moles of $\mathrm{H_3AsO_4}$ requires dividing the given 35.5 g by its molar mass of 142 g/mol, which equals 0.25 mol. Given that each molecule of $\mathrm{H_3AsO_4}$ contains one arsenic atom and each unit of $\mathrm{As_2S_5}$ contains two, the conservation relationship is defined as:

$1 \times n_{\mathrm{H_3AsO_4}} = 2 \times n_{\mathrm{As_2S_5}}$

Substituting the calculated moles into this expression results in $1 \times 0.25 = 2 \times n_{\mathrm{As_2S_5}}$, which determines the final amount of $\mathrm{As_2S_5}$ to be 0.125 mol.

The minimum molecular weight of a compound containing a specific element occurs when the molecule contains exactly one atom of that element. Since 8% of the compound's mass is sulphur and the atomic mass of sulphur is 32 g, we establish that 32 g must correspond to 8% of the total molar mass, $M$. This relationship is defined by the following equation:

$\frac{32}{M} \times 100 = 8$

Solving for $M$ by rearranging the expression to $M = \frac{32 \times 100}{8}$ results in a minimum molecular weight of $400\,\mathrm{g}\,\mathrm{mol}^{-1}$.

Ion exchange operations are governed by stoichiometric principles, specifically the requirement to maintain charge neutrality when a divalent cation replaces monovalent ions on the resin backbone. When calcium ions, which carry a $+2$ charge, are removed from water by the resin $\mathrm{C_8H_7SO_3Na}$, they displace two sodium ions, meaning two moles of the resin are needed to bind a single mole of $\mathrm{Ca}^{2+}$. Since the molar mass of the resin is 206 g/mol, the total mass of resin required to capture one mole of $\mathrm{Ca}^{2+}$ is calculated as $2 \times 206 = 412$ g. Dividing the one mole of calcium by the required mass of the resin reveals the uptake capacity to be $\frac{1}{412}$ moles of $\mathrm{Ca}^{2+}$ per gram of resin.

The number of molecules in a gas sample is directly proportional to its molar amount, which is determined by dividing the mass of the substance by its molar mass. Since the mass ratio of oxygen to nitrogen is 1:4, we can express the mass of oxygen as w and nitrogen as 4w, using the respective molar masses of 32 g/mol and 28 g/mol for these diatomic gases.

Calculating the mole ratio requires the equation $\frac{n_{\text{O}_2}}{n_{\text{N}_2}} = \frac{w/32}{4w/28}$. Simplifying this expression by rewriting it as $\frac{w}{32} \times \frac{28}{4w}$ allows the mass variable to cancel out entirely. Carrying out the arithmetic yields a final molecular ratio of 7:32.

The stability of an ionic compound is governed by the principle of electrical neutrality, which demands that the total positive charge from cations perfectly balances the total negative charge from anions. In the formula $\mathrm{M}_{0.98}\mathrm{O}$, the oxygen atom carries a charge of $-2$, meaning the combined charge of the metal atoms must equal $+2$ to maintain a neutral state.

If we define $x$ as the number of $\mathrm{M}^{3+}$ ions, then the number of $\mathrm{M}^{2+}$ ions must be $0.98 - x$. This allows us to construct the charge balance equation $3x + 2(0.98 - x) = 2$. Solving this expression gives $3x + 1.96 - 2x = 2$, which simplifies to $x = 0.04$. The percentage of the total metal existing as $\mathrm{M}^{3+}$ is calculated by dividing this fraction of ions by the total amount of metal, resulting in $\frac{0.04}{0.98} \times 100$, which equals $4.08\%$.

When blending two solutions of the same substance, the final molarity is determined by the ratio of the total moles of solute to the combined volume of the mixture. This principle follows the conservation of moles, where the total molarity is calculated using the following equation:
$M = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$
The total moles are found by summing the product of 0.5 and 750 with the product of 2 and 250, which equals 875 millimoles. Dividing this value by the total volume of 1000 mL yields a final concentration of 0.875 M.

Complete combustion of any hydrocarbon ensures that all constituent carbon atoms form $\mathrm{CO}_2$ and all hydrogen atoms form $\mathrm{H}_2\mathrm{O}$. Calculating the molar quantities from the provided masses involves dividing 3.08 g of $\mathrm{CO}_2$ by its molar mass of 44 g/mol to get 0.07 mol, and dividing 0.72 g of $\mathrm{H}_2\mathrm{O}$ by 18 g/mol to get 0.04 mol. Because each mole of $\mathrm{CO}_2$ contains one mole of carbon and each mole of $\mathrm{H}_2\mathrm{O}$ contains two moles of hydrogen, the sample contains 0.07 mol of carbon and 0.08 mol of hydrogen. Relating these values creates a molar ratio of 7 carbon atoms for every 8 hydrogen atoms, establishing the empirical formula $\mathrm{C}_7\mathrm{H}_8$.

Molarity is defined as the number of moles of solute present per liter of solution, which can be represented by the formula $M = \frac{n}{V}$. To find the number of moles of urea, divide its mass of 120 g by the molar mass of 60 u, which results in 2 moles. Because the solution comprises both solute and solvent, its total mass is the sum of 120 g and 1000 g, equaling 1120 g. Using the provided density of 1.15 g/mL, the volume of the solution is determined by the equation $\frac{1120}{1.15} = 973.9$ mL, which is equivalent to 0.9739 L. Dividing the 2 moles of solute by the 0.9739 L volume yields a final molarity of 2.05 M.

Molality measures the concentration of a solute relative to the mass of the solvent, whereas mole fraction compares the moles of the solute to the total moles in the solution. We can relate these two quantities by considering that molality is the number of moles of solute present in 1000 grams of solvent. Since the solvent is water, its molar mass is 18 g/mol, which allows us to use the conversion relationship:

$m = \frac{X}{1 - X} \times \frac{1000}{M_{\text{solvent}}}$

Here, $m$ is the molality, $X$ is the mole fraction of the solute methyl alcohol, and $M_{\text{solvent}}$ is the molar mass of water. Substituting the known value of 5.2 for $m$ provides the following equation:

$5.2 = \frac{X}{1 - X} \times \frac{1000}{18}$

Solving for $X$ requires isolating the mole fraction by rearranging the equation. By multiplying both sides by the denominator and grouping the terms containing $X$, the final calculation yields a value of 0.086.

Molarity quantifies concentration in terms of moles per liter, while percentage by mass expresses the solute mass relative to the total mass of the solution. Because mass and volume are linked by density, this property acts as the essential conversion factor to relate these two concentration units. The relationship is represented by the formula:

$M = \frac{10 \times (\%\ \mathrm{w/w}) \times d}{\mathrm{M}_{\text{solute}}}$

Substituting the provided parameters of $3.60$ for molarity, $29$ for the mass percentage, and $98$ for the molar mass gives the equation $3.60 = \frac{10 \times 29 \times d}{98}$. Multiplying $3.60$ by $98$ yields $352.8$, which is equal to $290 \times d$. Dividing $352.8$ by $290$ results in a density of $1.22\ \mathrm{g\ mL}^{-1}$.

The stoichiometry of the balanced chemical equation
$2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 6\text{Cl}^-(aq) + 3\text{H}_2(g)$
dictates the relationship between reactants and products. Specifically, the reaction indicates that $6$ moles of $\text{HCl}$ are required to produce $3$ moles of $\text{H}_2$ gas, which simplifies to a ratio where $1$ mole of $\text{HCl}$ produces $\frac{3}{6}$, or $0.5$, moles of $\text{H}_2$ gas.

By applying Avogadro's hypothesis, we know that one mole of an ideal gas occupies $22.4\text{ L}$ at standard temperature and pressure (STP). Multiplying the $0.5$ moles of $\text{H}_2$ produced per mole of $\text{HCl}$ by the molar volume of $22.4\text{ L}$ results in a volume of $11.2\text{ L}$. Thus, the consumption of one mole of $\text{HCl}$ consistently results in the liberation of $11.2\text{ L}$ of hydrogen gas at STP.

The chemical formula $\mathrm{Mg}_3(\mathrm{PO}_4)_2$ indicates that each unit contains 8 oxygen atoms, establishing that 1 mole of the compound contains 8 moles of oxygen atoms. Given a target of 0.25 moles of oxygen atoms, the necessary amount of compound, defined as $n$, satisfies the relationship $n \times 8 = 0.25$. Solving this equation results in $n = 3.125 \times 10^{-2}$.

Molality is defined as the number of moles of solute dissolved in one kilogram of solvent, which requires us to isolate the mass of the solvent from the total solution mass. Since molarity relates moles to volume, assuming a sample size of 1 L allows us to directly calculate the mass of the acetic acid and the corresponding solvent mass.

In 1 L of this solution, there are 2.05 moles of acetic acid. Using the given density of 1.02 g/mL, the total mass of the 1000 mL solution is $1000 \times 1.02 = 1020$ g. Because the molar mass of acetic acid is 60 g/mol, the mass of the solute in this volume is $2.05 \times 60 = 123$ g, while subtracting this from the total solution mass yields a solvent mass of $1020 - 123 = 897$ g, or 0.897 kg. Finally, dividing the moles of solute by the mass of the solvent in kilograms gives the molality, $\frac{2.05}{0.897} = 2.28$ mol kg^-1.

Relative atomic mass is defined by the ratio $\frac{\text{mass of an atom}}{\frac{1}{12} \times \text{mass of a } C^{12} \text{ atom}}$. If we redefine the atomic mass unit as $\frac{1}{6} \times \text{mass of a } C^{12} \text{ atom}$, the divisor in our calculation becomes twice as large as the original reference. Since the calculated mass value is inversely proportional to the magnitude of the unit used, replacing the standard in this manner results in the mass of one mole of any substance decreasing twice.

The molarity of a mixture is calculated by determining the total amount of solute present and dividing that by the total volume of the resulting solution. This relationship is defined by the formula $M_{final} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$. The solute quantity from the first solution is 1.5 times 480, which equals 720, and from the second solution is 1.2 times 520, which equals 624. Summing these values gives $720 + 624 = 1344$ as the total solute amount. The combined volume is 480 plus 520, totaling 1000 ml. Dividing the total solute by the total volume results in the final molarity calculation of $1344 / 1000 = 1.344$ M.

Molarity measures concentration as the number of moles of solute per liter of solution. Calculate the number of moles of urea by dividing the provided number of molecules by the Avogadro constant: $\frac{6.02 \times 10^{20}}{6.02 \times 10^{23}} = 0.001\ \mathrm{mol}$. Because molarity requires the volume to be expressed in liters, convert the solution volume from 100 ml to 0.1 L. Dividing the calculated moles by this volume in liters determines the final concentration: $\frac{0.001\ \mathrm{mol}}{0.1\ \mathrm{L}} = 0.01\ \mathrm{M}$.