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

Converting molality into molarity requires connecting the mass of solute in a fixed amount of solvent to the total volume occupied by the resulting solution. Consider a basis of 1000 g of solvent, which contains 6.50 moles of KOH. Given the atomic masses, the molar mass of KOH is calculated as $39.0 + 16.0 + 1.0 = 56.0\ \mathrm{g\ mol}^{-1}$, so the mass of the solute is $6.50 \times 56.0 = 364\ \mathrm{g}$. Combining the solute and solvent yields a total solution mass of $364 + 1000 = 1364\ \mathrm{g}$.

The density of the solution, 1.89 g/cm³, allows for calculating the volume occupied by this mass. Dividing the total solution mass of 1364 g by the density gives a volume of $V = \frac{1364}{1.89} = 721.69\ \mathrm{cm}^3$, or 0.72169 dm³. Molarity is defined as moles of solute per unit volume of solution in dm³, so the final concentration is $M = \frac{6.50}{0.72169} = 9.0065\ \mathrm{mol\ dm}^{-3}$. Rounding this value to the nearest integer provides an answer of 9.

The synthesis of benzyl trimethyl ammonium bromide from benzylamine involves exhaustive methylation, where the nitrogen atom participates in nucleophilic substitution to replace its hydrogen atoms with methyl groups. The molar mass of the product, $\mathrm{C_6H_5CH_2N^+(CH_3)_3Br^-}$, is determined by summing the atomic masses of its constituents: 10 carbons (120.0), 16 hydrogens (16.0), 1 nitrogen (14.0), and 1 bromine (80.0), which yields 230.0 g/mol. With 23 g of the product obtained, the number of moles formed is 23 divided by 230, which equals 0.1 mole.

The stoichiometry of this process is represented by the chemical equation $\mathrm{C_6H_5CH_2NH_2 + 3CH_3Br \rightarrow C_6H_5CH_2N^+(CH_3)_3Br^-}$. According to this relationship, 1 mole of benzylamine requires 3 moles of bromomethane for the complete substitution to form the quaternary ammonium salt. Therefore, 0.1 mole of benzylamine consumes 0.1 multiplied by 3, which is 0.3 moles of bromomethane. Expressing this result as $n \times 10^{-1}$ reveals that n equals 3.

In combustion analysis, we utilize the principle of conservation of mass to relate the combustion products back to the original organic sample. Because all hydrogen atoms in the organic compound are converted into water, we can determine the mass of hydrogen present by examining the stoichiometry of water. Knowing that 18 grams of water contain 2 grams of hydrogen, the mass of hydrogen in the 210 grams of water produced is calculated as $\frac{2}{18} \times 210 = \frac{70}{3}$ grams.

To find the percentage composition of hydrogen in the original 750 grams of the organic compound, we divide the mass of the hydrogen by the mass of the organic compound and multiply by 100. This is represented by the formula $\frac{70/3}{750} \times 100$, which yields a result of 3.11 percent. When rounded to the nearest integer, this percentage value becomes 3.

Percentage yield is determined by comparing the actual quantity of the product formed to the theoretical maximum possible amount under the given reaction conditions. Calculating the actual amount of amide yields $0.00077\ \text{mol}$ from the measured $0.210\ \text{g}$ mass and $273\ \text{g/mol}$ molar mass. Based on the reactant quantities of $0.001\ \text{mol}$ and $0.0023\ \text{mol}$ derived from their respective masses and molar masses, the reaction stoichiometry dictates a theoretical maximum yield of $0.001\ \text{mol}$. Applying the relationship $\% \;\text{ yield } = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$ leads to the calculation $\frac{0.00077}{0.001} \times 100$, which results in a value of $77 \%$.

The combustion of propane is governed by the stoichiometry shown in the balanced equation $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$. We first convert the mass of the reactants into moles using their respective molar masses, finding that 100 g of propane corresponds to 2.27 mol, while 1000 g of oxygen corresponds to 31.25 mol. Comparing these values, we observe that since 1 mol of propane requires 5 mol of oxygen, the reaction consumes 11.35 mol of oxygen. Because the available 31.25 mol of oxygen exceeds the required 11.35 mol, propane acts as the limiting reagent, leaving 19.90 mol of oxygen unreacted.

The amount of products formed depends entirely on the consumption of the limiting reagent, resulting in $3 \times 2.27 = 6.81$ mol of $\mathrm{CO_2}$ and $4 \times 2.27 = 9.08$ mol of steam. The final mixture within the flask consists of 19.90 mol of unreacted $\mathrm{O_2}$, 6.81 mol of $\mathrm{CO_2}$, and 9.08 mol of steam, which sums to a total of 35.79 mol. The mole fraction of $\mathrm{CO_2}$ is calculated by the ratio $\frac{6.81}{35.79} = 0.19$. Given that this mole fraction equals $x \times 10^{-2}$, it follows that $x$ is 19.

Chemical reactions involving solutions depend entirely on the stoichiometric proportions of the reactants, where the species in the shortest supply limits the overall yield. Determining the outcome requires calculating the actual amount of each substance available by multiplying its molar concentration by the total solution volume.

For this specific transformation, lead nitrate reacts with chromic sulphate according to the balanced equation $3\ \mathrm{Pb}(\mathrm{NO}_3)_2 + \mathrm{Cr}_2(\mathrm{SO}_4)_3 \rightarrow 3\ \mathrm{PbSO}_4 + 2\ \mathrm{Cr}(\mathrm{NO}_3)_3$. Applying the provided values, the quantity of lead nitrate present is $35\ \mathrm{mL} \times 0.15\ \mathrm{M} = 5.25\ \mathrm{mmol}$, while the quantity of chromic sulphate is $20\ \mathrm{mL} \times 0.12\ \mathrm{M} = 2.4\ \mathrm{mmol}$.

Based on the stoichiometry, the reaction requires 3 moles of lead nitrate for every 1 mole of chromic sulphate. Assessing the availability, the $5.25\ \mathrm{mmol}$ of lead nitrate would only require $1.75\ \mathrm{mmol}$ of chromic sulphate for a complete reaction. Since the system holds $2.4\ \mathrm{mmol}$ of chromic sulphate, there is a surplus, which confirms that lead nitrate acts as the limiting reagent.

Because the limiting reagent governs the formation of products, the yield depends entirely on the amount of lead nitrate consumed. The balanced equation demonstrates a 3:3, or 1:1, molar ratio between lead nitrate consumed and lead sulphate precipitate formed. Therefore, the $5.25\ \mathrm{mmol}$ of lead nitrate produces $5.25\ \mathrm{mmol}$ of lead sulphate, which equates to $525 \times 10^{-5}\ \mathrm{moles}$.

Molality is defined as the amount of solute in moles per kilogram of solvent. Start by considering 1000 cm^3 of the solution. Using the given density of 1.2 g cm^-3, the total mass is calculated as $1000 \times 1.2 = 1200$ grams. By assuming the solute volume is negligible, the mass of the solvent is taken as 1000 grams, which is equal to 1 kilogram.

The mass of the NaOH solute is $1200 - 1000 = 200$ grams. The molar mass of NaOH is calculated from the atomic masses as $23.0 + 16.0 + 1.0 = 40.0$ grams per mole. The number of moles of NaOH is determined by the equation $\frac{200}{40.0} = 5$ moles. Since molality is the number of moles per kilogram of solvent, the final value is $\frac{5}{1} = 5$ molal.

Balanced chemical equations provide the stoichiometric mole ratios necessary to calculate the mass of reactants needed for a specific yield of products, adhering to the law of conservation of mass. The complete combustion of butane is represented by the equation:
$\mathrm{C}_4\mathrm{H}_{10} + \frac{13}{2}\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 5\mathrm{H}_2\mathrm{O}$
Calculating the moles of water produced by dividing 72.0 g by its molar mass of 18 g/mol yields 4 moles. Because the stoichiometry dictates that 1 mole of butane produces 5 moles of water, the combustion requires one-fifth of that quantity, or 0.8 moles of butane. Multiplying these 0.8 moles by the molar mass of butane of 58 g/mol gives a total mass of 46.4 g. Expressing 46.4 g in terms of a value scaled by $10^{-1}$ g results in 464.

When determining the precision of a number, exponential factors like $10^{-3}$ are excluded because they do not contribute to the count of significant figures. In the coefficient $50000.020$, all non-zero digits are significant, and the four internal zeros placed between the initial $5$ and the $2$ are significant as well according to established rules. Because the final trailing zero appears after a decimal point, it is also considered significant, which results in a final count of 7 significant figures.

The complete combustion of ethane follows the balanced equation $C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O$, which establishes that every mole of ethane produces three moles of water. We determine the moles of ethane by dividing the mass of 3 g by the molar mass of 30 g/mol, resulting in 0.1 mol of ethane. According to the reaction stoichiometry, this quantity yields 0.3 mol of water. To calculate the number of molecules, we use the expression $0.3 \times 6.023 \times 10^{23} = 1.8069 \times 10^{23}$, which is equivalent to $18.069 \times 10^{22}$. Rounding this value to the nearest integer gives 18.

The law of equivalence states that in any balanced redox reaction, the gram equivalents of the oxidizing agent must equal the gram equivalents of the reducing agent. Gram equivalents are calculated by multiplying the molarity, the volume, and the n-factor, which represents the total change in oxidation state per mole of reactant.

For $\mathrm{CrO}_4^{2-}$, the chromium atom undergoes a reduction from an oxidation state of +6 to +3 in $\mathrm{Cr}(\mathrm{OH})_4^{-}$, resulting in an n-factor of 3. For $\mathrm{S}_2\mathrm{O}_3^{2-}$, the sulfur atoms oxidize from +2 to +6 in the $\mathrm{SO}_4^{2-}$ product. Because each thiosulfate ion contains two sulfur atoms, the total change in oxidation state is $2 \times (6 - 2)$, which gives an n-factor of 8.

Applying the relationship between equivalents, we set up the following equation using the given concentrations and volume:
$0.154 \times V_1 \times 3 = 0.25 \times 40 \times 8$
To determine the unknown volume $V_1$, we rearrange the terms:
$V_1 = \frac{0.25 \times 40 \times 8}{0.154 \times 3}$
This expression results in approximately 173.16. Rounding to the nearest integer, the required volume of the chromate solution is 173 mL.

When substances are present in an equimolar mixture, they contribute to the total mass based on an equal number of moles, allowing us to express the total weight through the sum of their individual molar masses multiplied by that common mole count. Given that the molar mass of $\mathrm{NaOH}$ is 40 g/mol and $\mathrm{Na}_2\mathrm{CO}_3$ is 106 g/mol, if we designate the shared mole quantity as $n$, the relationship is established by the equation:

$40n + 106n = 4$

Simplifying this expression gives $146n = 4$, which leads to the mole count $n = \frac{4}{146}$. Since the mass of $\mathrm{NaOH}$ is determined by multiplying $n$ by its molar mass, the calculation for $x$ becomes $x = \frac{4}{146} \times 40$, which equals approximately 1.095 g. Expressed as the nearest integer, this value is 1 g.

The balanced reaction $2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{g})$ dictates that two units of sulfur dioxide consume one unit of oxygen to produce two units of sulfur trioxide. Since 250 mbar of sulfur dioxide requires only 125 mbar of oxygen for complete reaction, sulfur dioxide serves as the limiting reactant. As the reaction completes, the sulfur dioxide is entirely consumed, while the initial 750 mbar of oxygen is reduced by 125 mbar to leave 625 mbar. Simultaneously, the stoichiometry generates 250 mbar of sulfur trioxide, and summing the remaining 625 mbar of oxygen with the 250 mbar of sulfur trioxide yields a final total pressure of 875 mbar.

Combustion analysis relies on the principle of conservation of mass, asserting that every carbon atom in the organic sample forms carbon dioxide, and every hydrogen atom forms water. By determining the mass of these elements individually, you can isolate the mass of the oxygen present in the original compound by subtracting the other masses from the initial total.

Calculating the carbon content begins by finding the moles of carbon dioxide produced: $\frac{2.64}{44} = 0.06$. Since each mole of carbon dioxide contains one mole of carbon, this accounts for a mass of $0.06 \times 12 = 0.72$ grams of carbon. The hydrogen content is determined by analyzing the water produced, which gives $\frac{1.08}{18} = 0.06$ moles of water molecules. Because each water molecule contains two hydrogen atoms, this results in $0.06 \times 2 = 0.12$ moles of hydrogen atoms, which amounts to $0.12$ grams of hydrogen.

Subtracting the combined mass of carbon and hydrogen from the total sample mass of $1.80$ grams leaves the mass of oxygen as $1.80 - 0.72 - 0.12 = 0.96$ grams. Dividing this oxygen mass by the total sample weight determines the final percentage: $\frac{0.96}{1.80} \times 100 = 53.33$.

Avogadro's Law dictates that for gases at constant temperature and pressure, the stoichiometric coefficients in a balanced chemical equation are directly proportional to the volumes of the reactants and products. The general equation for the complete combustion of a hydrocarbon is $\mathrm{C}_{x}\mathrm{H}_{y} + (x + \frac{y}{4}) \mathrm{O}_{2} \rightarrow x \mathrm{CO}_{2} + \frac{y}{2} \mathrm{H}_{2}\mathrm{O}$. Because the volume of carbon dioxide produced is four times the volume of the original hydrocarbon, we can determine that the stoichiometric coefficient $x$ is 4.

Given that the volume of oxygen required is six times the volume of the hydrocarbon, the coefficient for oxygen, $(x + \frac{y}{4})$, must be equal to 6. Substituting the known value of $x = 4$ into this relation gives $4 + \frac{y}{4} = 6$. Subtracting 4 from both sides leaves $\frac{y}{4} = 2$, which leads directly to the result $y = 8$.

Molarity defines the concentration of a solution as the number of moles of a solute divided by the solution volume in liters. Determining the number of moles of compound A involves dividing the mass of 4.5 g by its molecular weight of 90 g/mol, yielding $\frac{4.5}{90} = 0.05$ mol, while converting the solution volume from 250 mL into liters provides 0.25 L. Dividing the amount of solute by the volume in liters gives the molarity:
$M = \frac{0.05}{0.25} = 0.2$
Since this concentration is expressed in the form $x \times 10^{-1}$, equating $0.2$ to $x \times 10^{-1}$ shows that $x = 2$.

Redox titrations rely on the principle of equivalence, where the total milli-equivalents of the substance being oxidized must exactly balance the total milli-equivalents of the substance being reduced. The number of milli-equivalents is calculated by finding the product of molarity, volume in milliliters, and the n-factor, which represents the number of electrons transferred per mole of reactant. In this reaction, the $\mathrm{Fe}^{2+}$ ion is oxidized to $\mathrm{Fe}^{3+}$, resulting in an n-factor of 1, while $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ acts as the oxidizing agent with an n-factor of 6.

Equating the milli-equivalents of both species provides the relationship $M \times 10 \times 1 = 0.02 \times 15 \times 6$. Simplifying the right side results in 1.8, leading to the equation $10 \times M = 1.8$. Dividing both sides by 10 yields a molarity of 0.18 M, which can be expressed as $18 \times 10^{-2} \text{ M}$. Therefore, the value of $x$ is 18.

Molarity represents the concentration of a solution defined as the number of moles of solute per liter of solution. Begin by determining the molar mass of $\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2}\mathrm{O}$ by summing the constituent atomic masses: $63.54 + 32 + (4 \times 16) + 5 \times (2 \times 1 + 16)$, which results in $249.54\,\mathrm{g/mol}$.

Divide the mass of the solute, $80\,\mathrm{g}$, by this molar mass to find that there are approximately $0.32\,\mathrm{mol}$ present. Dividing this quantity by the total volume of $5\,\mathrm{L}$ yields a molarity of $0.06411\,\mathrm{mol/L}$. Expressing this value in the requested format of $x \times 10^{-3}\,\mathrm{mol/L}$ involves shifting the decimal point three places to the right, which gives $64.11 \times 10^{-3}\,\mathrm{mol/L}$, identifying $x$ as 64.

The stoichiometric relationship between a dissolved salt and its ionic components allows us to equate the molar quantity of the parent compound with that of the resulting ions. Since $\mathrm{NaNO}_3$ dissociates into one mole of sodium ions for every mole of the salt, the number of moles of sodium required determines the amount of solute needed.

The total mass of sodium ions in the 50 mL solution comes from multiplying the concentration of 70.0 mg per mL by the volume, resulting in 3500 mg, or 3.5 g of sodium. Dividing this mass by the atomic weight of sodium, 23 g per mol, gives the number of moles of sodium present. Because the mole ratio is one-to-one, this value also represents the required moles of $\mathrm{NaNO}_3$. Multiplying these moles by the molar mass of $\mathrm{NaNO}_3$, which is 85 g per mol, provides the total mass of the solute as follows:
$\text{Mass} = \frac{3.5}{23} \times 85 \approx 12.934$
Rounding this result to the nearest integer yields 13 g.

Molarity is defined as the concentration of a solution expressed as the number of moles of solute per liter of solution. To determine this for oxalic acid dihydrate $\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 \cdot 2\mathrm{H}_2\mathrm{O}$, start by calculating its molar mass by summing the atomic weights of all constituent atoms, which is $2(1.0) + 2(12.0) + 6(16.0) + 4(1.0) = 126 \text{ g/mol}$. Dividing the given mass of $6.3 \text{ g}$ by this molar mass yields $0.05 \text{ mol}$ of oxalic acid. Converting the solution volume from $250 \text{ mL}$ to liters results in $0.25 \text{ L}$. The molarity is then found through the following calculation:
$\text{Molarity} = \frac{0.05 \text{ mol}}{0.25 \text{ L}} = 0.2 \text{ mol L}^{-1}$
Because the problem defines molarity as $x \times 10^{-2}$, expressing $0.2$ in scientific notation gives $20 \times 10^{-2}$, which identifies $x$ as $20$.