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

The reaction between an alcohol and a Grignard reagent such as methylmagnesium iodide proceeds by the displacement of the hydroxyl hydrogen, which is released as methane gas. Because the stoichiometry is 1:1, the moles of methane gas evolved correspond directly to the moles of the alcohol sample.

At standard temperature and pressure, where 1 mole of gas occupies 22.7 L, the 3.1 mL of methane gas represents $\frac{3.1}{22700}$ moles. By equating this quantity to the moles of alcohol, calculated as the mass of 4.5 mg divided by its molecular weight M, we obtain the expression $\frac{4.5 \times 10^{-3}}{M} = \frac{3.1}{22700}$. Solving this relation gives a molecular weight M of approximately 33 g/mol.

The complete combustion of methane is represented by the balanced chemical equation $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$, which indicates that one mole of methane produces two moles of water. To determine the methane required for 81 g of water, we first calculate the total moles of water produced by dividing the given mass by the molar mass of water, 18 g/mol, which yields 4.5 mol. According to the stoichiometric ratio, the required amount of methane is half the number of moles of water, so we calculate $4.5 / 2 = 2.25$ mol. Expressing this result in the requested format gives $2.25 = 225 \times 10^{-2}$ mol, identifying the final value as 225.

When a Grignard reagent such as ethylmagnesium bromide reacts with an acidic proton source like methanol, the ethyl group abstracts a proton to generate ethane gas. We can quantify the amount of gas produced at standard temperature and pressure by using the molar volume of an ideal gas. Converting the provided volume of 2.24 mL into liters gives $2.24 \times 10^{-3}$ liters. Dividing this by the standard molar volume of 22.4 liters per mole shows that the reaction releases $10^{-4}$ moles of ethane. Because the molar mass of ethane is 30 grams per mole, the total mass in grams is calculated as $10^{-4} \times 30$, which results in 3 mg.

The Carius method relies on the principle that bromine atoms in an organic compound can be quantitatively precipitated as silver bromide when treated with silver nitrate. Since every mole of silver bromide contains one mole of bromine, we first determine the chemical amount of the precipitate. Dividing the mass of silver bromide, 0.36 g, by its molar mass, 188 g mol-1, gives the number of moles.

Multiplying this value by the atomic mass of bromine, 80 g mol-1, provides the total mass of bromine, calculated as $\frac{0.36}{188} \times 80 = 0.1532$ g. Finally, to find the percentage by mass, we divide this bromine content by the initial mass of the organic compound, 0.45 g, and multiply by 100. This calculation results in a bromine content of 34.04%.

The Kjeldahl method determines the nitrogen content of organic compounds by converting nitrogen into ammonia, which is then quantified through acid-base titration. The stoichiometry of the process dictates that the amount of nitrogen in the sample is directly proportional to the ammonia produced during digestion and subsequent distillation. From the provided data, the ammonia evolved is captured by an acid solution, and the resulting neutralization requires 30.0 mL of 0.25 N $\text{NaOH}$, which corresponds to 7.5 millimoles of $\text{NH}_3$.

Since every mole of $\text{NH}_3$ contains one mole of nitrogen, the sample contains $7.5 \times 10^{-3}$ moles of nitrogen. Using the atomic mass of 14 g/mol, the total mass of nitrogen is $7.5 \times 10^{-3} \times 14$ g. To calculate the mass percentage, we divide the mass of nitrogen by the total sample mass of 0.166 g and multiply by 100. This calculation is represented as:

$\frac{7.5 \times 10^{-3} \times 14}{0.166} \times 100 \simeq 63\%$

The mass of any chemical substance is determined by the product of its number of moles and its molar mass. Because the problem specifies that 0.15 moles of $A_2 B$ and 0.15 moles of $A B_3$ weigh the same, the molar masses of these two compounds must be identical. Designating the atomic weight of $A$ as $x$ and that of $B$ as $y$, the molar mass for $A_2 B$ is $2x + y$, whereas the molar mass for $A B_3$ is $x + 3y$. Equating these two molar masses gives $2x + y = x + 3y$. Simplifying this relationship by subtracting $x$ and $y$ from both sides leaves $x = 2y$, demonstrating that the atomic weight of $A$ is 2 times the atomic weight of $B$.

The principle of combustion analysis rests on the fact that all carbon atoms within an organic compound are converted into carbon dioxide during complete combustion, allowing the carbon content of the original substance to be calculated directly from the produced $\mathrm{CO}_2$. Because the molar mass of $\mathrm{CO}_2$ is $44\ \mathrm{g/mol}$, the $0.20\ \mathrm{g}$ of carbon dioxide produced corresponds to $\frac{0.20}{44}$ moles of the gas.

Since each molecule of carbon dioxide contains exactly one carbon atom, the number of moles of carbon is equivalent to the moles of carbon dioxide. Multiplying this mole value by the atomic mass of carbon, which is $12\ \mathrm{g/mol}$, yields a carbon mass of approximately $0.0545\ \mathrm{g}$. The percentage of carbon is determined by calculating the ratio of this carbon mass to the total original sample mass of $0.30\ \mathrm{g}$ and multiplying by $100$, resulting in $18.16\%$, which rounds to the nearest integer of $18\%$.

Stoichiometry relies on the balanced chemical equation to establish the exact molar ratio between reactants and products. To find the required amount of $HNO_3$, start by calculating the molar mass of $KNO_3$, which is $39 + 14 + 3 \times 16 = 101$ g/mol. Dividing the given mass of 110.0 g by this molar mass reveals that approximately $110.0 / 101 = 1.089$ mol of $KNO_3$ needs to be produced.

According to the balanced reaction, the ratio of $HNO_3$ consumed to $KNO_3$ produced is $4/3$. Applying this ratio, the moles of $HNO_3$ necessary are calculated as $(4/3) \times 1.089 = 1.452$ mol. Since the molar mass of $HNO_3$ is $1 + 14 + 3 \times 16 = 63$ g/mol, multiplying these moles by the molar mass yields the final requirement of $1.452 \times 63 = 91.5$ g.

The minimum molar mass of a protein containing a specific percentage of a component is determined by assuming there is exactly one unit of that component in the total molecular mass of the protein. Since glycine comprises 0.30% of the total mass and has a molar mass of 75 g mol^-1, we relate these quantities using the equation
$\frac{0.30}{100} \times M = 75$
where $M$ is the molar mass of the protein. Solving for $M$ yields 25000 g mol^-1, which is equivalent to 25 multiplied by 10^3 g mol^-1.

Molarity is defined as the concentration of a solution expressed as the number of moles of solute present in one liter of the solution. To determine this value for a commercial acid solution characterized by a percentage by mass, assuming a standard 100 g sample provides an efficient way to relate the mass of the solute to the total mass of the solution.

Since the solution is 35% HCl by mass, a 100 g sample contains 35 g of HCl. Using the combined atomic masses of hydrogen and chlorine, the molar mass of HCl is 36.5 g/mol, which means the amount of solute is 35 / 36.5 moles. The volume of this 100 g sample is determined by its density of 1.46 g/mL, resulting in 100 / 1.46 mL, which is converted to liters by dividing by 1000. Applying the formula for molarity, we divide the moles of solute by the volume in liters:

$M = \frac{35 / 36.5}{100 / (1.46 \times 1000)} = 14$

Stoichiometry dictates that the quantities of reactants consumed and products created are defined by the balanced chemical equation. The combustion reaction for the fuel is:
$\mathrm{C}_{15}\mathrm{H}_{30} + \frac{45}{2}\mathrm{O}_{2} \rightarrow 15\mathrm{CO}_{2} + 15\mathrm{H}_{2}\mathrm{O}$
Given 1 L of fuel with a density of 0.756 g/mL, the total mass of the fuel is calculated as $1000 \times 0.756 = 756$ grams. Using the molar mass of the fuel, which is $15(12) + 30 = 210$ g/mol, the amount of fuel in moles is $756 / 210 = 3.6$ moles.

According to the molar ratios, each mole of fuel reacts with 22.5 moles of oxygen. Therefore, 3.6 moles of fuel require $3.6 \times 22.5 = 81$ moles of oxygen. Since the molar mass of oxygen gas is 32 g/mol, the required mass of oxygen is $81 \times 32 = 2592$ grams. Similarly, the equation indicates that 1 mole of fuel produces 15 moles of carbon dioxide. The combustion of 3.6 moles of fuel yields $3.6 \times 15 = 54$ moles of carbon dioxide. Given that the molar mass of carbon dioxide is 44 g/mol, the total mass released is $54 \times 44 = 2376$ grams.

When performing multiplication and division, the final result must not exceed the precision of the measurement with the fewest significant figures. In the expression $\frac{0.02858 \times 0.112}{0.5702}$, the values 0.02858 and 0.5702 contain four significant figures, while 0.112 provides the limiting precision with only three significant figures. Because this smallest value determines the accuracy of the calculation, the final output needs to be rounded to three significant figures. Performing the arithmetic results in 0.005613..., which truncates to 0.00561 as the appropriate result.

Determining the count of specific atoms within a sample requires converting mass to moles, identifying the stoichiometric ratio of the target element within the molecule, and applying Avogadro's number to quantify the particles. The molar mass of $\mathrm{C}_7 \mathrm{H}_5 \mathrm{N}_3 \mathrm{O}_6$ is calculated as $(7 \times 12) + (5 \times 1) + (3 \times 14) + (6 \times 16) = 227$ g/mol. Dividing the given mass of 681 g by this molar mass confirms that the sample contains 3 moles of the compound.

Because every molecule of this substance contains 3 nitrogen atoms, the sample holds $3 \times 3 = 9$ moles of nitrogen atoms. To find the total count of atoms, this molar amount is multiplied by Avogadro's constant, $6.02 \times 10^{23}$ mol$^{-1}$, resulting in $9 \times 6.02 \times 10^{23} = 54.18 \times 10^{23}$ nitrogen atoms. Expressing this value in the required form of $x \times 10^{21}$ involves shifting the decimal point, which transforms $54.18 \times 10^{23}$ into $5418 \times 10^{21}$. Therefore, the value of $x$ is 5418.

Combustion analysis relies on the principle of conservation of mass, as all carbon and hydrogen atoms present in the organic sample are fully converted into $\mathrm{CO_2}$ and $\mathrm{H_2O}$ during the reaction. By calculating the mass of these specific elements, we can determine the remaining mass contributed by oxygen.

The mass of carbon is derived from the molar mass of $\mathrm{CO_2}$, which is 44 g/mol. Since 0.793 g of $\mathrm{CO_2}$ is produced, the carbon mass is calculated as $\frac{0.793}{44} \times 12 = 0.216$ g. Similarly, the molar mass of $\mathrm{H_2O}$ is 18 g/mol, meaning the hydrogen mass from 0.442 g of $\mathrm{H_2O}$ is $\frac{0.442}{18} \times 2 = 0.05$ g.

The mass of oxygen is obtained by subtracting the combined mass of carbon and hydrogen from the total mass of the compound, represented by $0.492 - 0.216 - 0.05 = 0.226$ g. The percentage composition of oxygen is then found using the expression $\frac{0.226}{0.492} \times 100 \approx 45.9$. Rounding to the nearest integer, the percentage is 46.

The relationship between the volumes of reacting gases is governed by Avogadro's law, which dictates that volume is directly proportional to the number of moles, meaning stoichiometric coefficients also represent the ratio of volumes. According to the balanced chemical equation $\mathrm{N}_2(\mathrm{g}) + 3 \mathrm{H}_2(\mathrm{g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{g})$, 1 volume of nitrogen gas reacts to produce 2 volumes of ammonia.

Since 20 L of ammonia was produced, we can calculate the nitrogen consumed by using this 1:2 ratio, which indicates that $20 / 2 = 10$ L of nitrogen participated in the reaction. With 56.0 L of nitrogen gas available at the start, the volume remaining after the reaction is calculated as $56.0 - 10 = 46$ L.

The hydrolysis of sulfuryl chloride yields a mixture of sulfuric acid and hydrochloric acid according to the chemical equation $\mathrm{SO}_2\mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{HCl}$. Stoichiometric analysis shows that the one mole of $\mathrm{H}_2\mathrm{SO}_4$ generated requires 2 moles of $\mathrm{NaOH}$ for neutralization, and the two moles of $\mathrm{HCl}$ produced require another 2 moles of $\mathrm{NaOH}$. This means every mole of $\mathrm{SO}_2\mathrm{Cl}_2$ results in an acidic mixture that consumes a total of 4 moles of $\mathrm{NaOH}$. Dividing the 16 moles of $\mathrm{NaOH}$ provided by the requirement of 4 moles per reaction unit reveals that 4 moles of $\mathrm{SO}_2\mathrm{Cl}_2$ were used.

Molality is defined as the number of moles of solute dissolved in one kilogram of solvent. Determining this value for the given solution requires calculating the mass of the glucose solute and the mass of the water solvent separately. Since the total mass of the solution is 250 g and it contains 10.8% carbon by weight, the total mass of carbon present is $250 \times 0.108 = 27$ grams.

The molar mass of glucose, $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$, is determined by summing the atomic weights of its constituents: $(6 \times 12) + (12 \times 1) + (6 \times 16) = 180$. Within one mole of glucose, carbon contributes 72 grams to the total molar mass of 180 grams. By setting up the ratio of carbon mass to glucose mass, we find the total mass of the glucose solute:
$\frac{72}{180} = \frac{27}{\text{mass of glucose}}$
Solving this equation gives a glucose mass of 67.5 grams. Subtracting this from the total solution weight provides the solvent mass: $250 - 67.5 = 182.5$ grams, which converts to 0.1825 kilograms.

To find the number of moles of glucose, we divide its mass by its molar mass:
$\frac{67.5}{180} = 0.375$
Finally, molality is determined by dividing the moles of glucose by the mass of the solvent in kilograms:
$\frac{0.375}{0.1825} \approx 2.06$
The molality of the solution is approximately 2.06 mol/kg.

Determining the mass of the product in this reaction requires identifying the limiting reagent, which is the reactant that is fully consumed and thus dictates the theoretical yield. First, calculate the molar masses of the reactants using the provided atomic masses: 143 g/mol for $\mathrm{Cu}_2\mathrm{O}$ and 159 g/mol for $\mathrm{Cu}_2\mathrm{S}$. Converting the given masses into moles yields 20 moles of $\mathrm{Cu}_2\mathrm{O}$ and 30 moles of $\mathrm{Cu}_2\mathrm{S}$.

The balanced chemical equation $2 \mathrm{Cu}_2\mathrm{O} + \mathrm{Cu}_2\mathrm{S} \rightarrow 6 \mathrm{Cu} + \mathrm{SO}_2$ indicates that two moles of $\mathrm{Cu}_2\mathrm{O}$ react with one mole of $\mathrm{Cu}_2\mathrm{S}$. With 20 moles of $\mathrm{Cu}_2\mathrm{O}$ available, only 10 moles of $\mathrm{Cu}_2\mathrm{S}$ are required for a complete reaction. Since 30 moles of $\mathrm{Cu}_2\mathrm{S}$ are present, $\mathrm{Cu}_2\mathrm{S}$ is \in excess and $\mathrm{Cu}_2\mathrm{O}$ is the limiting reagent.

Based on the stoichiometry of the reaction, two moles of $\mathrm{Cu}_2\mathrm{O}$ produce six moles of copper, which simplifies to a ratio of one mole of $\mathrm{Cu}_2\mathrm{O}$ producing three moles of copper. Applying this to the 20 moles of the limiting reagent results in the formation of 60 moles of copper. Multiplying this amount by the atomic mass of copper, 63.5, gives a total mass of 3810 g.

Determining the concentration of the resulting product relies on understanding the stoichiometry of the neutralization reaction. The balanced chemical equation is $\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}$ which establishes that one mole of sodium sulfate is formed for every two moles of sodium hydroxide consumed. Calculating the initial quantities, we find there are $2 × 0.2 = 0.4$ moles of sulfuric acid and $2 × 0.1 = 0.2$ moles of sodium hydroxide available. Because sodium hydroxide is the limiting reagent, the total amount of product formed corresponds to half the moles of the limiting reactant, or $0.2 / 2 = 0.1$ moles of sodium sulfate. With the total volume of the solution being $2 + 2 = 4$ liters, the molarity is determined by the ratio $\frac{0.1}{4} = 0.025 \mathrm{M}$. Converting this concentration into millimolar units involves multiplying by $1000$, resulting in 25 millimolar.

The empirical formula of a substance is determined by calculating the relative molar ratio of each element present, which is found by dividing the mass of each component by its corresponding atomic mass. For this substance, we convert the masses of $7.5\ \mathrm{g}$ of hydrogen, $60\ \mathrm{g}$ of oxygen, and $48.5\ \mathrm{g}$ of carbon into molar amounts by dividing them by their respective atomic weights of $1$, $16$, and $12$.

These calculations yield molar proportions that simplify to a ratio of roughly two hydrogen atoms and one oxygen atom for every carbon atom, establishing an empirical formula of $\mathrm{CH}_2\mathrm{O}$. By comparing this result to the provided options, we find that $\mathrm{CH}_3\mathrm{COOH}$ represents $\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2$, which reduces to this same empirical ratio, and $\mathrm{HCHO}$ matches the empirical formula directly. Because the remaining structures possess different atomic proportions, the provided analytical data aligns with exactly two of the given chemical formulae.