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

The number of molecules in a sample is proportional to the number of moles, which is determined by $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$ for condensed substances or $\text{moles} = \frac{\text{volume}}{22.4}$ for gases at standard temperature and pressure. For 18 mL of water, the mass is 18 g, resulting in $\frac{18}{18} = 1$ mole, which gives $N_A$ molecules. For 0.18 g of water, there are 0.01 moles, yielding $10^{-2} N_A$ molecules. The option providing 0.001 moles results in $10^{-3} N_A$ molecules. Finally, for 0.00224 L of water vapor, the calculation $\frac{0.00224}{22.4} = 10^{-4}$ shows there are $10^{-4}$ moles, resulting in $10^{-4} N_A$ molecules. These calculations confirm that 18 mL of water contains the highest number of molecules.

Concentrated $\mathrm{H_2SO_4}$ acts as a dehydrating agent that decomposes formic acid and oxalic acid into their respective gaseous products. The 2.3 g of formic acid corresponds to 0.05 mol and produces 0.05 mol of $\mathrm{CO}$, while the 4.5 g of oxalic acid, also representing 0.05 mol, decomposes to release 0.05 mol of $\mathrm{CO}$ along with 0.05 mol of $\mathrm{CO_2}$. Because KOH pellets selectively absorb acidic $\mathrm{CO_2}$ but leave the neutral $\mathrm{CO}$ gas untouched, only carbon monoxide remains. Summing the moles of $\mathrm{CO}$ from both reactions yields a total of 0.1 mol, which corresponds to a final mass of 2.8 g when multiplied by the molar mass of 28 g/mol.

The molar mass of a chemical compound is fundamentally determined by the ratio of the total mass of a sample to the number of moles, and this total molar mass is equivalent to the sum of the atomic weights of all atoms within the molecular formula. For the compound $XY_2$, dividing the 10 g mass by 0.1 mole gives a molar mass of 100 g/mol, which establishes the algebraic equation $X + 2Y = 100$. Similarly, for the second compound $X_3Y_2$, dividing the 9 g mass by 0.05 mole yields a molar mass of 180 g/mol, resulting in the equation $3X + 2Y = 180$. By subtracting the first equation from the second, the variable $Y$ cancels out, leaving $2X = 80$, which confirms that the atomic weight of X is 40. Substituting this value back into the first equation leads to $40 + 2Y = 100$, simplifying to $2Y = 60$ and ultimately providing an atomic weight of 30 for Y.

The number of molecules in any substance is determined by the amount in moles, as one mole of any substance contains exactly $6.023 \times 10^{23}$ molecules. To determine which quantity is the largest, we can evaluate each option by converting it into a common unit of measure. Converting 1.8 g of water into moles gives $1.8 / 18 = 0.1 \text{ moles}$, which results in $0.1 \times 6.023 \times 10^{23} = 6.023 \times 10^{22}$ molecules. Comparing these, 18 g of water equals 1 mole or $6.023 \times 10^{23}$ molecules, 18 molecules is simply a count of 18, and 18 moles represents $18 \times 6.023 \times 10^{23}$ molecules, making it evident that the 18-mole quantity contains the highest number of water molecules.

The mole is fundamentally a counting unit that dictates how many individual particles are grouped together to represent a standard chemical quantity. Because the mass of an individual carbon atom is a fixed physical property, redefining the number of atoms that constitute a single mole forces the total mass of that collection to shift proportionally. If the Avogadro number were reduced from $6.022 \times 10^{23}$ to $6.022 \times 10^{20}$, the new mass for one mole of carbon would be:
$\text{Mass} = \frac{12 \text{ g} \times 6.022 \times 10^{20}}{6.022 \times 10^{23}} = 12 \times 10^{-3} \text{ g}$
Since the chemical formulas of compounds and the stoichiometric coefficients in balanced equations are determined by the relative counts of atoms rather than the absolute mass of a mole, those relationships remain entirely unchanged.

Converting mass ratios to molar ratios requires dividing the mass of each substance by its molar mass, which is defined by the formula $n = \frac{w}{M}$. Given the mass ratio of $H_{2}$ to $O_{2}$ is 1:4, the mole calculation involves dividing the relative mass by the molar mass of each gas, using 2 g/mol for hydrogen and 32 g/mol for oxygen. This process yields the expression $\frac{1/2}{4/32}$, which evaluates to $\frac{1}{2} \times \frac{32}{4} = \frac{16}{4}$, resulting in a final molar ratio of 4:1.

The precipitation of silver chloride in this double decomposition reaction is governed by the stoichiometry of $\mathrm{AgNO}_3 + \mathrm{NaCl} \rightarrow \mathrm{AgCl} \downarrow + \mathrm{NaNO}_3$. Calculating the reactant masses from the solutions yields $\frac{16.9}{100} \times 50 = 8.45 \text{ g}$ of $\mathrm{AgNO}_3$ and $\frac{5.8}{100} \times 50 = 2.9 \text{ g}$ of $\mathrm{NaCl}$. Converting these to moles using their respective molar masses-$107.8 + 14 + (3 \times 16) = 169.8 \text{ g/mol}$ for $\mathrm{AgNO}_3$ and $23 + 35.5 = 58.5 \text{ g/mol}$ for $\mathrm{NaCl}$-results in approximately $0.0497$ moles and $0.0495$ moles, respectively. Because the reaction occurs in a one-to-one ratio, the amount of $\mathrm{AgCl}$ produced is defined by these limiting values, which is approximately $0.049 \text{ moles}$. Multiplying this by the molar mass of $\mathrm{AgCl}$, calculated as $107.8 + 35.5 = 143.3 \text{ g/mol}$, gives a final precipitate mass of $0.049 \times 143.3 \approx 7.02 \text{ g}$, which matches the mass of $7 \text{ g}$.

At standard temperature and pressure (STP), the molar volume of any gas is 22.4 litres, meaning 22.4 litres of $H_2$ represents 1 mole, while 11.2 litres of $Cl_2$ corresponds to 0.5 moles. According to the balanced chemical equation $H_2 + Cl_2 \rightarrow 2HCl$, the reactants combine in a 1:1 molar ratio, making chlorine the limiting reagent because there is less of it present relative to the hydrogen. Since the stoichiometry dictates that 1 mole of chlorine produces 2 moles of hydrogen chloride, the 0.5 moles of chlorine available will react fully to generate 1 mole of $HCl$.

Chemical stoichiometry requires reactants to combine in fixed proportions, specifically 24 g of magnesium for every 16 g of oxygen. By applying this ratio to the available 0.56 g of oxygen, the amount of magnesium required for the reaction is found through the calculation $\frac{24}{16} \times 0.56 = 0.84$, which means 0.84 g of magnesium is consumed. Since 1.0 g of magnesium is present, subtracting the 0.84 g utilized during the process leaves 0.16 g of unreacted material, confirming that magnesium is the reactant in excess.

Avogadro's law states that at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles, which means the ratio of volumes is equivalent to the ratio of the number of moles. For identical masses $w$ of $\mathrm{H}_2$, $\mathrm{O}_2$, and $\mathrm{CH}_4$, the moles for each are calculated using the formula $n = \frac{w}{M}$ with molar masses of 2, 32, and 16. This establishes the mole ratio as $\frac{w}{2} : \frac{w}{32} : \frac{w}{16}$. Simplifying this expression by multiplying through by $\frac{32}{w}$ results in a final volume ratio of 16:1:2.

Precipitation reactions between silver nitrate and metal chlorides follow fixed stoichiometric ratios where every silver ion reacts with a single chloride ion. Calculating the millimoles reveals that the chloride solution contains $0.05 \times 10 = 0.5$ millimoles, whereas the silver nitrate solution provides $0.1 \times 10 = 1$ millimole. Since 1 millimole of silver nitrate reacts with 0.5 millimoles of the salt, the ratio of silver nitrate to chloride salt is 2:1. This stoichiometry indicates that each molecule of the chloride salt must contain two chloride ions to precipitate with two silver ions, represented by the equation $X\mathrm{Cl}_{2} + 2\mathrm{AgNO}_{3} \rightarrow 2\mathrm{AgCl} + X(\mathrm{NO}_{3})_{2}$ which identifies the formula as $X\mathrm{Cl}_{2}$.

Molarity measures the concentration of a solution by finding the number of moles of solute dissolved in exactly one liter of solution. To find the molar concentration, first convert the count of urea molecules into moles by dividing 6.02 x 10^20 by Avogadro's number, 6.02 x 10^23, which yields 0.001 moles. Next, convert the solution volume from 100 mL into 0.1 L to maintain the required units. Using the formula $M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$, dividing 0.001 moles by 0.1 L results in a concentration of 0.01 M.

The number of molecules in any sample is directly proportional to the number of moles present, which is determined by the formula $n = \frac{m}{M}$. Since the total number of molecules is simply the mole count multiplied by Avogadro's constant, the sample with the highest number of moles will necessarily contain the most molecules.

Calculating the moles for each substance reveals that 44 grams of $\mathrm{CO}_2$, with a molar mass of 44 grams per mole, equals 1 mole. Similarly, 48 grams of $\mathrm{O}_3$ and 64 grams of $\mathrm{SO}_2$ each contain 1 mole, as their given masses match their respective molar masses. In contrast, 8 grams of $\mathrm{H}_2$ corresponds to 4 moles, given its molar mass of 2 grams per mole. Because this represents the highest mole count among the samples, the hydrogen sample contains the maximum number of molecules.

Molarity is defined as the ratio of solute moles to the volume of the solution in liters, providing a baseline to determine the concentration of specific ions within that solution. By calculating the moles of $\mathrm{Na}_2\mathrm{CO}_3$ from the mass of $25.3\,\text{g}$ using the molar mass of $106\,\text{g}\,\text{mol}^{-1}$ and dividing by the volume of $0.250\,\text{L}$, the molarity of the sodium carbonate solution is established as $0.955\,\text{M}$.

When sodium carbonate undergoes complete dissociation in water, the reaction follows the equation $\mathrm{Na}_2\mathrm{CO}_3 \rightarrow 2\mathrm{Na}^{+} + \mathrm{CO}_3^{2-}$. This stoichiometric relationship indicates that every unit of the salt yields two sodium ions and one carbonate ion. Consequently, the concentration of $\mathrm{Na}^{+}$ is calculated by multiplying the solution's molarity by two to obtain $1.910\,\text{M}$, while the concentration of $\mathrm{CO}_3^{2-}$ remains equal to the molarity of the salt at $0.955\,\text{M}$.

The total number of atoms in a gas sample is determined by the product of the number of moles, Avogadro's constant, and the atomicity of the gas molecules. For this sample of 0.1 moles, multiplying by Avogadro's constant gives the total number of molecules, while a final factor of three accounts for the fact that the gas is triatomic.

$\text{Total atoms} = 0.1 \times 6.023 \times 10^{23} \times 3 = 1.806 \times 10^{23}$

This problem relies on identifying the limiting reagent, which is the reactant that is completely consumed first, thereby dictating the maximum amount of product that can be formed. We begin by converting the given masses into moles using the molar masses of hydrogen and oxygen. With 10 g of hydrogen, we have 5 moles, while 64 g of oxygen provides 2 moles.

The balanced chemical equation for the combustion of hydrogen is $2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$. This indicates that 1 mole of oxygen reacts with 2 moles of hydrogen to produce 2 moles of water. Comparing our supply of 2 moles of oxygen and 5 moles of hydrogen, we see that the oxygen will be exhausted first because 2 moles of oxygen requires only 4 moles of hydrogen, leaving 1 mole of hydrogen in excess. Because oxygen is the limiting reagent, the amount of water produced depends solely on the 2 moles of oxygen available. Since the stoichiometry shows a 1:2 ratio between oxygen and water, the reaction generates 4 moles of water.

The extent of a chemical reaction is determined by the limiting reagent, which is the reactant that is fully consumed and restricts the total yield of the product. The balanced chemical equation for this process is:
$\mathrm{PbO} + 2\mathrm{HCl} \rightarrow \mathrm{PbCl}_2 + \mathrm{H}_2\mathrm{O}$
By dividing the mass of each reactant by its respective molar mass, we determine that 6.5 g of $\mathrm{PbO}$ provides 0.029 moles, while 3.2 g of $\mathrm{HCl}$ provides 0.087 moles. The stoichiometric coefficients indicate that 1 mole of $\mathrm{PbO}$ reacts with 2 moles of $\mathrm{HCl}$. Thus, 0.029 moles of $\mathrm{PbO}$ would only require 0.058 moles of $\mathrm{HCl}$ to react completely. Because the available 0.087 moles of $\mathrm{HCl}$ is greater than the required 0.058 moles, the $\mathrm{HCl}$ is in excess and $\mathrm{PbO}$ is the limiting reagent. Since the stoichiometry shows a one-to-one ratio between $\mathrm{PbO}$ and $\mathrm{PbCl}_2$, the reaction produces 0.029 moles of $\mathrm{PbCl}_2$.

Gay-Lussac's Law of Combining Volumes states that when gases react at constant temperature and pressure, they combine in volume ratios that match the stoichiometric coefficients in a balanced chemical equation. The combustion of propane is represented by the following equation:

$\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}$

Because the balanced equation shows that 1 mole of propane reacts with 5 moles of oxygen, we can directly apply this ratio to the gas volumes under identical conditions. This implies that 1 volume of propane requires 5 volumes of oxygen to burn completely, meaning 1 L of propane consumes 5 L of oxygen.

Finding an empirical formula requires identifying the simplest whole-number ratio of atoms based on their constituent mass percentages. Because the compound consists of carbon, hydrogen, and oxygen, the oxygen percentage is determined by subtracting the carbon and hydrogen values from 100%, leaving 51.62%. Dividing each element's percentage by its atomic mass yields the relative molar amounts: $38.71/12 = 3.22$ for carbon, $9.67/1 = 9.67$ for hydrogen, and $51.62/16 = 3.22$ for oxygen. Dividing these results by the smallest value, 3.22, establishes a simple atomic ratio of 1:3:1 for carbon, hydrogen, and oxygen, resulting in the empirical formula $\mathrm{CH}_3\mathrm{O}$.

To determine the weighted average atomic mass of an element consisting of multiple isotopes, you multiply the mass of each isotope by its percentage abundance, sum those values, and divide the total by 100. Using the provided isotopic composition, the calculation is represented by the expression $\frac{200 \times 90 + 199 \times 8 + 202 \times 2}{100}$. Solving the numerator yields $\frac{18000 + 1592 + 404}{100}$, which results in an average atomic mass of $199.96$ amu. Rounding this figure to the nearest whole number gives a final value of $200$ amu.