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

Phosphorous acid is a diprotic acid, meaning it reacts with potassium hydroxide in a stoichiometric ratio of 1:2 as shown by the reaction $\text{H}_3\text{PO}_3 + 2\text{KOH} \rightarrow \text{K}_2\text{HPO}_3 + 2\text{H}_2\text{O}$ Since neutralization depends on this specific mole ratio, calculate the amount of acid using the relationship $n = M × V$, where $n$ is the number of moles, $M$ is molarity, and $V$ is volume. Multiplying the 0.1 M concentration by the 20 mL volume yields 2 millimoles of acid. Because each mole of acid requires 2 moles of base for full neutralization, you need 4 millimoles of KOH. Dividing these 4 millimoles by the 0.1 M concentration of the base results in a volume of 40 mL.

This problem utilizes the Kjeldahl method for nitrogen estimation, where organic nitrogen is converted into ammonia and captured in a specific quantity of sulfuric acid. By determining how much acid remains unreacted after capturing the ammonia, we can perform a back-titration to calculate the nitrogen content of the original sample.

Initially, the system contains $100 \mathrm{mL}$ of $0.1 \mathrm{M}$ sulfuric acid, providing $0.01$ moles of $\mathrm{H}_2 \mathrm{SO}_4$. During the back titration, $20 \mathrm{mL}$ of $0.5 \mathrm{M}$ sodium hydroxide is added, which corresponds to $0.01$ moles of $\mathrm{NaOH}$. Because the reaction $2 \mathrm{NaOH} + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4 + 2 \mathrm{H}_2 \mathrm{O}$ requires two moles of base per mole of acid, the sodium hydroxide neutralizes $0.005$ moles of the sulfuric acid. Subtracting this from the initial amount leaves $0.005$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ that must have reacted with the evolved ammonia.

The reaction between the ammonia and the sulfuric acid follows the stoichiometry $2 \mathrm{NH}_3 + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow (\mathrm{NH}_4) _2 \mathrm{SO}_4$, implying that $0.005$ moles of acid consume $0.01$ moles of $\mathrm{NH}_3$. This amount of ammonia contains $0.01$ moles of nitrogen atoms, which equals $0.01 \times 14 = 0.14 \mathrm{g}$ of nitrogen. Calculating the nitrogen percentage in the $0.30 \mathrm{g}$ sample gives $(0.14 / 0.30) \times 100$, which is $46.6 \%$. Comparing this value to the nitrogen content of the given options-urea at $46.6 \%$, benzamide at $11.5 \%$, acetamide at $23.4 \%$, and thiourea at $36.8 \%$-clearly identifies the organic compound as urea.

In any neutralization reaction, the stoichiometric relationship between reactants is fundamental to calculating concentrations. The reaction between barium hydroxide and hydrochloric acid is described by the equation
$\mathrm{Ba}(\mathrm{OH})_2 + 2\mathrm{HCl} \rightarrow \mathrm{BaCl}_2 + 2\mathrm{H}_2\mathrm{O}$
This balanced equation indicates that one mole of the base is neutralized by two moles of the acid. Consequently, at the point of equivalence, the ratio of the amount of base to its stoichiometric coefficient must equal the ratio of the amount of acid to its coefficient.

Applying this principle leads to the equation
$\frac{M \times 25}{1} = \frac{0.1 \times 35}{2}$
where $M$ represents the molarity of the barium hydroxide solution. Evaluating the right side of the expression gives 3.5 divided by 2, resulting in 1.75. Dividing this value by 25 effectively isolates $M$, resulting in a final molarity of 0.07.

Stoichiometry allows us to predict the amount of reactants required based on the balanced chemical equation representing the transformation of boron trichloride into elemental boron. The balanced reaction is
$2\,\mathrm{BCl}_3 + 3\,\mathrm{H}_2 \rightarrow 2\,\mathrm{B} + 6\,\mathrm{HCl}$
which establishes a molar ratio where 3 moles of hydrogen gas are consumed for every 2 moles of elemental boron produced.

First, determine the number of moles of boron formed by dividing the target mass of 21.6 g by the atomic mass of boron, which is 10.8 g/mol, resulting in exactly 2 moles of boron. Because the stoichiometric coefficients dictate that producing 2 moles of boron requires exactly 3 moles of hydrogen, the reaction must consume 3 moles of hydrogen. At standard conditions of 273 K and 1 atm, one mole of any ideal gas occupies 22.4 L. Therefore, the total volume of hydrogen gas required is calculated by the expression $3 \times 22.4 = 67.2$ L.

Temperature fluctuations influence the physical volume of a liquid solution because substances generally expand upon heating, while mass and mole counts remain entirely unaffected. Molarity is defined as $\text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution}}$ and is therefore susceptible to changes in temperature due to this volumetric shift. In contrast, molality, mole fraction, and weight fraction rely exclusively on mass or mole quantities, ensuring these values remain stable even as the thermal environment changes.

To find the molecular formula, you first determine the relative number of moles for each element by dividing their given mass contribution by their respective atomic masses. Dividing the weight proportions of $9$ for carbon, $1$ for hydrogen, and $3.5$ for nitrogen by their atomic weights of $12$, $1$, and $14$ yields values of $0.75$, $1$, and $0.25$. Converting these into the simplest whole-number ratio by dividing each by the smallest value of $0.25$ results in $3$ for carbon, $4$ for hydrogen, and $1$ for nitrogen, which establishes the empirical formula as $\mathrm{C}_3 \mathrm{H}_4 \mathrm{N}$.

The empirical formula mass is calculated as $(3 \times 12) + (4 \times 1) + 14$, totaling $54 \, \mathrm{g \, mol^{-1}}$. Dividing the compound's actual molecular weight of $108 \, \mathrm{g \, mol^{-1}}$ by this empirical formula weight provides a factor of $n = 108 / 54 = 2$. Multiplying the empirical formula $\mathrm{C}_3 \mathrm{H}_4 \mathrm{N}$ by this factor results in the final molecular formula $\mathrm{C}_6 \mathrm{H}_8 \mathrm{N}_2$.

The number of atoms in a sample is determined by dividing the mass by the molar atomic mass and multiplying the result by Avogadro's number $N_{A}$. Dividing the 558.5 g iron sample by its atomic weight of 55.85 g/mol results in 10 moles, which means the sample contains $10 N_{A}$ atoms. Similarly, dividing the 60 g carbon sample by its atomic mass of 12 g/mol gives 5 moles, resulting in $5 N_{A}$ atoms. Because 10 is twice 5, the iron sample contains double the number of atoms found in the carbon sample.