Some Basic Concepts Of Chemistry – NEET UG ChemistryPractice Questions & PYQs

According to Avogadro's law, at a constant temperature and pressure, the volume of any gas is directly proportional to the number of molecules it contains. Since the provided conditions indicate that 1 L of oxygen holds $N$ molecules, doubling the volume to 2 L for sulfur dioxide under the same conditions results in a proportional increase in the molecule count. Thus, 2 L of sulfur dioxide contains $2N$ molecules, as the identity and molecular weight of the gas are irrelevant to this relationship when temperature and pressure remain fixed.

Calculating the total valence electrons in a chemical sample requires finding the number of moles present and multiplying that by the electron count per particle. Each nitrogen atom contributes 5 valence electrons, so the $\mathrm{N}_{3}^{-}$ ion possesses a total of $3 \times 5 + 1 = 16$ valence electrons. With a molar mass of 42 g/mol for the azide ion, a 4.2 g sample equates to 0.1 moles of the ion. Consequently, the total valence electron content is determined by $0.1 \times 16 \times \mathrm{N}_{A} = 1.6 \mathrm{N}_{A}$.

The principle of equivalence provides a direct way to solve this problem, as the number of gram-equivalents of metal oxide reacting must equal the gram-equivalents of hydrogen consumed. First, express the mass of hydrogen in grams, which is 0.006 g, and knowing that the equivalent weight of hydrogen is 1, the reaction involves 0.006 equivalents of hydrogen. This means that 0.1596 g of the metal oxide $\mathrm{Z}_{2} \mathrm{O}_{3}$ also represents 0.006 equivalents.

The equivalent weight of the metal oxide is calculated using the relation $\frac{0.1596}{0.006} = 26.6$. Because the equivalent weight of a compound is the sum of the equivalent weights of its constituents, and the equivalent weight of oxygen is 8, we can determine the equivalent weight of metal Z by calculating $26.6 - 8 = 18.6$. Given the formula $\mathrm{Z}_{2} \mathrm{O}_{3}$, the metal must have a valency of 3. Using the relationship $\text{Atomic Weight} = \text{Equivalent Weight} \times \text{Valency}$, the atomic weight is found by computing $18.6 \times 3 = 55.8$.

The even-odd rule dictates that when the digit to be removed is 5, the preceding digit remains constant if even and increments if odd. For 17.0145, the fourth decimal digit is 5, and the preceding digit 4 is even, so it remains unchanged, leading to $Result = 17.014$. For 21.0235, the fourth decimal digit is 5, and the preceding digit 3 is odd, so it increases to 4, leading to $Result = 21.024$.

Relating the mass of a substance to its molar mass via the mole concept allows us to determine the number of particles present. The molar mass of $\text{CO}_2$ is 44 g/mol, so 4.4 g of $\text{CO}_2$ equals $4.4 / 44 = 0.1$ moles. Given that one mole contains $6 \times 10^{23}$ molecules, the sample contains $0.1 \times 6 \times 10^{23} = 6 \times 10^{22}$ molecules. Because every molecule of $\text{CO}_2$ contains two oxygen atoms, the total count is $2 \times 6 \times 10^{22} = 1.2 \times 10^{23}$ atoms.

Pressure represents a distribution of force over a specific area, resulting in the dimensional formula $ML^{-1}T^{-2}$ when we divide force ($MLT^{-2}$) by area ($L^2$). Similarly, when we determine the dimensions of energy per unit volume, we divide the dimensions of energy ($ML^2T^{-2}$) by volume ($L^3$), which simplifies to the same $ML^{-1}T^{-2}$. This mathematical equality confirms that both physical quantities share identical dimensional representation.

Combustion problems are governed by the stoichiometry of a balanced chemical equation, which establishes the precise mass relationships between reactants. The reaction for ethylene is represented as follows:
$\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}$
This equation indicates that 28 g of ethylene requires 96 g of oxygen for complete combustion. Because the stoichiometric ratio remains constant, the oxygen needed for 2.8 kg of ethylene is determined by the following calculation:
$\text{Mass}_{\text{O}_2} = \frac{96 \text{ g}}{28 \text{ g}} \times 2.8 \text{ kg} = 9.6 \text{ kg}$
The total mass of oxygen required for the reaction is 9.6 kg.

The balanced chemical reaction,
$4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_{2} \mathrm{O}_{(l)}$
demonstrates that 4 moles of ammonia require 5 moles of oxygen to react fully. Since 1 mole of ammonia requires $\frac{5}{4} = 1.25$ moles of oxygen and only 1 mole of oxygen is provided, the oxygen acts as the limiting reactant. Consequently, the reaction will exhaust all available oxygen before all the ammonia can be consumed.

Vapour density is defined as the ratio of the mass of a specific volume of a gas to the mass of an equal volume of hydrogen at the same temperature and pressure. To find the result, first convert the volume 45 mL to 0.045 L. Then, multiply this volume by the given hydrogen density of 0.089 g/L to obtain the mass of the hydrogen, which equals 0.004005 g. Finally, divide the mass of the volatile gas, 0.24 g, by the mass of the hydrogen to determine the vapour density:

$\text{Vapour Density} = \frac{0.24}{0.004005} = 59.93$

Determining the density of a gas at standard temperature and pressure involves dividing its molar mass by the fixed volume that one mole of gas occupies under these conditions. The molar mass of $\mathrm{CCl}_{4}$ is calculated by summing the mass of one carbon atom and four chlorine atoms, resulting in $12 + (4 \times 35.5) = 154 \, \text{g/mol}$. Since one mole of an ideal gas occupies $22.4 \, \text{L}$ at STP, the density is found by calculating the ratio $\frac{154}{22.4} \, \text{g/L}$, which yields $6.875 \, \text{g/L}$.

According to Gay-Lussac's Law, when gases react, the volumes involved maintain simple whole-number ratios corresponding to the stoichiometric coefficients in the balanced chemical equation, which is $\mathrm{CO}_2(g) + \mathrm{C}(s) \rightarrow 2\mathrm{CO}(g)$. This implies that for every unit volume of carbon dioxide consumed, two volume units of carbon monoxide are produced at constant temperature and pressure.

If we represent the volume of carbon dioxide that reacts as $x$, then the volume of carbon monoxide generated is $2x$. The final gaseous mixture contains both the carbon monoxide produced and the unreacted carbon dioxide. Given the initial volume was 1 dm^3, the volume of unreacted carbon dioxide is $1 - x$. The total final volume is defined by the sum $(1 - x) + 2x = 1 + x$. Setting this equal to the observed final volume of 1.4 dm^3 yields $1 + x = 1.4$, which means $x = 0.4$. Consequently, the volume of carbon monoxide formed is $2 \times 0.4 = 0.8$ dm^3, and the volume of carbon dioxide remaining is $1 - 0.4 = 0.6$ dm^3.

The ratio of specific heats, represented by the formula $\frac{C_p}{C_V} = 1.4$, confirms that the gas in question is diatomic. According to Avogadro's principle, one mole of any ideal gas occupies 22.4 liters at normal temperature and pressure and contains 6.02 x 10^23 molecules. Therefore, 11.2 liters of this gas corresponds to exactly 0.5 moles, which contains 3.01 x 10^23 molecules. Because the gas is diatomic, each molecule consists of two atoms, so the total number of atoms, denoted as $N_{atoms}$, is calculated through the following equation:

$N_{atoms} = 2 \times 3.01 \times 10^{23} = 6.02 \times 10^{23}$

This calculation reveals that there are 6.02 x 10^23 atoms present in the given volume.

The fundamental principle is that chemical substances react in equal numbers of equivalents, meaning one equivalent of washing soda is necessary to precipitate exactly one equivalent of magnesium ions. To determine the equivalent mass of $\mathrm{Mg}^{2+}$, one divides its atomic mass by its valency, calculated as $24 / 2 = 12$ grams per equivalent.

Since one gram equivalent is equivalent to 1000 milli-equivalents and 12 grams is equal to 12000 milligrams, it follows that 1000 milli-equivalents of washing soda react with 12000 milligrams of $\mathrm{Mg}^{2+}$. By simplifying this ratio, it is established that 1 milli-equivalent of washing soda is sufficient to remove 12 milligrams of $\mathrm{Mg}^{2+}$. Because the water sample contains exactly 12 milligrams of $\mathrm{Mg}^{2+}$, it requires 1 milli-equivalent of washing soda to remove the hardness.

Avogadro's law states that equal volumes of gases at the same temperature and pressure contain the same number of molecules, with the standard molar volume of any gas at NTP established as $22400\,\text{cc}$. Since $1.00\,\text{L}$ is equivalent to $1000\,\text{cc}$, we can determine the properties of this specific gas sample by using its ratio to the molar volume.

The number of molecules present in this volume is calculated as $\frac{6.02 \times 10^{23}}{22400}$, which directly matches the second option. For nitrous oxide, each molecule comprises three atoms, so we calculate the total atoms by multiplying the molecular count by three, yielding $\frac{3 \times 6.02 \times 10^{23}}{22400}$, which simplifies to $\frac{1.8 \times 10^{22}}{224}$ atoms as indicated in the first option.

Finally, determining the total number of electrons requires summing the atomic numbers of the constituent elements, where two nitrogen atoms contribute $14$ and one oxygen atom contributes $8$, totaling $22$ electrons per molecule. Multiplying this by the molecular ratio provides $\frac{6.02 \times 10^{23}}{22400} \times 22$, which simplifies to $\frac{1.32}{224} \times 10^{23}$ electrons, confirming that every calculation provided in the choices is accurate.

The conversion from molecule count to mole count relies on Avogadro's constant, $N_A = 6.02 \times 10^{23}$. Dividing the given number of carbon monoxide molecules by this constant yields 10 moles of CO. As each CO molecule contains one oxygen atom, this amount corresponds to 10 moles of oxygen atoms. Because a gram molecule of oxygen is defined as one mole of $O_2$ molecules, and each $O_2$ molecule contains two oxygen atoms, the 10 moles of oxygen atoms are equivalent to 5 gram molecules of oxygen.

To solve this problem, we first determine the quantity of urea in moles by dividing the mass of the sample by its molar mass. Since chemical formulas provide the ratio of atoms within a single molecule, we can identify that urea, represented as $\mathrm{CO(NH_2)_2}$, contains four hydrogen atoms per molecule. This relationship allows us to calculate the total number of hydrogen atoms by first determining the moles of urea, calculating the moles of hydrogen, and then scaling by Avogadro's constant.

Dividing the given mass of 5.4 g by the molar mass of 60 g mol⁻¹ yields 0.09 mol of urea. Because each molecule contains four hydrogen atoms, the total amount of hydrogen present is given by the equation $0.09 \times 4 = 0.36 \text{ mol}$. Multiplying this molar amount by the constant $N_A = 6.022 \times 10^{23} \text{ particles mol}^{-1}$ provides the final count, expressed by the equation $0.36 \times 6.022 \times 10^{23} = 2.168 \times 10^{23}$ hydrogen atoms.

Normality relates to molarity through the n-factor of the substance, expressed as $N = M \times \text{n-factor}$. Since sulfuric acid acts as a diprotic acid, its n-factor is 2, converting the initial 5 M concentration into 10 N. Because the total number of equivalents stays constant during dilution, we use the equation $N_1V_1 = N_2V_2$ to find the new concentration. Plugging in the initial values of 10 N and 1 litre against the final volume of 10 litres, we get $10 \times 1 = N_2 \times 10$, which simplifies to 1 N.

Significant figures define the precision of a measurement, and the established rule dictates that zeros appearing to the left of the first non-zero digit serve only as placeholders for decimal placement and are never significant. When evaluating 161 cm, 0.161 cm, and 0.0161 cm, we find that the leading zeros in the decimal values do not contribute to the count of significant digits. Because the only non-zero digits defining the precision in each case are 1, 6, and 1, the total count of significant figures for each of these three numbers is $3$.

The combustion of $n$-butane relies on the stoichiometry defined by the balanced equation:
$C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$
where one mole of butane yields four moles of carbon dioxide. Calculating the molar masses reveals that $n$-butane is 58 g/mol ($4 \times 12 + 10 \times 1$), while carbon dioxide is 44 g/mol ($12 + 2 \times 16$).

Using 116 grams of $n$-butane, the number of moles is determined by dividing this mass by the molar mass, which is calculated as $116 / 58 = 2$ moles. Because the reaction produces four moles of carbon dioxide for every mole of butane, two moles of butane generate 8 moles of carbon dioxide. Multiplying these 8 moles by the molar mass of 44 grams per mole yields a total of 352 grams of carbon dioxide.

According to the molar volume principle, one mole of any ideal gas occupies 22.4 liters, or 22400 milliliters, at standard conditions. Since air is 21 percent oxygen by volume, a one-liter sample contains 210 milliliters of pure oxygen. Dividing the volume of this gas by the standard molar volume provides the total moles of oxygen present:
$\frac{210}{22400} = 0.0093$
This calculation indicates that 0.0093 mol of oxygen exists in the sample.