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

The complete combustion of methane is described by the chemical equation $\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$. To find the mass of methane required, start by calculating the moles of carbon dioxide produced, which is 22 g divided by 44 g/mol, resulting in 0.5 mol. Because the stoichiometry of the reaction requires a 1:1 molar ratio of methane to carbon dioxide, you need 0.5 mol of methane. Multiplying this amount by the molar mass of methane, which is 16 g/mol, gives the final result of 8 g of methane.

Molality expresses concentration as moles of solute per kilogram of solvent, making it distinct from molarity which is volume-based. Assuming a $1\ \text{L}$ volume of $0.8\ \text{M}$ $\text{H}_2\text{SO}_4$ solution provides $0.8\ \text{mol}$ of solute, which has a mass of $0.8 \times 98 = 78.4\ \text{g}$ based on the molar mass of $98\ \text{g/mol}$. The total mass of this solution is $1060\ \text{g}$ because its density is $1.06\ \text{g}\ \text{cm}^{-3}$. Subtracting the solute mass from the total mass leaves $981.6\ \text{g}$, or $0.9816\ \text{kg}$, of solvent. Dividing $0.8\ \text{mol}$ of solute by $0.9816\ \text{kg}$ of solvent results in a molality of $0.815\ \text{m}$, which is expressed as $815 \times 10^{-3}\ \text{m}$.

Molarity is defined as the amount of solute in moles per liter of solution. To calculate the concentration, first determine the molar mass of sodium chloride by adding the atomic masses, which yields $23 + 35.5 = 58.5$ grams per mole. Convert the given mass of the solute into moles by using the equation $5.85 / 58.5 = 0.1$ moles.

Since the volume of the solution is $500$ milliliters, convert this to $0.5$ liters. Apply these values to the molarity formula by dividing the calculated moles by the volume in liters:
$M = \frac{0.1}{0.5} = 0.2$
The resulting concentration of the solution is $0.2$ M.

Determining the limiting reagent is essential for predicting the exact amount of product formed in any reaction. For the reaction $3\mathrm{PbCl}_2 + 2(\mathrm{NH}_4)_3\mathrm{PO}_4 \rightarrow \mathrm{Pb}_3(\mathrm{PO}_4)_2 + 6\mathrm{NH}_4\mathrm{Cl}$, we identify the limiting reagent by dividing the available millimoles of each reactant by its stoichiometric coefficient.

For $\mathrm{PbCl}_2$, dividing 72 mmol by 3 results in 24, while for $(\mathrm{NH}_4)_3\mathrm{PO}_4$, dividing 50 mmol by 2 results in 25. Since 24 is less than 25, $\mathrm{PbCl}_2$ acts as the limiting reagent. Because the balanced equation shows that 3 moles of $\mathrm{PbCl}_2$ produce 1 mole of $\mathrm{Pb}_3(\mathrm{PO}_4)_2$, we calculate the total yield by dividing the 72 mmol of available $\mathrm{PbCl}_2$ by 3, which equals 24 mmol of $\mathrm{Pb}_3(\mathrm{PO}_4)_2$.

A 3 M solution contains 3 moles of NaCl within exactly 1 liter of solution. The molar mass of NaCl is 58.5 g/mol, so the solute mass is 3 × 58.5 = 175.5 g. Given the density of 1.25 g/mL, 1000 mL of this solution weighs 1250 g. Subtracting the solute mass from this total weight provides the solvent mass of 1074.5 g, which is 1.0745 kg. Dividing the 3 moles by this mass in kilograms gives the molality, expressed as $m = \frac{3}{1.0745} = 2.79\ \text{m}$.

Thermal decomposition of metal carbonates is a process where heating the solid sample releases carbon dioxide gas, leaving behind the corresponding metal oxide as a stable residue. Because the total mass of the mixture is conserved before and after ignition, the difference between the initial weight and the final weight represents the carbon dioxide lost. By quantifying the mass of the oxides produced in terms of the initial mass of the carbonates, we can establish a relationship that accounts for the observed weight of the residue.

Designating the mass of $\mathrm{CaCO}_3$ as $x \mathrm{g}$ means the mass of $\mathrm{MgCO}_3$ is $(2.21 - x) \mathrm{g}$. Ignition transforms $\mathrm{CaCO}_3$ into $\mathrm{CaO}$ with a molar mass of $56 \mathrm{g mol}^{-1}$, while $\mathrm{MgCO}_3$ converts into $\mathrm{MgO}$ with a molar mass of $40 \mathrm{g mol}^{-1}$. Since the moles of carbonate consumed correspond directly to the moles of oxide produced, the mass balance equation is given by $\frac{x}{100} \times 56 + \frac{2.21 - x}{84} \times 40 = 1.152$. Solving this linear equation reveals $x = 1.1886 \mathrm{g}$ for the mass of $\mathrm{CaCO}_3$, which results in $1.0214 \mathrm{g}$ for the mass of $\mathrm{MgCO}_3$. These calculated values identify the specific composition of the mixture.

The stoichiometry of a chemical reaction, dictated by the balanced chemical equation, allows us to determine the precise amount of reactants needed to produce a specific quantity of product. The combustion of methane is represented by the equation:
$\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$
This equation establishes that 1 mole of methane is required to yield 1 mole of carbon dioxide. Knowing the molar mass of carbon dioxide is 44 g/mol, we find the number of moles produced by dividing 11 g by this molar mass, which equals 0.25 mol. Since the reaction consumes methane and produces carbon dioxide in a one-to-one ratio, the combustion must have required 0.25 mol of methane to generate that specific quantity.

The relationship connecting molarity $M$ and molality $m$ relies on the density $d$ of the solution to bridge the gap between the total volume of the solution and the mass of the solvent. This connection is mathematically expressed by the formula $m = \frac{1000 \times M}{1000 \times d - M \times Mw}$.

Substituting $m = 3$, $d = 1.12$ g/mL, and $Mw = 40$ g/mol into the expression gives $3 = \frac{1000x}{1000 \times 1.12 - 40x}$, which simplifies to $3 = \frac{1000x}{1120 - 40x}$. Multiplying both sides by the denominator results \in $3(1120 - 40x) = 1000x$, leading to $3360 - 120x = 1000x$. Combining the $x$ terms yields $1120x = 3360$, and dividing by 1120 confirms $x = 3$.

When a solution experiences a change in temperature, its volume alters due to thermal expansion or contraction, even though the total mass of the components remains constant. Molarity, expressed as $Molarity = \frac{\text{Moles of solute}}{\text{Volume of solution}}$, directly involves this temperature-sensitive volume in its denominator, which causes the final concentration value to shift. Conversely, mass-based expressions like molality, mole fraction, and mass percentage do not depend on volume, ensuring their values remain invariant when temperature fluctuates.

Molarity defines the concentration of a solution by relating the moles of solute to the volume of the solution in liters, represented by the formula:
$M = \frac{n}{V}$
Begin by determining the total mass of the solution by multiplying the density of 1.54 g/cc by the volume of 1000 ml, which follows the equation $1.54 \times 1000 = 1540$ g. Since the phosphoric acid is 70% pure by weight, the mass of the pure solute is found as $0.7 \times 1540 = 1078$ g. Converting this mass into moles by dividing by the molar mass of 98 g/mol leads to the calculation $1078 / 98 = 11$ moles. Because this amount of solute is dissolved in 1 L of solution, the final molarity is 11 M.

Combustion of methane follows the balanced chemical equation
$\mathrm{CH}_{4(\mathrm{g})} + 2 \mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})} + 2 \mathrm{H}_2 \mathrm{O}_{(\ell)}$
which establishes that one mole of methane is required to produce one mole of carbon dioxide. By dividing the given 22 g of carbon dioxide by its molar mass of 44 g/mol, we determine that 0.5 moles of carbon dioxide are formed. Given the 1:1 molar ratio, the same amount of methane, 0.5 moles, must be consumed. Converting this value to the format $x \times 10^{-2}$ leads to $50 \times 10^{-2}$, meaning the value of $x$ is 50.

Stoichiometric calculations rely on the fixed molar mass relationship defined by a balanced chemical equation, which establishes that a specific mass of reactant yields a predictable mass of product. In the conversion of aniline to acetanilide, the reaction stoichiometry dictates that $93\,\mathrm{g}$ of aniline is required to synthesize $135\,\mathrm{g}$ of acetanilide.

The yield for a sample of $6.55\,\mathrm{g}$ of aniline is determined by scaling the theoretical mass using the ratio of these molar masses. Performing the calculation $\frac{135}{93} \times 6.55$ results in a total acetanilide mass of $9.5\,\mathrm{g}$. Converting this value into the specified notation of $\times 10^{-1}\,\mathrm{g}$ produces $95 \times 10^{-1}\,\mathrm{g}$.

Determining molality from molarity requires partitioning the total mass of the solution into the mass of the solute and the mass of the solvent. Because the solution has a concentration of $0.2\,\text{M}$ in $500\,\text{mL}$, it contains $0.1\,\text{moles}$ of $\mathrm{CuSO}_4$, which equates to a mass of $15.95\,\text{g}$ given the molar mass of $159.5\,\text{g/mol}$.

Multiplying the total solution volume by the density gives a total mass of $625\,\text{g}$. Subtracting the solute mass from this total provides the solvent mass of $609.05\,\text{g}$, which is $0.60905\,\text{kg}$. Calculating molality through the standard relationship:
$m = \frac{\text{moles of solute}}{\text{mass of solvent \in kg}}$
yields $m = \frac{0.1}{0.60905} \approx 0.164\,\text{m}$. This value expressed in the requested scientific notation is $164 \times 10^{-3}\,\text{m}$.

The reaction between aniline and acetic anhydride occurs in a 1:1 stoichiometric ratio, meaning the moles of acetanilide produced are equal to the moles of aniline consumed. To find the resulting mass, first determine the moles of aniline by using its molar mass of 93 g mol⁻¹.

Calculate the moles of aniline using the formula $n = \frac{9.3}{93} = 0.1$ mol. Since the stoichiometry is 1:1, this confirms that 0.1 mol of acetanilide is formed, and with a molar mass of 135 g mol⁻¹, the final mass is found using $m = 0.1 \times 135 = 13.5$ g. Expressing this result as $13.5 = 135 \times 10^{-1}$ g shows that the numerical value required is 135.

The balanced chemical equation for the complete combustion of glucose is $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{O}_2 \rightarrow 6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O}$, which shows that every single mole of glucose requires 6 moles of oxygen gas to react fully. Since the molar mass of glucose is 180 g/mol, a 900 g sample corresponds to 5 moles of glucose. According to the reaction stoichiometry, these 5 moles of glucose will consume 30 moles of oxygen because 5 multiplied by the stoichiometric coefficient of 6 equals 30. Multiplying this molar requirement by the molar mass of oxygen, which is 32 g/mol, results in a total of 960 g of oxygen needed for the combustion.

Molality relates the number of moles of solute to the mass of the solvent in kilograms, whereas molarity relates the moles to the total solution volume. Converting between these concentration terms requires determining the mass of the solvent by subtracting the mass of the dissolved solute from the total mass of the solution.

Consider exactly 1 liter of the solution. Since the concentration is 3 M, there are 3 moles of NaCl present. Given the density of 1.0 g/mL, the total mass of the solution is 1000 grams. The mass of the NaCl solute is determined by multiplying its moles by its molar mass of 58.5 g/mol, resulting in $3 \times 58.5 = 175.5 \text{ g}$. Subtracting this solute mass from the total solution mass gives the mass of the solvent: $1000 - 175.5 = 824.5 \text{ g}$, which is 0.8245 kg. Dividing the moles of solute by the mass of the solvent \in kilograms provides the molality: $m = \frac{3}{0.8245} \approx 3.64 \text{ mol/kg}$. Expressing this result \in the required format yields $364 × 10^{-2}$ m.

Dilution problems rely on the principle that the total amount of solute remains constant, which is represented by the relationship $M_1V_1 = M_2V_2$. The molarity of the initial stock solution is found by dividing the moles of solute by the volume of the solution in liters. Because the molar mass of NaOH is 40 g/mol, 5 g of NaOH represents $\frac{5}{40}$ moles. With the solution volume at 450 mL, the initial molarity $M_1$ is $\frac{5}{40} \times \frac{1000}{450}$.

Substituting these values into the dilution equation provides $\left(\frac{5}{40} \times \frac{1000}{450}\right) \times V_1 = 0.1 \times 500$. Solving for $V_1$ gives the expression $V_1 = \frac{0.1 \times 500 \times 40 \times 450}{5 \times 1000}$, leading to a required volume of 180 mL.

Elemental analysis through combustion relies on the law of conservation of mass, where all carbon atoms in the sample are converted into $\mathrm{CO}_2$ and all hydrogen atoms into $\mathrm{H}_2\mathrm{O}$. To find the unknown mass of the original compound, we calculate the mass of carbon in the sample using the stoichiometry of combustion. Since the molar mass of $\mathrm{CO}_2$ is 44, the 0.220 g of $\mathrm{CO}_2$ produced contains $\frac{0.220}{44} \times 12 = 0.06$ g of carbon. Because this carbon accounts for 24% of the total mass, the total mass of the compound is found using $\frac{0.06}{24} \times 100 = 0.25$ g.

With the total mass of the sample established, we can determine the percentage of hydrogen. Since each mole of water contains two moles of hydrogen atoms, the 0.126 g of water produced corresponds to $\frac{0.126}{18} \times 2 = 0.014$ g of hydrogen. Dividing this mass by the total compound mass of 0.25 g yields the mass fraction $\frac{0.014}{0.25} = 0.056$. Converting this fraction to a percentage results in 5.6%. Expressing this value in the required format yields 56 $\times 10^{-1}$, identifying the required integer as 56.

Gravimetric sulfur estimation relies on the principle that the sulfur within an organic compound is quantitatively converted into barium sulfate. Because every mole of barium sulfate contains exactly one mole of sulfur atoms, the mass fraction of sulfur in the precipitate is determined by the ratio of the atomic mass of sulfur to the total molar mass of barium sulfate.

The molar mass of barium sulfate is 233 g/mol. The mass of sulfur present in the 1.4439 g of precipitate is derived by multiplying that mass by the ratio of the atomic mass of sulfur, which is 32, to the molar mass of the barium sulfate. To find the percentage of sulfur relative to the 0.471 g sample, we use the following relationship:

$\text{Percentage of Sulphur} = \frac{32}{233} \times \frac{1.4439}{0.471} \times 100$

Performing this calculation results in 42.10, which rounds to the nearest integer value of 42.

The non-stoichiometric nature of $\mathrm{Fe}_{0.93}\mathrm{O}$ is governed by the principle of charge neutrality, where the total positive charge from iron ions must balance the negative charge from oxide ions. By assigning $x$ moles to $\mathrm{Fe}^{2+}$ and $y$ moles to $\mathrm{Fe}^{3+}$, we establish the mass balance equation $x + y = 0.93$ and the charge balance equation $2x + 3y = 2$. Solving this system yields $x = 0.79$ moles of $\mathrm{Fe}^{2+}$ and $y = 0.14$ moles of $\mathrm{Fe}^{3+}$. These quantities directly determine the percentage composition, with $\frac{0.79}{0.93} \times 100 \approx 85\%$ for $\mathrm{Fe}^{2+}$ and the remainder, approximately $15\%$, for $\mathrm{Fe}^{3+}$.

To determine the equivalent weight, we calculate the $n$-factor, which represents the total change in oxidation state per mole of the formula unit during its conversion to $\mathrm{Fe}_{2}\mathrm{O}_{3}$. Since the average oxidation state of iron changes from $+\frac{2}{0.93}$ to $+3$, the $n$-factor is $\left(3 - \frac{2}{0.93}\right) \times 0.93$, which simplifies to $0.79$. Consequently, the equivalent weight is defined as the molecular weight divided by this $n$-factor of $0.79$. Finally, $\mathrm{Fe}_{0.93}\mathrm{O}$ is a well-characterized metal-deficient oxide, known as wüstite, which features a cubic closed packed arrangement of $\mathrm{O}^{2-}$ ions. Given these results, all four statements are correct.