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

The total number of atoms in a substance is determined by calculating the number of moles, converting that into molecules using Avogadro's constant, and then multiplying by the atomicity of the molecule. Ammonia, represented as $\mathrm{NH}_{3}$, has a molar mass of $14 + 3 = 17$ g/mol. By dividing the given sample mass by its molar mass, we find the number of moles is $4.25 / 17 = 0.25$ mol.

Multiplying these moles by Avogadro's constant, $6.023 \times 10^{23}$, gives the total number of molecules. Since each molecule of ammonia contains four atoms, the final calculation is $4 \times 0.25 \times 6.023 \times 10^{23} = 6.023 \times 10^{23}$. This total number of atoms is approximately $6 \times 10^{23}$.

To calculate the average atomic weight of an element existing as a mixture of isotopes, we determine the weighted average of their individual mass numbers based on their percentage abundance. You multiply each isotope's mass number by its percentage, sum these contributions, and divide the total by 100. For boron, the calculation is:

$\frac{19 \times 10 + 81 \times 11}{100} = 10.81$

This average atomic weight corresponds to 10.8.

The percentage composition by weight allows us to calculate the mass of a specific element within a molecule by applying its percentage to the total molecular weight. We find the total mass of iron in one molecule of hemoglobin by multiplying the total molecular weight of 67200 by 0.334%.

$\text{Mass of iron} = \frac{0.334}{100} \times 67200 = 224.45$

Once the total mass of iron is established, we divide this value by the atomic weight of a single iron atom, 56, to determine the total number of atoms present in the molecule.

$\text{Number of iron atoms} = \frac{224.45}{56} \approx 4$

This calculation demonstrates that there are 4 iron atoms in one molecule of hemoglobin.

Stoichiometry dictates that the molar mass of reactants relates directly to the volume of gaseous products under standard conditions. In the reaction
$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2$
one mole of zinc, weighing 65 g, reacts completely to release one mole of hydrogen gas. Because one mole of any gas occupies 22400 mL at STP, 65 g of zinc is responsible for producing 22400 mL of $\text{H}_2$. To find the mass required for 224 mL, we calculate $\frac{65}{22400} \times 224$, which results in 0.65 g.

To find the samples with an equal number of atoms, we calculate the total atoms in each case using the relationship $\frac{\text{mass}}{\text{molar mass}} \times \text{atomicity} \times N_A$. Starting with $212 \text{ g}$ of $\mathrm{Na}_2\mathrm{CO}_3$, we find $2 \text{ moles}$ of the compound, and because each formula unit contains 6 atoms, the total count is $2 \times 6 \times N_A = 12 N_A$. Similarly, $248 \text{ g}$ of $\mathrm{Na}_2\mathrm{O}$ provides $4 \text{ moles}$ of the substance, which, when multiplied by an atomicity of 3, results in $4 \times 3 \times N_A = 12 N_A$. In the case of $240 \text{ g}$ of $\mathrm{NaOH}$, we have $6 \text{ moles}$ of the compound, yielding $6 \times 3 \times N_A = 18 N_A$ total atoms. For $12 \text{ g}$ of $\mathrm{H}_2$, we calculate $6 \text{ moles}$ of molecules, and since each contains 2 atoms, this gives $6 \times 2 \times N_A = 12 N_A$. Finally, $220 \text{ g}$ of $\mathrm{CO}_2$ equates to $5 \text{ moles}$, which, with an atomicity of 3, results in $5 \times 3 \times N_A = 15 N_A$ atoms. Comparing these calculations shows that A, B, and D all possess an equal number of atoms.

The monochlorination of 2-methylbutane involves the substitution of hydrogen atoms from four distinct chemical environments, accounting for both structural and stereoisomers. The six equivalent hydrogen atoms on the two terminal methyl groups (the C1 methyl and the methyl substituent at C2) yield 1-chloro-2-methylbutane, which possesses a chiral center and exists as a pair of enantiomers, contributing 2 products. The single hydrogen atom at the C2 position yields achiral 2-chloro-2-methylbutane, contributing 1 product. The two methylene hydrogens at C3 yield 2-chloro-3-methylbutane, which contains a chiral center and exists as a pair of enantiomers, contributing 2 products. Finally, the three hydrogen atoms on the terminal methyl group at C4 yield achiral 1-chloro-3-methylbutane, contributing 1 product. Summing these possibilities gives a total of $2 + 1 + 2 + 1 = 6$ isomeric products.

John Dalton's atomic theory successfully provided a theoretical foundation for fundamental chemical principles, including the law of conservation of mass, the law of constant proportions, and the law of multiple proportions. Despite this success, it fell short in explaining Gay-Lussac's law of gaseous volumes, which describes how gases react in simple whole-number ratios of volume under uniform pressure and temperature. Because Dalton's model focused on individual atoms and their fixed weight ratios rather than the physical volume occupied by reactive gases, it lacked the necessary framework to accommodate the relationships defined by Gay-Lussac.

Molarity is defined as the number of moles of solute present in one liter of solution, which provides a direct relationship between concentration, mass, molar mass, and volume. This relationship is expressed by the formula $M = \frac{w}{M_m \times V}$, where $M$ is molarity, $w$ is the mass of the solute, $M_m$ is the molar mass, and $V$ is the volume in liters. Isolating the mass variable results in the equation $w = M \times M_m \times V$.

Substituting the specific values where $M = \frac{1}{20}$, the molar mass $M_m = 180$ g/mol, and the volume $V = 0.25$ L leads to the calculation $w = \frac{1}{20} \times 180 \times 0.25$. Performing the arithmetic, $9 \times 0.25$ yields a total of 2.25 g of glucose required to prepare the solution.

The number of molecules in a chemical sample is directly proportional to the number of moles, a value calculated by the relation $n = \frac{\text{mass}}{\text{molar mass}}$. For the reference sample of 1.0 g of $\mathrm{H}_2$, the molar mass is 2 g/mol, which yields $\frac{1}{2} = 0.5$ moles, or $0.5 \mathrm{N}_A$ molecules. Calculating the values for each substance shows that 14 g of $\mathrm{N}_2$ (molar mass 28 g/mol) provides $\frac{14}{28} = 0.5$ moles, which matches the reference sample. In contrast, 18 g of $\mathrm{H}_2\mathrm{O}$ contains 1 mole, 16 g of CO results in $\frac{4}{7}$ moles, and 28 g of $\mathrm{N}_2$ contains 1 mole, none of which equate to the 0.5 moles found in the initial sample.

Combustion analysis is rooted in the law of conservation of mass, where all carbon in an organic sample is oxidized to carbon dioxide and all hydrogen is oxidized to water. By utilizing the known molar ratios-12 parts carbon per 44 parts carbon dioxide and 2 parts hydrogen per 18 parts water-the percentage composition of each element within the original sample is determined by scaling the mass of the resulting combustion products relative to the total mass of the compound.

To determine the percentage of carbon, the mass of carbon dioxide produced is multiplied by the mass fraction of carbon and then normalized against the total mass of the compound using the expression $\frac{12 \times 0.2 \times 100}{44 \times 0.3}$, which results in 18.18%. A similar process applies to hydrogen using the mass of water and the corresponding mass fraction of 2 parts hydrogen per 18 parts water, calculated as $\frac{2 \times 0.1 \times 100}{18 \times 0.3}$, which yields 3.70%.

The empirical formula of a compound identifies the simplest whole-number ratio of its constituent atoms, which is determined by converting the mass percentage of each element into its respective number of moles. Since the total mass percentage of the compound must be 100%, the percentage of element C is $100 - 32 - 20 = 48\%$. By dividing the given percentage of each element by its respective atomic mass, we find the relative molar amount for each: A is $\frac{32}{64} = 0.5$, B is $\frac{20}{40} = 0.5$, and C is $\frac{48}{32} = 1.5$. Dividing each of these values by the smallest resulting number, 0.5, yields a simple integer ratio of $1 : 1 : 3$ for A, B, and C, confirming the empirical formula is $ABC_3$.

Determining the total number of helium atoms requires calculating the amount in moles for each provided sample, as the atom count is directly proportional to the number of moles multiplied by Avogadro's constant, $N_A$. The quantity 4 mol contains 4 $N_A$ atoms. A mass of 4 u represents the mass of a single helium atom, while 4 g of helium equals 1 mole, or 1 $N_A$ atoms, given that the molar mass of helium is 4 g/mol. Dividing the sample volume of 2.271098 L at STP by the standard molar volume of 22.71098 L/mol yields 0.1 mole, equivalent to 0.1 $N_A$ atoms. Comparing the resulting values of 4 $N_A$, 1 atom, 1 $N_A$, and 0.1 $N_A$, the 4 mol sample provides the largest number of atoms.

The fundamental principle governing this neutralization reaction is stoichiometry, which ensures that reactants combine in predictable mass ratios based on their molar masses. Since $\text{NaOH}$ and $\text{HCl}$ react according to the equation $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$, we first determine the mass of $\text{HCl}$ present in the provided solution by using its concentration and volume. With a volume of 25 mL and a concentration of 0.75 M, the mass of $\text{HCl}$ calculates to 0.684 g.

Once the available mass of $\text{HCl}$ is established, we use the mass ratio from the reaction to determine how much $\text{NaOH}$ is consumed. Given that 36.5 g of $\text{HCl}$ reacts precisely with 40 g of $\text{NaOH}$, we find that the 0.684 g of acid will consume 0.750 g of the base. Subtracting this consumed portion from the starting 1 g leaves 0.250 g, resulting in 250 mg of unreacted $\text{NaOH}$.

Molality relates the moles of solute to the mass of the solvent in kilograms, while molarity relates moles to the total solution volume. Because the density provides the conversion between volume and mass, you can determine the solvent mass by subtracting the solute mass from the total mass of the solution. Starting with a $1 \,\text{L}$ solution containing $1$ mole of solute, the total mass is $1000 \,\text{mL} \times 1.25 \,\text{g mL}^{-1}$, equaling $1250 \,\text{g}$. Subtracting the mass of the $1$ mole of solute, which is $85 \,\text{g}$, leaves a solvent mass of $1165 \,\text{g}$, or $1.165 \,\text{kg}$. The conversion formula $m = \frac{1000 \times M}{1000 \times d - M \times MM_w}$ leads to the final value $m = \frac{1000 \times 1}{1000 \times 1.25 - 1 \times 85} = \frac{1000}{1165} \approx 0.858 \,\text{m}$

Heating limestone triggers the thermal decomposition of calcium carbonate into calcium oxide and carbon dioxide, a reaction governed by stoichiometric principles. The decomposition proceeds according to the equation:

$\mathrm{CaCO}_3 \xrightarrow[]{1200 \text{K}} \mathrm{CaO} + \mathrm{CO}_2$

Based on molecular weights, 100 g of calcium carbonate yields 44 g of carbon dioxide.

Because the limestone sample is only 20% pure, the actual mass of $\mathrm{CaCO}_3$ undergoing the reaction is 20 g multiplied by 0.20, which equals 4 g. Applying the stoichiometric ratio derived from the reaction equation, every 100 g of pure $\mathrm{CaCO}_3$ generates 44 g of $\mathrm{CO}_2$. Therefore, the mass of $\mathrm{CO}_2$ produced from 4 g of reactant is calculated as $\frac{44}{100} \times 4$, which equals 1.76 g.

Stoichiometry allows us to determine the precise quantity of reactant needed for a reaction by establishing a mole balance between reactants. The number of moles of $\mathrm{HCl}$ available is calculated by multiplying its molarity, 0.5 M, by the volume in liters, 0.05 L, resulting in $0.5 \times 0.05 = 0.025$ moles. The reaction proceeds as follows:

$\mathrm{CaCO}_{3(s)} + 2\mathrm{HCl}_{(aq)} \rightarrow \mathrm{CaCl}_{2(aq)} + \mathrm{CO}_{2(g)} + 2\mathrm{H}_{2}\mathrm{O}_{(l)}$

Since the balanced equation dictates that one mole of $\mathrm{CaCO}_{3}$ reacts with two moles of $\mathrm{HCl}$, the amount of pure $\mathrm{CaCO}_{3}$ required is $0.025 / 2 = 0.0125$ moles. Using the molar mass of 100 g/mol, the necessary mass of pure calcium carbonate is $0.0125 \times 100 = 1.25$ grams. Because the sample is only 95% pure, the total mass needed is found by dividing the required pure mass by the purity factor, resulting in $1.25 / 0.95 = 1.3157...$ grams, which rounds to 1.32 grams.

Determining an empirical formula relies on finding the simplest whole-number ratio of atoms in a compound by converting mass percentages into molar ratios. Since the compound consists of 78% carbon and the remainder as hydrogen, the hydrogen percentage is $100\% - 78\% = 22\%$. We find the number of moles for each element by dividing their mass percentages by their respective atomic weights. For carbon, the calculation is $78 / 12 = 6.5$, and for hydrogen, it is $22 / 1 = 22$.

Calculating the simplest ratio involves dividing these molar values by the smallest mole count, which is 6.5. Performing this division gives $6.5 / 6.5 = 1$ for carbon. For hydrogen, the calculation $22 / 6.5 \approx 3$ identifies the proportional relationship between the elements. Combining these ratios yields the empirical formula $\mathrm{CH}_3$.

The number of atoms in a sample is found by dividing the mass by the molar mass, multiplying by the number of atoms in each formula unit, and then multiplying by Avogadro's constant, $N_A$. For 1 g of magnesium, the value is $\frac{1}{24} N_A$, whereas for 1 g of oxygen gas, the diatomic structure leads to $\frac{1}{32} \times 2 \times N_A$ atoms. Calculating for 1 g of lithium results in $\frac{1}{7} N_A$ atoms, and for 1 g of silver, the result is $\frac{1}{108} N_A$ atoms. Because $\frac{1}{7}$ is the largest fraction among these values, lithium contains the greatest number of atoms.

Molality defines concentration as the moles of solute per kilogram of solvent, which requires calculating the mass of the solvent from the known total volume and density. Because a 2 M solution means there are 2 moles of NaOH in 1 L of solution, we determine the total mass by using the equation $1000\,\text{mL} \times 1.28\,\text{g/cm}^3 = 1280\,\text{g}$.

Since the molecular mass of NaOH is 40 g/mol, the 2 moles of solute contribute $2\,\text{mol} \times 40\,\text{g/mol} = 80\,\text{g}$ to the total mass. Subtracting this from the solution mass gives the solvent mass, calculated as the equation $1280\,\text{g} - 80\,\text{g} = 1200\,\text{g}$, which is 1.2 kg. Dividing the moles of solute by the solvent mass in kilograms gives the final molality via the equation $2\,\text{mol} / 1.2\,\text{kg} = 1.67\,\text{m}$.

Stoichiometry relies on the fixed molar ratios established by balanced chemical equations to determine reactant requirements. The synthesis of ammonia via Haber's process, $\mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}$, indicates that 2 moles of ammonia are produced from 3 moles of hydrogen. Producing 20 moles of ammonia necessitates scaling this ratio by multiplying the desired output by the proportion of hydrogen to ammonia. Consequently, the calculation $\frac{3 \times 20}{2} = 30$ demonstrates that 30 moles of hydrogen are required to achieve this output.