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

The volume strength of hydrogen peroxide represents the volume of oxygen gas liberated at standard temperature and pressure from a single volume of the solution upon decomposition. Since two moles of hydrogen peroxide, equivalent to 68 g, yield one mole of oxygen gas, 22.7 L of oxygen is produced by the decomposition of 68 g of hydrogen peroxide. This ratio establishes that 11.35 L of oxygen corresponds to 34 g of the solute.

A 50 volume solution indicates that 50 L of oxygen are produced per liter of solution. By relating this liberated volume to the stoichiometric requirements, the concentration is determined using the expression $\text{Strength} = \frac{50}{11.35} \times 34$. Executing this arithmetic results in approximately 149.78, which rounds to the nearest integer of 150 g/L.

The combustion of any hydrocarbon follows the stoichiometric reaction:
$\mathrm{C}_x\mathrm{H}_y + \left(x + \frac{y}{4}\right)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O}$
Comparing this to the provided values, the water coefficient equation $\frac{y}{2} = 3$ indicates that $y = 6$. Substituting this into the oxygen requirement equation $x + \frac{y}{4} = 9.5$ leads to $x + 1.5 = 9.5$, which determines that $x = 8$, identifying the hydrocarbon as $\mathrm{C}_8\mathrm{H}_6$.

The balanced equation $2\mathrm{C}_{(s)} + \mathrm{O}_{2(g)} \rightarrow 2\mathrm{CO}_{(g)}$ defines the stoichiometric requirement of two moles of carbon for every one mole of oxygen, establishing a one-to-one relationship between the moles of carbon consumed and the moles of carbon monoxide produced. Converting the given masses into moles shows that $12\ \mathrm{g}$ of carbon equals $1\ \mathrm{mol}$, while $48\ \mathrm{g}$ of oxygen gas equals $1.5\ \mathrm{mol}$.

Because the reaction requires only $0.5\ \mathrm{mol}$ of oxygen for the $1\ \mathrm{mol}$ of available carbon, oxygen remains in excess and carbon acts as the limiting reagent. The entire $1\ \mathrm{mol}$ of carbon reacts to produce $1\ \mathrm{mol}$ of carbon monoxide, which at the provided molar volume of $22.7\ \mathrm{L\,mol}^{-1}$ results in a final volume of $227 \times 10^{-1}\ \mathrm{L}$.

Concentration in parts per million (ppm) by mass is defined as the mass of the solute divided by the total mass of the solution, multiplied by a factor of $10^6$. The molar mass of dichloromethane ($\mathrm{CH_2Cl_2}$) is calculated as $12 + 2(1) + 2(35.5) = 85\ \text{g/mol}$. Using the molarity of $2.6 \times 10^{-3}\ \text{M}$ and the volume of $0.671141\ \text{L}$, which corresponds to the given $671.141\ \text{mL}$, the mass of the dichloromethane is $2.6 \times 10^{-3}\ \text{mol/L} \times 0.671141\ \text{L} \times 85\ \text{g/mol}$, resulting in $0.148\ \text{g}$.

The mass of the chloroform solvent is determined by multiplying its density, $1.49\ \text{g/cm}^3$, by its volume, $671.141\ \text{cm}^3$. Dividing the solute mass of $0.148\ \text{g}$ by this total solvent mass of $1.49 \times 671.141\ \text{g}$ and multiplying by $10^6$ yields a concentration of $148\ \text{ppm}$.

When a carbonate reacts with hydrochloric acid, the balanced chemical equation
$\text{M}_2 \text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{H}_2 \text{O} + \text{CO}_2$
confirms a one-to-one stoichiometric ratio between the carbonate reactant and the carbon dioxide product. Because one mole of $\text{M}_2 \text{CO}_3$ yields exactly one mole of $\text{CO}_2$, the production of 0.01 mol of $\text{CO}_2$ indicates that 0.01 mol of the carbonate was initially consumed. Dividing the sample mass of 1 g by the quantity of 0.01 mol gives the molar mass of 100 g mol-1.

The reaction between barium chloride and sodium phosphate is defined by the stoichiometric equation $3\mathrm{BaCl}_2 + 2\mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3(\mathrm{PO}_4)_2 + 6\mathrm{NaCl}$, which dictates that 3 moles of $\mathrm{BaCl}_2$ react with 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$. Given that we have 5 moles of $\mathrm{BaCl}_2$ but only 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$, the latter is clearly the limiting reagent because there is insufficient sodium phosphate to consume all the barium chloride. Therefore, the quantity of product formed depends entirely on the 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$. Based on the molar ratio in the balanced equation, consuming 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$ results in the formation of exactly 1 mole of $\mathrm{Ba}_3(\mathrm{PO}_4)_2$.

Concentration units quantify the relative amount of solute within a given solvent or total solution, serving as a standard for comparing the strengths of chemical mixtures. Among the provided terms, mass percent, mole fraction, molarity, parts per million (ppm), and molality specifically function as measures of these quantitative relationships, whereas the mole acts merely as a fundamental unit for the quantity of a substance. Because mass percent, mole fraction, molarity, ppm, and molality are the only items in the list that define concentration, there are 5 such units present.

To determine the molecular formula of the metal chloride, we calculate the molar mass of the compound from its vapor density properties at standard conditions and then find the molar ratio of its constituent elements. The molar mass is derived from the vapor mass at STP using the relation $\text{Molar Mass} = \frac{0.57}{100} \times 22700 = 129.39 \text{ g/mol}$. Given that chlorine accounts for 55.0% of this weight, the mass contributed by chlorine \in one mole of the substance is calculated as $129.39 × 0.55 = 71.1645 \text{ g}$. Dividing this value by the atomic weight of 35.5 u gives the number of chlorine atoms per molecule via $\text{moles of Cl} = \frac{71.1645}{35.5} \approx 2$, confirming the molecular formula as $\mathrm{MCl}_2$.

The concentration of a mixture is determined by the ratio of the total mass of the solute to the total mass of the solution, multiplied by 100 to express it as a percentage. By calculating the absolute amount of sugar in each individual sample, you can sum these amounts to find the total solute and combine the solution weights to find the total denominator.

The first portion contains 25 percent of 200 g, yielding 50 g of sugar. Similarly, the second portion holds 40 percent of 500 g, contributing another 200 g of sugar. Adding these together provides a total sugar mass of 250 g within a combined total solution mass of 700 g. The resulting percentage is calculated as
$\frac{250}{700} \times 100 \approx 35.71$
Rounding this value to the nearest integer gives 36.

Combustion analysis relies on the principle of conservation of mass, asserting that all carbon atoms in the produced carbon dioxide originate from the carbon present in the initial organic compound. Since the molecular weight of $\mathrm{CO}_2$ is 44 and the atomic weight of carbon is 12, the carbon content by mass in the $\mathrm{CO}_2$ is defined by the ratio $\frac{12}{44}$. Multiplying this ratio by the measured mass of $\mathrm{CO}_2$, 0.792 g, reveals the total mass of carbon in the sample:

$\text{Mass of C} = \frac{12}{44} \times 0.792 = 0.216 \text{ g}$

To find the percentage of carbon in the original organic compound, we divide this specific carbon mass by the total weight of the starting material, 0.492 g, and multiply by 100. Performing the calculation $\frac{0.216}{0.492} \times 100$ yields 43.90%, which is 44 when rounded to the nearest integer.

The total heat generated is the sum of the energy released from the partial combustion of carbon into $\mathrm{CO}$ and its complete combustion into $\mathrm{CO_2}$. First, we identify the mass of pure carbon in the 0.6 kg coal sample at 60% purity, which is calculated as $600 \times 0.60 = 360$ grams. Converting this mass to moles using the molar mass of carbon, 12 g/mol, gives $360 / 12 = 30$ moles of carbon available for reaction.

The problem states that 60% of the carbon converts to $\mathrm{CO}$ and the remainder converts to $\mathrm{CO_2}$, meaning $0.60 \times 30 = 18$ moles produce $\mathrm{CO}$ and $0.40 \times 30 = 12$ moles produce $\mathrm{CO_2}$. Given that the formation of $\mathrm{CO}$ releases 100 kJ per mole and $\mathrm{CO_2}$ releases 400 kJ per mole, the energy contributed by each path is calculated as $18 \times 100 = 1800$ kJ for $\mathrm{CO}$ and $12 \times 400 = 4800$ kJ for $\mathrm{CO_2}$. Adding these together results in a total heat release of $1800 + 4800 = 6600$ kJ.

In chemical stoichiometry, the limiting reagent is identified by the reactant that is fully consumed first, which consequently restricts the total yield of the reaction. According to the equation $\mathrm{N}_2 + 3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3$, one mole of nitrogen gas, having a mass of 28 g, reacts with three moles of hydrogen gas, which totals 6 g. When working with an available supply of 20 g of $\mathrm{N}_2$ and 5 g of $\mathrm{H}_2$, we compare these values to the required ratios. Because the reaction requires 6 g of $\mathrm{H}_2$ for every 28 g of $\mathrm{N}_2$, the 20 g of $\mathrm{N}_2$ available would consume less than the 5 g of $\mathrm{H}_2$ present, confirming that $\mathrm{N}_2$ is the limiting reagent and will be completely depleted.

Since the formation of ammonia depends entirely on the amount of nitrogen available, we determine the product quantity based on the 20 g of $\mathrm{N}_2$. The balanced reaction indicates that each mole of $\mathrm{N}_2$ yields two moles of $\mathrm{NH}_3$. By calculating the moles of $\mathrm{N}_2$ as $\frac{20}{28}$ and applying the stoichiometric coefficient of two for the product, the reaction produces $2 \times \frac{20}{28}$ moles of $\mathrm{NH}_3$, which equates to approximately 1.42 moles.

The reaction of pent-1-ene with bromine is a standard addition reaction in which the pi bond of the alkene is broken to allow one molecule of bromine to add across the double bond, establishing a one-to-one molar relationship. To find the required mass of bromine, first calculate the molar mass of $\mathrm{C}_5\mathrm{H}_{10}$ by summing the atomic masses of the carbon and hydrogen atoms, which yields $(12 \times 5) + (1 \times 10) = 70 \text{ g/mol}$. Given a 5 g sample of pent-1-ene, the number of moles is determined by the ratio $\frac{5}{70}$.

Because the stoichiometry is balanced, this exact amount of moles of $\mathrm{Br}_2$ is needed for a complete reaction. Since the atomic mass of bromine is 80 g/mol, the molecular weight of $\mathrm{Br}_2$ is 160 g/mol. Multiplying the moles of pent-1-ene by the molar mass of bromine yields the total required mass: $\frac{5}{70} \times 160 = 11.428 \text{ g}$. Converting this value into the specified form results in $1142.8 \times 10^{-2} \text{ g}$, which rounds to the nearest integer of 1143.

The principle of redox titration relies on the fact that at the equivalence point, the number of equivalents of the oxidizing agent must equal the number of equivalents of the reducing agent. In an acidic medium, potassium dichromate acts as an oxidizing agent by accepting six electrons per mole, establishing an $n$-factor of $6$, while the oxidation of iron(II) ions involves the loss of one electron, resulting in an $n$-factor of $1$. Using the relation $M_1 \times V_1 \times n_1 = M_2 \times V_2 \times n_2$, we substitute the known values: $0.02 \times 20 \times 6 = \mathrm{M} \times 10 \times 1$. Simplifying this expression leads to $2.4 = 10 \times \mathrm{M}$, which results in a molarity of $0.24\,\mathrm{M}$. Converting this result to the required scientific notation gives $24 \times 10^{-2}\,\mathrm{M}$.

Determining the final molarity of the nitric acid solution begins with calculating the initial amount of solute. A solution with a volume of 800 mL at 0.5 M concentration contains $0.8 \times 0.5 = 0.4$ moles of nitric acid. Utilizing the provided molar mass of 63 g/mol, the total initial mass of the acid is $0.4 \times 63 = 25.2$ grams. The process of heating the solution leads to the evaporation of 11.5 grams of acid, which means $25.2 - 11.5 = 13.7$ grams remains in the beaker. With the volume halved during heating, the final volume of the solution is 400 mL, or 0.4 L. Dividing the remaining mass by the molar mass and then by the final volume yields the new molarity:

$M = \frac{13.7}{63 \times 0.4} = \frac{13.7}{25.2} \approx 0.54$

This concentration is equivalent to $54 \times 10^{-2}$ M, indicating that the value of $x$ is 54.

The total mass of iron in a sample is derived from its mass percentage composition. Since hemoglobin contains 0.34% iron, the mass of iron present in 3.3 grams of hemoglobin is determined by $\frac{0.34}{100} \times 3.3$ grams. Converting this mass into moles requires dividing it by the atomic mass of iron, which is 56 u.

Multiplying the resulting number of moles by Avogadro's number, $6.022 \times 10^{23} \mathrm{mol}^{-1}$, provides the total count of iron atoms. The entire calculation is represented by $\frac{0.34 \times 3.3}{100 \times 56} \times 6.022 \times 10^{23}$, which results in $1.21 \times 10^{20}$ atoms.

Finding a molecular formula from percentage composition relies on determining the simplest whole-number ratio of atoms present in the compound, which establishes the empirical formula. Once the empirical formula is known, comparing its mass to the provided molar mass reveals how many times this unit repeats in the actual molecule.

Assuming a 100 g sample, the masses of the elements are 74 g for Carbon, 8.7 g for Hydrogen, and 17.3 g for Nitrogen. Converting these masses to moles by dividing by their respective atomic masses yields 6.167 moles of Carbon, 8.7 moles of Hydrogen, and 1.23 moles of Nitrogen. Dividing each of these values by 1.23, the smallest amount, provides the molar ratio of 5:7:1, which corresponds to the empirical formula $C_5H_7N$. The formula mass for this unit is calculated as $5 \times 12 + 7 \times 1 + 14 = 81$ Dividing the total molar mass of 162 g mol^-1 by the empirical mass of 81 g mol^-1 gives a multiplier of 2, leading to the molecular formula $\mathrm{C}_{10}\mathrm{H}_{14}\mathrm{N}_2$.

The number of alcoholic hydrogen atoms in a polyhydric alcohol is determined by the stoichiometry of the reaction that releases hydrogen gas. We begin by calculating the amount of compound X in moles. Dividing the sample mass of 1.84 mg by the molar mass of 92.0 g/mol gives $2 \times 10^{-5}$ mol. Next, we convert the volume of hydrogen gas produced at standard temperature and pressure into moles. Dividing the 1.344 mL of gas by the molar volume of 22400 mL/mol results in $6 \times 10^{-5}$ mol. The number of alcoholic hydrogen atoms corresponds to the ratio of the moles of hydrogen gas to the moles of the alcohol compound, calculated as $\frac{6 \times 10^{-5}}{2 \times 10^{-5}}$, which equals 3.

In the combustion of an organic compound containing only carbon and hydrogen, the Law of Conservation of Mass ensures that all carbon in the original sample is converted into carbon dioxide, while all hydrogen is converted into water. By measuring the masses of these products, one can determine the exact mass of carbon and hydrogen present in the original sample and calculate their respective percentages.

The total mass of the compound is 120 g. With 330 g of $\mathrm{CO}_2$ produced, and knowing the molar mass of $\mathrm{CO}_2$ is 44 g/mol, the reaction yields:
$\frac{330}{44} = 7.5$
moles of $\mathrm{CO}_2$. Since each mole of $\mathrm{CO}_2$ contains 12 g of carbon, the total mass of carbon is $7.5 \times 12 = 90$ g. With 270 g of water produced and a molar mass of 18 g/mol, the reaction yields:
$\frac{270}{18} = 15$
moles of water. Because each mole of water contains 2 g of hydrogen, the total mass of hydrogen is $15 \times 2 = 30$ g. Calculating the percentage of each element relative to the 120 g sample gives carbon as $\frac{90}{120} \times 100 = 75\%$ and hydrogen as $\frac{30}{120} \times 100 = 25\%$.

The concentration unit of parts per million (ppm) for aqueous solutions corresponds directly to the mass of solute in milligrams per liter. With a concentration of $48\text{ ppm}$ in a total volume of $2\text{ L}$, the total mass of magnesium present is calculated as $48\text{ mg/L} \times 2\text{ L}$, which equals $96\text{ mg}$ or $0.096\text{ g}$. Dividing this mass by the atomic mass of magnesium, $24\text{ g mol}^{-1}$, yields $0.004\text{ mol}$ of the element.

To determine the number of atoms, we multiply the number of moles by Avogadro's number, $N_A = 6.02 \times 10^{23}\text{ mol}^{-1}$. This calculation results in $0.004 \times 6.02 \times 10^{23}$, which equals $24.08 \times 10^{20}$. Rounding this value to the nearest integer, we identify the value of $x$ as 24.