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

To determine the number of molecules in a sample, we utilize Avogadro's hypothesis, which states that one mole of any gas at STP occupies a volume of $22.4\text{ L}$ and contains $6.023 \times 10^{23}$ molecules. For gaseous samples provided in volume, we divide the given volume by $22.4\text{ L}$ to find the number of moles and then multiply by the Avogadro number. Applying this calculation to the $5\text{ L}$ of $\text{H}_{2}$ gas at STP yields $\frac{5 \times 6.023 \times 10^{23}}{22.4}$, which results in $1.344 \times 10^{23}$ molecules. When evaluating other substances, we calculate moles by dividing the given mass by the molar mass, as shown by the fact that $0.5\text{ g}$ of $\text{H}_{2}$ gas corresponds to $\frac{0.5}{2} \times 6.023 \times 10^{23}$, equaling $1.505 \times 10^{23}$ molecules. Despite the mass-based calculations for other compounds like $10\text{ g}$ of $\text{O}_{2}$ resulting in $1.882 \times 10^{23}$ molecules, the volume-based assessment of the specified gas sample remains the standard procedure for comparing molecular quantities at STP.

The number of molecules in a sample is directly proportional to its number of moles, which is determined using the relation $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$. For 2 g of $\mathrm{H}_{2}$, the number of moles is $\frac{2}{2} = 1 \text{ mol}$, while for 16 g of $\mathrm{NO}_{2}$ it is $\frac{16}{46} \approx 0.35 \text{ mol}$, and for 16 g of $\mathrm{O}_{2}$ it is $\frac{16}{32} = 0.5 \text{ mol}$. Since the total number of molecules is calculated by multiplying these mole values by Avogadro's number $N_{A}$, the substance with the highest mole count will contain the most molecules. Comparing these values, 1 mol is greater than 0.5 mol and 0.35 mol, confirming that 2 g of $\mathrm{H}_{2}$ contains the maximum number of molecules.

Molarity represents the number of moles of solute present in exactly one liter of solution. Since the provided density is $1.17 \text{ g/cc}$, one liter, which equals $1000 \text{ cc}$, contains a total mass of $1170 \text{ g}$. Calculating the molarity involves dividing this total mass by the molar mass of $\mathrm{HCl}$, which is $36.5 \text{ g/mol}$.

$\text{Molarity} = \frac{1.17 \times 1000}{36.5 \times 1} = 32.05 \text{ M}$

Determining the minimum molecular weight of a compound based on elemental composition relies on the principle that the molecule must contain at least one atom of the specified element. When a substance is defined as having a certain percentage of an element by weight, it implies that this specific amount of the element is distributed within 100 parts of the total molecular mass. Because the enzyme contains 0.5% Selenium, we know that every 0.5 grams of the element corresponds to 100 grams of the enzyme.

With the atomic weight of Selenium established at 78.4, the total mass of the enzyme required to host exactly one atom of the element is determined by the ratio of the atomic weight to the percentage. This is represented by the following calculation:
$\text{Minimum Molecular Weight} = \frac{100}{0.5} \times 78.4 = 1.568 \times 10^{4}$
This figure represents the smallest possible molecular weight for the anhydrous enzyme, satisfying the requirement that at least one atom of Selenium is integrated into the molecular structure.

Molecular weight represents the mass of one mole of a substance, which is determined by multiplying the mass of a single particle by Avogadro's number. The volume of the cylindrical virus is defined by the equation $V = \pi r^2 l$. Converting the given dimensions, we define the radius as $r = 7 \text{ Å} = 7 × 10^{-8} \text{ cm}$ and the length as $l = 10 \text{ Å} = 10 × 10^{-8} \text{ cm}$. Substituting these values into the volume equation provides $V = \frac{22}{7} \times (7 \times 10^{-8})^2 \times (10 \times 10^{-8}) = 154 \times 10^{-23} \text{ cc}$.

With a specific volume defined as $v_s = 6.02 \times 10^{-2} \text{ cc/g}$, the mass of one particle is obtained from the equation $m = \frac{V}{v_s} = \frac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \text{ g}$. Multiplying this mass by Avogadro's constant, $N_A = 6.02 × 10^{23}$, yields the molecular weight $M = m \times N_A = 15400 \text{ g/mol}$. Converting this result to kilograms provides a final molecular weight of 15.4 kg/mol.

The decomposition of $\mathrm{BaCO}_{3} \rightarrow \mathrm{BaO} + \mathrm{CO}_{2}$ establishes that one mole of barium carbonate yields one mole of carbon dioxide gas. Since one mole of gas occupies 22.4 L at N.T.P. and the molar mass of $\mathrm{BaCO}_{3}$ is 197.34 g/mol, the volume produced is directly proportional to the amount of reactant.

The number of moles of $\mathrm{BaCO}_{3}$ in 9.85 g is calculated as $\frac{9.85}{197.34} \approx 0.05$ mol. Because the reaction follows a 1:1 molar ratio, the decomposition of this quantity produces an equivalent amount of $\mathrm{CO}_{2}$. Calculating the final volume using these values, $\frac{9.85}{197.34} \times 22.4 = 1.118$ L, which corresponds to 1.12 L.

A chemical compound is electrically neutral, which requires the sum of the oxidation numbers of all constituent atoms to equal zero. For the formula $A_{3}(BC_{4})_{2}$, we calculate the total charge contributed by $A$ as $3 \times (+2) = +6$. The polyatomic group $BC_4$ carries a charge of $(+5) + 4 \times (-2) = -3$, and multiplying this by the subscript $2$ results in $2 \times (-3) = -6$, making the total sum $+6 + (-6) = 0$.