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

The amount of product formed in this combustion reaction is strictly limited by the reagent that is consumed first, known as the limiting reagent. We begin by converting the mass of each reactant into moles using their respective molar masses: 100 g of methane ($\mathrm{CH}_4$) equals 6.25 moles (as $100/16 = 6.25$), and 208 g of oxygen ($\mathrm{O}_2$) equals 6.5 moles (as $208/32 = 6.5$). The balanced chemical equation for the combustion is $\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$, which tells us that 1 mole of methane requires 2 moles of oxygen to react completely. Consequently, the 6.25 moles of methane present would require 12.5 moles of oxygen, which far exceeds the 6.5 moles we actually have. This confirms that oxygen is the limiting reagent and determines the maximum possible product formation. Based on the stoichiometry, 2 moles of oxygen produce 1 mole of carbon dioxide, so our 6.5 moles of oxygen will yield 3.25 moles of carbon dioxide. Multiplying this amount by the molar mass of carbon dioxide 44 g/mol results in a total of 143 g of carbon dioxide produced.

The total quantity of solute in a final mixture is determined by the principle of additivity, where the absolute number of moles is conserved regardless of changes in solvent volume during dilution. When a specific volume fraction of an initial solution is taken, the number of moles transferred is proportional to that fraction, assuming the solution is homogeneous. This quantity serves as the base amount, which remains unchanged upon adding water. Any subsequent addition of solute increases the total mole count linearly, allowing for a direct summation of the transferred and added amounts.

Applying this logic, the initial solution contains $0.02\ \mathrm{m\ mol}$ of $\mathrm{H}_2\mathrm{SO}_4$. Taking $0.50\%$ of this solution transfers an amount equal to $0.02\ \mathrm{m\ mol} \times 0.005 = 0.0001\ \mathrm{m\ mol}$. When the additional $0.01\ \mathrm{m\ mol}$ of $\mathrm{H}_2\mathrm{SO}_4$ is introduced to this aliquot, the total content becomes $0.0001\ \mathrm{m\ mol} + 0.01\ \mathrm{m\ mol} = 0.0101\ \mathrm{m\ mol}$. To express this final quantity in the required format of $\times 10^3\ \mathrm{m\ mol}$, the total is divided by $10^3$, resulting in $0.0000101 \times 10^3\ \mathrm{m\ mol}$.

The stoichiometry of the reaction $X + Y + 3Z \rightleftharpoons XYZ_3$ dictates that the limiting reagent determines the total amount of product formed. With one mole of both $X$ and $Y$ available alongside only 0.05 moles of $Z$, it is evident that $Z$ is the limiting reagent because it will be fully consumed long before the other reactants. Since the balanced equation indicates that 3 moles of $Z$ are required to produce 1 mole of $XYZ_3$, the 0.05 moles of $Z$ will generate $\frac{0.05}{3}$ moles of the product. To find the yield in grams, we first calculate the molar mass of $XYZ_3$ as $10 + 20 + 3 \times 30 = 120\ \mathrm{g/mol}$. Multiplying the amount of product in moles by this molar mass results in $\frac{0.05}{3} \times 120$, which equals 2 grams.

When an organic compound undergoes complete combustion, its carbon content is fully oxidized to carbon dioxide, and its hydrogen content is converted into water. By utilizing stoichiometric ratios, the mass percentage of each element in the original sample can be determined from the combustion products. The mass percentage of carbon is calculated by finding the proportion of carbon in carbon dioxide relative to the total sample, represented by $\frac{12}{44} \times \frac{0.7938\,\mathrm{g}}{0.492\,\mathrm{g}} \times 100$, which equals $43.47\%$.

Similarly, the hydrogen mass percentage is derived from the proportion of hydrogen in water relative to the sample, using $\frac{2}{18} \times \frac{0.4428\,\mathrm{g}}{0.492\,\mathrm{g}} \times 100$, which results in $9.9\%$. Since the compound consists entirely of carbon, hydrogen, and oxygen, the oxygen content is determined by subtracting the combined percentages of carbon and hydrogen from the total mass percentage of $100\%$. Performing this calculation, $100 - (43.47 + 9.9)$, yields approximately $46\%$ for the oxygen composition.

When dealing with chemical reactions where specific amounts of reactants are mixed, the quantity of product formed is dictated by the limiting reagent, which is the substance completely consumed first. To identify this reagent for the reaction $\mathrm{Fe}_3\mathrm{O}_4(\text{s}) + 4\,\mathrm{CO}(\text{g}) \rightarrow 3\,\mathrm{Fe}(\text{l}) + 4\,\mathrm{CO}_2(\text{g})$, we must first convert the mass of each reactant into moles. Given that the molar mass of $\mathrm{Fe}_3\mathrm{O}_4$ is $232\,\text{g/mol}$, the $4.640\,\text{kg}$ provided equals $20\,\text{mol}$. Similarly, for $\mathrm{CO}$, which has a molar mass of $28\,\text{g/mol}$, the $2.520\,\text{kg}$ provided equates to $90\,\text{mol}$.

Determining the limiting reagent involves comparing the available moles relative to their stoichiometric coefficients in the balanced equation. For $\mathrm{Fe}_3\mathrm{O}_4$, dividing its $20\,\text{mol}$ by its coefficient of $1$ yields $20$. For $\mathrm{CO}$, dividing its $90\,\text{mol}$ by its coefficient of $4$ results in $22.5$. Because the ratio for $\mathrm{Fe}_3\mathrm{O}_4$ is lower, it acts as the limiting reagent and defines the maximum amount of iron that can be synthesized.

According to the balanced equation, $1\,\text{mol}$ of $\mathrm{Fe}_3\mathrm{O}_4$ yields $3\,\text{mol}$ of elemental iron. Therefore, the consumption of $20\,\text{mol}$ of $\mathrm{Fe}_3\mathrm{O}_4$ produces $60\,\text{mol}$ of iron. Multiplying this molar quantity by the atomic mass of iron, which is $56\,\text{g/mol}$, gives a total mass of $60 \times 56$, resulting in $3360\,\text{g}$ of iron.

When determining significant figures, leading zeros serve strictly as placeholders to establish the decimal position and do not possess significance. Conversely, all non-zero digits are counted, along with any trailing zeros that appear to the right of the decimal point. Applying these rules to 0.00340, we disregard the leading zeros and count the remaining digits 3, 4, and 0, which results in a total of 3 significant figures.

The principle of chemical equivalence states that at the point of complete reaction, the milliequivalents of the oxidizing agent must match those of the reducing agent. We determine these values by calculating the product of molarity, volume, and the valency factor, where the valency factor represents the total change in oxidation state per mole of the reactant. Dichromate acts as an oxidant, reducing chromium from +6 to +3, which across two chromium atoms results in a valency factor of 6. Conversely, iron oxidizes from +2 to +3, providing a valency factor of 1. Equating the milliequivalents for both reactants leads to the following expression:

$(0.03 \times 20) \times 6 = (M \times 15) \times 1$

Solving this equation yields $3.6 = 15M$, which results in a molarity of 0.24. Expressed in the requested format, the value is 24 × 10^-2 molar.

Neutralisation depends on the basicity of the acids, which dictates the molar ratio between the acid and the base, $\mathrm{NaOH}$. Phosphorous acid, $\mathrm{H}_3\mathrm{PO}_3$, acts as a diprotic acid by providing two ionizable hydrogen atoms per molecule, requiring two moles of $\mathrm{NaOH}$ for every mole of acid. With 50 mL of a 1 M solution, there are 50 millimoles of the acid, necessitating 100 millimoles of $\mathrm{NaOH}$; at a concentration of 1 M, this requires 100 mL of the base.

Hypophosphorous acid, $\mathrm{H}_3\mathrm{PO}_2$, acts as a monoprotic acid because only one of its hydrogen atoms is replaceable in aqueous solution. Given 100 mL of 2 M solution, the total quantity is 200 millimoles of the acid, which consumes an equivalent amount of $\mathrm{NaOH}$ due to the $1:1$ reaction stoichiometry. Consequently, 200 millimoles of $\mathrm{NaOH}$ at a concentration of 1 M necessitates a total volume of 200 mL.

The energy required for daily biological functions is derived from the combustion of glucose, which provides a specific amount of energy per mole. By utilizing the enthalpy of combustion, which is 2700 kJ mol^-1, we can determine the necessary molar quantity by dividing the total daily energy requirement of 10000 kJ by this enthalpy value, expressed by the calculation $\frac{10000}{2700} \approx 3.704$ mol.

To obtain the corresponding mass of glucose, we multiply this molar quantity by the molar mass of 180.0 g mol^-1. The resulting computation of $3.704 \times 180.0 \approx 666.66$ g allows us to conclude that approximately 667 g of glucose are needed to meet the daily energy demand.

Balancing redox equations requires ensuring that the number of electrons lost during oxidation equals the number of electrons gained during reduction. The oxidation half-reaction involving the oxalate ion proceeds as $\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \longrightarrow 2\mathrm{CO}_{2} + 2e^{-}$, while the reduction of the permanganate ion is represented by $\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \longrightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}$. To equalize the electron transfer between these two processes, multiply the oxidation half-reaction by five and the reduction half-reaction by two, resulting in ten electrons exchanged in each case.

Combining these balanced half-reactions yields the overall balanced equation:

$16\mathrm{H}^{+} + 5\mathrm{C}_{2}\mathrm{O}_{4}^{2-} + 2\mathrm{MnO}_{4}^{-} \longrightarrow 10\mathrm{CO}_{2} + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_{2}\mathrm{O}$

By comparing the coefficient of the hydrogen ions in this balanced equation to the original expression, the value of $c$ is 16.

Chemical stoichiometry relies on the conservation of mass and molar equivalence, where the conversion of aniline into acetanilide proceeds in a one-to-one molar ratio. The moles of aniline are determined by dividing the given 1.86 g by its molar mass of 93 g/mol:
$n = \frac{1.86}{93} = 0.02 \, \text{mol}$
Because every mole of aniline generates one mole of acetanilide, the reaction theoretically produces 0.02 moles of the final product. Multiplying this amount by the molar mass of acetanilide, which is 135 g/mol, yields the total mass expected in the absence of any losses:
$m_{\text{theoretical}} = 0.02 \times 135 = 2.70 \, \text{g}$
Since 10% of the product is lost during purification, only 90% of the theoretical quantity remains available. Applying this efficiency to the theoretical mass provides the actual yield:
$m_{\text{actual}} = 2.70 \times 0.90 = 2.43 \, \text{g}$
Expressing this final mass as $243 \times 10^{-2} \, \text{g}$ confirms that the required numerical component is 243.

Stoichiometry principles dictate that the quantity of product formed is directly proportional to the amount of reactant consumed, as defined by the balanced chemical equation $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$.

Determine the molar mass of sodium oxide by summing the atomic weights of two sodium atoms and one oxygen atom, which equals 62.0 g/mol. Given 20.0 g of sodium oxide, the number of moles is calculated by dividing this mass by the molar mass. Because the balanced equation shows one mole of sodium oxide produces two moles of sodium hydroxide, the total moles of product is double the moles of the reactant.

The concentration is determined by dividing the moles of sodium hydroxide by the volume of the solution in liters, where 500 mL is equivalent to 0.5 L. The calculation is:
$\frac{20.0}{62.0} \times 2 \div 0.5 = 1.29$
This concentration is 1.29 M, which is equivalent to 12.9 x 10^-1 M. Rounding 12.9 to the nearest integer yields 13.

Calculating the total number of atoms requires finding the amount of substance in moles. This is achieved by dividing the sample mass of 8 grams by the molar mass of 23 grams per mole. Multiplying the resulting mole count by Avogadro's constant, 6.02 x 10^23, provides the total quantity of atoms in the sample.

The relationship is given by:
$\text{Number of atoms} = \frac{8}{23} \times 6.02 \times 10^{23} \approx 2.09 \times 10^{23}$
When matching this result to the expression $x \times 10^{23}$, we find that x is approximately 2.09. Rounding to the nearest integer yields 2.

The molarity of a solution is determined by the ratio of the number of moles of the solute to the volume of the solution in liters. Begin by finding the moles of sodium atoms by dividing the given mass of $3.45 \text{g}$ by the molar mass of $23.0 \text{g/mol}$, resulting in $0.15 \text{mol}$ of sodium.

Because each molecule of $\mathrm{Na}_3\mathrm{PO}_4$ contains three sodium atoms, the molar quantity of the compound is one-third of the sodium molar quantity, which is $\frac{0.15}{3} = 0.05 \text{mol}$ of $\mathrm{Na}_3\mathrm{PO}_4$. With a solution volume of $100 \text{mL}$, equivalent to $0.1 \text{L}$, the molarity calculation is as follows:
$\text{Molarity} = \frac{0.05 \text{ mol}}{0.1 \text{ L}} = 0.5 \text{ mol L}^{-1}$
Converting this concentration into the requested units of $10^{-2} \text{mol L}^{-1}$ requires multiplying by $100$, yielding a final value of $50$.

The stoichiometry of the dehydration reaction indicates that one mole of 3-hydroxypropanal converts into one mole of acrolein. First, determine the number of moles of acrolein produced by dividing the actual mass of 7.8 grams by its molar mass of 56 grams per mole. Because the process operates at a 64 percent yield, the theoretical moles of reactant required must be higher, calculated by dividing the actual moles by 0.64. Since the molar ratio is 1:1, this value is equivalent to the number of moles of 3-hydroxypropanal needed. To calculate the final mass, multiply these moles by the molar mass of 3-hydroxypropanal, 74 grams per mole, which is expressed as:
$x = \frac{7.8}{56 \times 0.64} \times 74$
This calculation results in approximately 16.10 grams, which rounds to an integer value of 16.

Molarity is defined as the number of moles of solute per liter of solution, which is calculated by dividing the mass concentration of the solute by its molar mass. The molar mass of glucose, $\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}$, is found by summing the atomic masses of its constituent atoms: $(6 \times 12) + (12 \times 1) + (6 \times 16) = 180 \mathrm{g \ mol^{-1}}$. Dividing the concentration of $0.72 \mathrm{g \ L^{-1}}$ by the molar mass of $180 \mathrm{g \ mol^{-1}}$ yields a result of $0.004 \mathrm{M}$, which is expressed as $4 \times 10^{-3} \mathrm{M}$.

To determine the molarity of a solution, one must define the relationship between the moles of solute and the total volume of the solution. Since molality is defined as moles of solute per kilogram of solvent, a concentration of $3.30\ \mathrm{mol\ kg^{-1}}$ implies that $3.30\ \mathrm{moles}$ of $\mathrm{KCl}$ are dissolved in $1000\ \mathrm{g}$ of solvent. The mass of this solute is calculated as $3.30\ \mathrm{mol} \times 74.5\ \mathrm{g\ mol^{-1}}$, which equals $245.85\ \mathrm{g}$. Combining the mass of the solute and the mass of the solvent gives a total solution mass of $1245.85\ \mathrm{g}$.

With the density of the solution given as $1.20\ \mathrm{g\ mL^{-1}}$, the total volume is found using the relation $V = \frac{\text{mass}}{\text{density}}$, resulting in $V = \frac{1245.85\ \mathrm{g}}{1.20\ \mathrm{g\ mL^{-1}}} = 1038.2\ \mathrm{mL}$. Converting this volume into liters gives $1.0382\ \mathrm{L}$. Molarity is then calculated by dividing the number of moles of solute by the volume of the solution in liters, $M = \frac{3.30\ \mathrm{mol}}{1.0382\ \mathrm{L}}$, which equals approximately $3.17\ \mathrm{mol\ L^{-1}}$. Rounding to the nearest integer, the molarity of the solution is $3$.

The principle of stoichiometric neutralization governs this calculation, where the number of moles of acid and base reacting is determined by their molarity and volume. The balanced chemical equation is $\mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O}$. Since the reaction proceeds in a 1:1 molar ratio, the limiting reactant dictates how much of the excess reagent is consumed.

Calculating the initial amounts, 250 mL of 0.5 M NaOH provides 125 mmol, while 500 mL of 1 M HCl provides 500 mmol. Because NaOH is the limiting reagent, 125 mmol of HCl is consumed, which leaves 375 mmol of HCl unreacted. Converting this to moles gives 0.375 moles. Multiplying this value by Avogadro's constant of $6.022 \times 10^{23}$ yields approximately $225.8 \times 10^{21}$ molecules, which rounds to $226 \times 10^{21}$.

Mole fraction represents the ratio of the number of moles of a specific component to the total number of moles present in a solution. In a 100 molal aqueous solution, this definition implies there are 100 moles of solute dissolved in 1000 grams of the solvent, water. Given that water has a molar mass of 18 grams per mole, the solvent component contributes $\frac{1000}{18} = 55.5$ moles to the system.

To determine the mole fraction of the solute, we divide the solute moles by the total moles in the mixture, which is calculated as $\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$. Substituting the known quantities results in $\frac{100}{100 + 55.5}$, or $\frac{100}{155.5}$, which equals approximately 0.643. Expressing this result as a value scaled by $10^{-2}$ gives $64.3 \times 10^{-2}$, which rounds to 64 when considering the nearest integer requirement.

The Carius method quantifies bromine by converting it completely into silver bromide, allowing us to determine the element's percentage composition based on the stoichiometry of the precipitate. Given the atomic masses of silver as 108 and bromine as 80, the molar mass of silver bromide is $188 \text{ g/mol}$. This confirms that for every 188 mass units of the precipitate, 80 units correspond to bromine.

To find the mass of bromine in the $0.188 \text{ g}$ of silver bromide obtained, multiply the mass of the precipitate by the mass fraction of bromine, which is $0.188 \times \frac{80}{188} = 0.08 \text{ g}$. Calculating the final percentage involves comparing this mass to the original $0.2 \text{ g}$ sample as shown below:

$\text{Percentage of Br} = \frac{0.08}{0.2} \times 100 = 40\%$