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

Mass percentage describes the concentration of a solute by identifying how much of the substance exists in a specific total weight of the solution. Because a 75% concentration by mass indicates that 75 g of nitric acid is present in every 100 g of the solution, the total mass of the solution required to house 30 g of the acid is calculated by dividing 30 by 0.75, which equals 40 g of total solution. The volume of this solution is then determined using the formula $V = \frac{m}{d}$, where the mass of 40 g is divided by the density of 1.25 g/mL. Carrying out this division results in a final volume of 32 mL.

The number of atoms in a given sample is governed by the relationship $N = \frac{m}{M} \times N_A$, where $m$ is the mass, $M$ is the molar mass, and $N_A$ is Avogadro's constant. Since the mass is identical for each element, the count of atoms is inversely proportional to the atomic mass, meaning the element with the smallest molar mass will yield the highest number of atoms. Comparing the atomic masses provided, Pr is 140, Pt is 195, Pb is 207, and Po is 209. Because Pr has the lowest atomic mass, it will contain the highest number of atoms among the given elements.

Combustion analysis relies on the conservation of mass, implying that all carbon in the original compound is converted into carbon dioxide. Since the molar mass of carbon dioxide is 44 g/mol and the atomic mass of carbon is 12 g/mol, the fraction of carbon in $\mathrm{CO}_2$ is determined by the ratio $\frac{12}{44}$. We apply this ratio to the 1.46 g of carbon dioxide produced to find the mass of carbon in grams:
$\text{Mass of C} = \frac{12}{44} \times 1.46 = 0.398$
After determining the mass of carbon, we calculate the percentage composition by dividing this mass by the 0.5 g of the original compound and multiplying by 100:
$\text{Percentage of C} = \frac{0.398}{0.5} \times 100 = 79.63$
Rounding this percentage value to the nearest integer yields 80.

The combustion of an organic compound converts all its hydrogen content into water, allowing for the calculation of the elemental mass from the products. Dividing the mass of water by its molar mass gives $\frac{0.9\ \mathrm{g}}{18\ \mathrm{g/mol}} = 0.05\ \mathrm{mol}$ of water. Since each water molecule carries two hydrogen atoms, the $0.01\ \mathrm{mol}$ of the organic compound necessarily contained $0.05\ \mathrm{mol} \times 2 = 0.1\ \mathrm{mol}$ of hydrogen atoms.

Converting this quantity to mass, we multiply by the atomic weight of hydrogen to get $0.1\ \mathrm{g}$. Scaling this up to a full mole of the compound by dividing by $0.01$ reveals that $1\ \mathrm{mol}$ of the substance contains $10\ \mathrm{g}$ of hydrogen. Given that hydrogen makes up $10\%$ of the total molar mass $M$, the relationship is established as $\frac{10}{M} \times 100 = 10$ which simplifies to $M = 100\ \mathrm{g/mol}$.

Combustion analysis relies on the principle of mass conservation, where the mass of carbon in the original compound corresponds to the carbon present in the resulting $\mathrm{CO}_2$, while the hydrogen is found within the generated $\mathrm{H}_2\mathrm{O}$. Applying this principle, 1.46 g of $\mathrm{CO}_2$ accounts for 0.398 g of carbon, and 0.567 g of $\mathrm{H}_2\mathrm{O}$ accounts for 0.063 g of hydrogen. Subtracting these values from the initial 1.0 g sample leaves 0.539 g of oxygen.

Turning these mass values into percentages of the original sample yields 39.8% carbon, 6.3% hydrogen, and 53.9% oxygen. Dividing each percentage by the respective atomic mass provides relative molar amounts, leading to the simple ratio of $1:2:1$ for carbon, hydrogen, and oxygen, respectively. This ratio defines the empirical formula $\mathrm{CH}_2\mathrm{O}$, and summing the individual atomic masses within this formula results in an empirical formula mass of 30 g.

Determining the limiting reagent is the fundamental step when reactants are provided in specific masses, as it dictates the maximum extent to which the reaction can proceed. First, convert the given masses of butane and oxygen into moles using their respective molar masses. For butane ($\mathrm{C}_4\mathrm{H}_{10}$), the molar mass is $4 \times 12 + 10 \times 1 = 58\ \mathrm{g\ mol}^{-1}$, meaning $174\ \mathrm{kg}$ translates to $3,000\ \mathrm{mol}$. For oxygen ($\mathrm{O}_2$), the molar mass is $32\ \mathrm{g\ mol}^{-1}$, making $320\ \mathrm{kg}$ equivalent to $10,000\ \mathrm{mol}$.

Comparing these amounts to the stoichiometric coefficients in the reaction $\mathrm{C}_4\mathrm{H}_{10} + \frac{13}{2}\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 5\mathrm{H}_2\mathrm{O}$ reveals the limiting reagent. Since one mole of butane requires $6.5\ \mathrm{mol}$ of oxygen, the $3,000\ \mathrm{mol}$ of butane would theoretically require $19,500\ \mathrm{mol}$ of oxygen. Because only $10,000\ \mathrm{mol}$ of oxygen are available, oxygen acts as the limiting reagent, and all calculations for the product must be based on this amount. According to the reaction, $\frac{13}{2}\ \mathrm{mol}$ of oxygen yields $5\ \mathrm{mol}$ of water, so $10,000\ \mathrm{mol}$ of oxygen produces approximately $7,692.3\ \mathrm{mol}$ of water. Multiplying this by the molar mass of water ($18\ \mathrm{g\ mol}^{-1}$) results in a total mass of $138,461.5\ \mathrm{g}$. Given the density of $1\ \mathrm{g\ mL}^{-1}$, this mass corresponds to $138,461.5\ \mathrm{mL}$, or $138.46\ \mathrm{L}$, which rounds to 138.

Defining mass as the quantity of matter and weight as the gravitational force exerted on an object clarifies why the first two descriptions provided are inverted and therefore incorrect. Volume is accurately characterized as the space occupied by a substance, while precision refers to the consistency and closeness between multiple measurements of the same quantity. Finally, although temperatures below 0 degrees Celsius are standard, the Kelvin scale operates on an absolute basis where negative temperatures are physically impossible.

The Dumas method determines nitrogen content by oxidizing an organic compound to liberate nitrogen gas, denoted as $\mathrm{N}_2$. For the compound $\mathrm{C}_4\mathrm{H}_{10}\mathrm{N}_2$, the molar mass is 86 g/mol; dividing the sample mass of 0.42 g by this value shows that 0.00488 moles of the compound are present. Since each molecule releases one molecule of $\mathrm{N}_2$, we also have 0.00488 moles of nitrogen, and using the standard molar volume of gas at STP, the calculation $0.00488 \times 22.4 = 0.1093$ gives the volume in liters. This result of 0.1093 L corresponds to 109.3 mL, which rounds to 109 mL.

Mass percentage represents the concentration of a component in a mixture by comparing the mass of the solute to the total mass of the solution, expressed as a percentage. Since the solution consists of 1 mole of ethyl alcohol and 9 moles of water, calculate the individual masses by multiplying the number of moles by the respective molar mass. The ethyl alcohol contributes $1 \times 46 = 46$ grams to the mixture, while the water contributes $9 \times 18 = 162$ grams.

Adding these together gives a total solution mass of $46 + 162 = 208$ grams. Determining the mass percent requires dividing the solute mass by this total mass:

$\text{Mass percent} = \frac{46}{208} \times 100 \approx 22.11$

Rounding this result to the nearest integer yields 22.

According to Avogadro's law, at constant temperature and pressure, the volume of gaseous reactants and products is directly proportional to their stoichiometric coefficients in a balanced chemical equation. The combustion of a gaseous hydrocarbon follows the reaction
$\mathrm{C}_x\mathrm{H}_y + \left(x + \frac{y}{4}\right)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O}$
where $10\ \text{mL}$ of hydrocarbon produces $40\ \text{mL}$ of $\mathrm{CO}_2$ and $50\ \text{mL}$ of water vapour. Equating the volumes based on the stoichiometry leads to $10x = 40$ for carbon, yielding $x = 4$, and $10(\frac{y}{2}) = 50$ for hydrogen, yielding $y = 10$. The resulting hydrocarbon $\mathrm{C}_4\mathrm{H}_{10}$ contains 4 carbon atoms and 10 hydrogen atoms, resulting in a total of 14 atoms.

Molarity expresses concentration as the number of moles of solute per liter of solution. For a mixture that is 31.4% $\mathrm{H}_2\mathrm{SO}_4$ by mass with a density of 1.25 g/mL, we can simplify the math by considering a 100 g sample. In this sample, the mass of sulfuric acid is 31.4 g, while the total volume is determined by dividing the total mass by density, resulting in $\frac{100}{1.25}$ mL.

The number of moles of solute is found by dividing the mass of sulfuric acid by its molar mass of 98 g/mol, giving $\frac{31.4}{98}$. Since molarity requires the volume in liters, we use the formula $M = \frac{n \times 1000}{V_{\text{mL}}}$ to arrive at $M = \frac{31.4 / 98}{100 / 1.25} \times 1000$. The result is 4.005, which approximates to 4.

Molality defines concentration as the number of moles of solute dissolved in exactly $1\text{ kg}$ of solvent. Consequently, a molality of $4.44\text{ m}$ implies that $4.44$ moles of urea are present in $1000\text{ g}$ of water. Determining the mole fraction requires calculating the total moles within the mixture, which involves finding the number of moles of water. Given that water has a molar mass of $18\text{ g/mol}$, the $1000\text{ g}$ of water equates to $\frac{1000}{18}$ moles.

The mole fraction is the ratio of the moles of urea to the total sum of urea and water moles:
$X_{\text{urea}} = \frac{4.44}{4.44 + \frac{1000}{18}}$
Performing this calculation results in $0.074$. Expressing this result in the requested format of $x \times 10^{-3}$ yields $74 \times 10^{-3}$, which indicates that the integer value of $x$ is $74$.

Molarity relates the concentration of a solution to the amount of solute in moles per liter. Since the target volume is 250 mL, converting this to 0.25 L allows for the determination of the required moles by multiplying the volume by the given molarity of 0.35 M.

$n = 0.35 \times 0.25 = 0.0875$

The total mass is then obtained by multiplying these moles by the molar mass of 82.02 g/mol. This calculation results in a product of approximately 7.18 g, which, when rounded to the nearest whole number, yields a final required mass of 7 g.

$\text{Mass} = 0.0875 \times 82.02 \approx 7.18$

The molecular formula of a compound represents the specific ratio of its constituent elements, which dictates both its molar mass and the percentage composition of each element. Identifying the remaining percentage of oxygen involves subtracting the carbon and hydrogen values from the whole, calculated as $100\% - (42.1\% + 6.4\%) = 51.5\%$. Calculating the molar mass for $\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}$ involves summing the atomic masses of its components, resulting in $(12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342$. This result matches the given molecular weight, and because this configuration perfectly aligns with the required percentage distribution of $42.1\%$ carbon, $6.4\%$ hydrogen, and $51.5\%$ oxygen, $\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}$ represents the molecular formula for this compound.

Concentrated $\mathrm{H}_2\mathrm{SO}_4$ reacts with $\mathrm{NaBr}$, $\mathrm{NaNO}_3$, and $\mathrm{KI}$ to release coloured vapours, which consist of bromine, nitrogen dioxide, and iodine, respectively. In contrast, $\mathrm{CaF}_2$ does not evolve any gas when treated with concentrated $\mathrm{H}_2\mathrm{SO}_4$, ensuring that no coloured vapours are produced. By summing the atomic mass of 40 for calcium and two fluorine atoms with an atomic mass of 19 each, we find the molar mass of $\mathrm{CaF}_2$ to be 78 g mol^-1.

To determine the number of atoms in the silver deposit, we must convert the physical dimensions of the layer into its total mass and then calculate the number of moles. First, convert the area from square meters to square centimeters; since one square meter equals 10,000 square centimeters, the area is 500 square centimeters. Multiplying this area by the thickness of 0.05 centimeters yields a total volume of 25 cubic centimeters.

Given the density of 7.9 grams per cubic centimeter, the total mass of the silver is $25 \times 7.9 = 197.5$ grams. By dividing this mass by the atomic mass of 108 grams per mole, we find the number of moles of silver. Finally, multiplying the moles by Avogadro's number, approximately $6.023 \times 10^{23}$, provides the total count of silver atoms. The calculation $\frac{197.5}{108} \times 6.023 \times 10^{23}$ results in approximately $11.01 \times 10^{23}$, identifying 11 as the required value.

Neutralization reactions operate on the principle that the number of equivalents of an acid must exactly match the number of equivalents of the base, as described by $M_1 V_1 n_1 = M_2 V_2 n_2$. For a diprotic acid like oxalic acid with an n-factor of 2 reacting with a monoprotic base like sodium hydroxide, the equation becomes $0.5 \times 50 \times 2 = M_{\text{NaOH}} \times 25 \times 1$. This calculation reveals a molarity of 2 M for the sodium hydroxide solution. The mass in 50 mL is then found by multiplying this molarity by the volume in liters and the molar mass of 40 g/mol, which equals 4 g.

The mole fraction of a specific component in a solution is defined as the ratio of its own number of moles to the total number of moles of all constituents present in the mixture. When calculating this for component $C$ in a system containing $A$, $B$, and $C$, the denominator must account for the combined sum of moles from all three substances. Consequently, the mole fraction of $C$ is determined by the following expression:

$\frac{n_C}{n_A+n_B+n_C}$

Molarity defines concentration by relating the moles of solute to the total solution volume in liters, described by the relationship $M = \frac{n}{V}$. Dividing the mass of 84 g by the formula weight of 40 g mol⁻¹ provides 2.1 moles of sodium hydroxide.

Applying the molarity values to the relationship results in the equation $3 = \frac{2.1}{V}$, which simplifies to a volume of 0.7 liters. Given that a liter is equivalent to a cubic decimeter, the volume corresponds to $7 \times 10^{-1} \text{ dm}^3$.

The reaction of a primary aliphatic amine like ethylamine with nitrous acid, generated from $\mathrm{NaNO}_2$ and $\mathrm{HCl}$, produces nitrogen gas in a direct one-to-one stoichiometric ratio. To determine the initial amount of reactant, we first calculate the moles of nitrogen gas evolved at standard temperature and pressure, where one mole of any gas occupies $22.4\,\mathrm{L}$. Dividing the given volume of $2.24\,\mathrm{L}$ by this molar volume results in $0.1\,\mathrm{mol}$ of nitrogen gas. This confirms that $0.1\,\mathrm{mol}$ of ethylamine was consumed in the reaction. Given that the molar mass of ethylamine, $\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2$, is $45\,\mathrm{g/mol}$, we find the required mass by multiplying $0.1\,\mathrm{mol}$ by $45\,\mathrm{g/mol}$, which equals $4.5\,\mathrm{g}$. Expressing this mass in the required format of $X \times 10^{-1}\,\mathrm{g}$ leads to $45 \times 10^{-1}\,\mathrm{g}$, meaning the value of $X$ is $45$.