Class 11 Chemistry (Chemistry Part-I) Chapter 1 Some Basic Concepts of Chemistry

Key Concept: Precision vs. Accuracy

a) Precision refers to the repeatability of measurements; accuracy refers to correctness relative to a standard.

[Solution Description] Precision refers to the closeness of measurements to each other, indicating consistency. Accuracy refers to how close a measurement is to the true or accepted value.

Key Concept: Conceptual Understanding

c) It explains the existence of diatomic molecules.

[Solution Description] Avogadro’s hypothesis posits that equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. This allows chemists to infer that diatomic molecules such as hydrogen and oxygen exist because their reactions follow predictable volume ratios based on molecule count rather than atomic mass or other factors.

Key Concept: Unit Identification

d) Newton

[Solution Description]

Weight is measured as a force, and the standard SI unit for force is the Newton (N).

The other options are units typically associated with measuring mass or volume, not force.

Therefore, Newton is the correct choice for measuring weight.

Key Concept: Derived Units

d) Pascal

[Solution Description]

A derived unit is one that can be expressed as a combination of base SI units. The unit for pressure, Pascal (Pa), is defined as $1 \, \text{N/m}^2$ which involves the base units meter (m), kilogram (kg), and second (s). It cannot be expressed entirely in terms of base units.

Key Concept: Conceptual Understanding

c) It can be observed without changing the substance’s identity

[Solution Description] Melting point is considered a physical property because it can be measured without changing the identity of the substance. During melting, only the state changes from solid to liquid, but the molecular composition remains unchanged.

Key Concept: Solution Concentration

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion correctly states the definition of molarity, which is a way to measure the concentration of a solution by expressing it as moles of solute per liter of solution. Meanwhile, the reason accurately defines molality, which differs from molarity as it is expressed in terms of moles of solute per kilogram of solvent. Both statements are true but describe different concepts; thus, the reason does not explain the assertion.

Key Concept: Critical Analysis, Real-World Application

c) Gay Lussac’s Law of Gaseous Volumes

[Solution Description] The observation that 2 volumes of gas A react with 1 volume of gas B to produce 2 volumes of gaseous product can be explained by Gay Lussac’s Law of Gaseous Volumes. This law states that gases combine in simple whole number ratios by volume when measured at the same temperature and pressure, which perfectly aligns with the given scenario where 2 volumes of A combine with 1 volume of B to yield 2 volumes of product.

Key Concept: Significant Figures in Calculations

a) 43

[Solution Description]

Multiply the given measurements:

$$12.67 \times 3.4 = 43.078$$

The result must be expressed with the least number of significant figures among the factors. Here, 3.4 has 2 significant figures. Thus, the product should also have 2 significant figures.

Therefore, rounding 43.078 to 2 significant figures gives 43.

Key Concept: Law Comparison

c) Avogadro’s Law

[Solution Description] Gay-Lussac’s Law of Gaseous Volumes is directly related to Avogadro’s work, who explained that equal volumes of gases at the same temperature and pressure contain an equal number of molecules. This relates Gay-Lussac’s Law to Avogadro’s Law in terms of understanding gaseous reactions by volume.

Key Concept: Real-World Application, Complex Compound Analysis

c) It is polar

[Solution Description] Water’s ability to dissolve many substances stems from its polar nature. The molecule has a bent shape that creates a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom. This polarity allows it to interact with various ionic and covalent compounds, effectively breaking them apart and allowing them to dissolve.

Key Concept: Interpretation of Scientific Notation, Real-World Application

a) $1.6 \times 10^{10}$ cells

[Solution Description]

Each hour, the population doubles. So, after 6 hours, the number of cells can be calculated as follows:

After 1 hour: $2 \times 2.5 \times 10^8 = 5 \times 10^8$

After 2 hours: $2 \times 5 \times 10^8 = 10 \times 10^8 = 1 \times 10^9$

After 3 hours: $2 \times 1 \times 10^9 = 2 \times 10^9$

After 4 hours: $2 \times 2 \times 10^9 = 4 \times 10^9$

After 5 hours: $2 \times 4 \times 10^9 = 8 \times 10^9$

After 6 hours: $2 \times 8 \times 10^9 = 16 \times 10^9 = 1.6 \times 10^{10}$

Hence, after 6 hours, there are $1.6 \times 10^{10}$ cells.

Key Concept: State Interconversion

b) It becomes a liquid

[Solution Description] When a solid is heated, its particles gain energy and start vibrating more vigorously, leading to an increase in the distance between them. Eventually, at the melting point, the solid turns into a liquid because the particles have enough energy to overcome the forces holding them together.

Key Concept: Percentage to Formula, Complex Calculation

d) $C_4H_9Cl$, $C_8H_{18}Cl_2$

[Solution Description]

Convert mass percentage to grams:

– H: $9.09 \text{ g}$

– C: $54.55 \text{ g}$

– Cl: $36.36 \text{ g}$

Convert grams to moles:

– Moles of H = $\frac{9.09 \text{ g}}{1.01 \text{ g/mol}} = 9.00$

– Moles of C = $\frac{54.55 \text{ g}}{12.01 \text{ g/mol}} = 4.54$

– Moles of Cl = $\frac{36.36 \text{ g}}{35.45 \text{ g/mol}} = 1.03$

Divide by the smallest number of moles to derive a ratio:

$\text{H} : \text{C} : \text{Cl} = \frac{9.00}{1.03} : \frac{4.54}{1.03} : \frac{1.03}{1.03} \approx 8.74 : 4.41 : 1$

Multiply each by conversion factor 2 (rounding):

$8.74 \times 2 \approx 17,\; 4.41 \times 2 \approx 9,\; 1 \times 2 = 2$

Empirical Formula: $C_9H_{18}Cl_2$

Calculate the empirical formula mass:

$9 \times 12.01 + 18 \times 1.01 + 2 \times 35.45 = 185.26 \text{ g/mol}$

Determine n:

$n = \frac{132 \text{ g/mol}}{185.26 \text{ g/mol}} \approx 0.71 \rightarrow 1$

Molecular Formula: Cannot be formed directly from empirical due to discrepancy, check calculation again for key mistake or misreport in original parameters.

Finalize choice based on adjusted balance: Could adjust base empirical to fit within close margins or reallocate initial distribution assumptions.

Key Concept: Problem Solving, Complex Calculation

c) 250 g

[Solution Description] Use the mass per cent to determine the mass of Cu in the alloy. Given:

Mass of alloy = 500 g,

Mass per cent of Cu = 50%.

Calculate the mass of Cu:

$\text{Mass of Cu} = \left( \frac{50}{100} \right) \times 500 = 250 \text{ g}$

So, 250 g of copper is present in the 500 g sample of the alloy.

Key Concept: Contributions of Scientists

c) Discovery of mercury sulphide and soap

[Solution Description] Chakrapani is credited with significant contributions to the field of chemistry, including the discovery of mercury sulphide and the preparation of soap.

Key Concept: Atomic Mass Concept

b) Carbon-12

[Solution Description] The isotope Carbon-12 $(^{12}\text{C})$ is used as the standard for atomic mass units. It is assigned an atomic mass of exactly 12 atomic mass units (amu), providing a baseline for measuring the atomic masses of other elements relative to carbon.

Key Concept: Limiting Reagent in Industrial Processes, Stoichiometry

a) 1250 kg of $\text{SO}_3$

[Solution Description]

– Calculate the moles of $\text{SO}_2$ and $\text{O}_2$. Molar mass of $\text{SO}_2 = 64 \, \text{g/mol}$, $\text{O}_2 = 32 \, \text{g/mol}$.

$\text{Moles of } \text{SO}_2 = \frac{1000 \times 1000}{64} = 15625 \, \text{mol}$

$\text{Moles of } \text{O}_2 = \frac{500 \times 1000}{32} = 15625 \, \text{mol}$

– According to the balanced equation, 2 mol $\text{SO}_2$ react with 1 mol $\text{O}_2$.

$\text{Required moles of } \text{O}_2 = \frac{15625}{2} = 7812.5 \, \text{mol}$

– Since available $\text{O}_2$ is more than required, $\text{SO}_2$ is not the limiting reagent.

– Calculate $\text{SO}_3$ produced using $\text{SO}_2$ as limiting:

$\text{Mass of } \text{SO}_3 = 15625 \times 80 = 1250000 \, \text{g} = 1250 \, \text{kg}$

Key Concept: Scientific Notation Operations

a) $2.77 \times 10^5$

[Solution Description] To solve this, first express both numbers with the same power of ten:

Convert $4.8 \times 10^4$ to $48.0 \times 10^3$, which equals $0.48 \times 10^5$.

Now subtract:

$3.25 \times 10^5 – 0.48 \times 10^5 = (3.25 – 0.48) \times 10^5 = 2.77 \times 10^5$

The result is expressed as $2.77 \times 10^5$.

Key Concept: Calculation with Negligible Isotopes, Advanced Reasoning

a) 204.0396 u

[Solution Description] Start by calculating the initial average atomic mass:

$A_{initial} = \frac{(200 \times 0.0002) + (202 \times 0.0003) + (204 \times 0.9895) + (206 \times 0.0100)}{1} = 204.0198 \, \text{u}$

Doubled abundance of $206\, \text{u}$ implies it becomes $2 \times 1% = 2%$. The excess 1% comes from $204\, \text{u}$, making its abundance $$98.95% – 1% = 97.95%$.

Re-calculate with adjusted abundances:

$A_{new} = \frac{(200 \times 0.0002) + (202 \times 0.0003) + (204 \times 0.9795) + (206 \times 0.0200)}{1} = 204.0396 \, \text{u}$

Therefore, the new average atomic mass would be approximately 204.0396 u.

Key Concept: Biochemical Processes

b) Food processing

[Solution Description] Enzymes are used extensively in the food industry for processes like fermentation, cheese production, and brewing, where they act as catalysts to speed up biochemical reactions without being consumed.

Key Concept: Real-world Application, Complex Rounding

d) 25.36 meters

[Solution Description] The measurement has an uncertainty of ±0.005 m, indicating that the last digit is uncertain. Therefore, the length should be reported up to the same decimal place as the uncertainty: three decimal places. According to the rules of rounding significant figures, since the digit to be removed (7) is more than 5, we increase the preceding number by one.

Length = 25.357 rounds to 25.36 meters because:

– Remove ‘7’ which is greater than 5

– Increase ’35’ to ’36’

Hence, the final reported length is 25.36 meters.

Key Concept: Prefix Application

b) 0.003 grams

[Solution Description] To convert milligrams to grams, use the prefix “milli,” which means $10^{-3}$. Therefore, 3 milligrams is calculated as follows:

$3 \, \text{mg} = 3 \times 10^{-3} \, \text{g} = 0.003 \, \text{g}$

Hence, 3 milligrams equals 0.003 grams.

Key Concept: Real-world Application, Temperature and Density

c) Increase by 5%

[Solution Description] Since the density $\rho$ is inversely proportional to the volume V, the percentage change in density is given by the formula:

\[
\%\Delta\rho = -\%\Delta V
\]

Given that $%\Delta V = +5%$, then

\[
\%\Delta\rho = -5\%
\]

This means the density increases by approximately 5%.

Key Concept: Reaction Optimization, Complex Stoichiometry

c) Increase L

[Solution Description]

First reaction: $6 \text{K} + 9 \text{L} \rightarrow 5 \text{M}$

– For K: 30 moles gives $\frac{30}{6}\times5 = 25$ moles of M.

– For L: 40 moles gives $\frac{40}{9}\times5 \approx 22.22$ moles of M.

L is the limiting reagent and allows the creation of roughly $22.22$ moles of M.

Second reaction: $5 \text{M} + \text{N} \rightarrow 2 \text{J}$

– From 22.22 moles of M, $\frac{22.22}{5}\times2 \approx 8.888$ moles of J.

– From 25 moles of N, $25\times2 = 50$ moles of J theoretically possible.

M limits to approximately $8.888$ moles of J; hence, increasing L maximizes J output.

Key Concept: Identifying Significant Figures

c) 3

[Solution Description] The zeros before the first non-zero digit are not counted as significant figures. Hence, only `450` counts. In this number, there are three significant figures: `4`, `5`, and the trailing zero, which is considered significant because it follows a decimal point.

Key Concept: Real-World Application

c) 1:2

[Solution Description] The balanced chemical reaction for the synthesis of ammonia from nitrogen and hydrogen is $N_2 + 3H_2 \rightarrow 2NH_3$. This implies that 1 volume of nitrogen produces 2 volumes of ammonia. Given 200 mL of nitrogen results in 400 mL of ammonia, the volume ratio of nitrogen to ammonia is 1:2.

Key Concept: Volume Calculation

c) 4 liters

[Solution Description] According to Avogadro’s Law, the volumes of gases involved in a reaction at the same temperature and pressure are proportional to the number of moles (or molecules). The balanced equation for the reaction between nitrogen and hydrogen gas to produce ammonia is:

$N_2 + 3H_2 \rightarrow 2NH_3$

From the equation, it can be seen that 1 liter of $N_2$ produces 2 liters of $NH_3$. Therefore, 2 liters of $N_2$ would produce:

$2 \, \text{liters } N_2 \times \frac{2 \, \text{liters } NH_3}{1 \, \text{liter } N_2} = 4 \, \text{liters } NH_3$

Thus, 4 liters of ammonia is produced.

Key Concept: Complex Empirical Formula, Molar Mass and Empirical Formula

b) $\text{C}_4\text{H}_8$

[Solution Description]

Convert the percentage composition into moles using atomic masses: C (12), H (1).

$\text{Moles of C} = \frac{85.7}{12} = 7.14$

$\text{Moles of H} = \frac{14.3}{1} = 14.3$

Divide by the smallest number of moles:

$\text{Ratio for C : } \frac{7.14}{7.14} = 1$

$\text{Ratio for H : } \frac{14.3}{7.14} \approx 2$

Empirical formula: $\text{CH}_2$.

Calculate empirical formula mass:

$\text{Empirical formula mass} = 12 + (2 \times 1) = 14 \text{ g/mol}$

Determine n:

$n = \frac{56}{14} = 4$

Therefore, the molecular formula is $\text{C}_4\text{H}_8$.

Key Concept: Limiting Reagent, Multi-Step Problem

b) Chlorine

[Solution Description]

Calculate moles of hydrogen:

$\text{Moles of H}_2 = \frac{10 \, \text{g}}{2 \times 1.008 \, \text{g/mol}} \approx 4.960 \, \text{mol}$

Calculate moles of chlorine:

$\text{Moles of Cl}_2 = \frac{158 \, \text{g}}{2 \times 35.45 \, \text{g/mol}} \approx 2.230 \, \text{mol}$

The balanced equation is:

$\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$

According to the equation, 1 mole of $\text{H}_2$ reacts with 1 mole of $\text{Cl}_2$. Since we have fewer moles of

$\text{Cl}_2$ compared to $\text{H}_2$, $\text{Cl}_2$ is the limiting reagent.

Key Concept: Contributions of Ancient India

b) Rasopanishada

[Solution Description] The “Rasopanishada” is an ancient Indian text that discusses numerous chemical processes such as distillation, sublimation, and calcination, highlighting India’s contributions to early chemical knowledge.

Key Concept: Basic Units

d) kelvin

[Solution Description] The SI unit for temperature is the kelvin, which is one of the seven base units in the International System of Units.

Key Concept: Explanation of Chemical Laws

b) Law of Definite Proportions

[Solution Description] Dalton’s Atomic Theory was able to explain the Law of Definite Proportions by stating that atoms of different elements join in specific ratios to form compounds. Each type of molecule will be composed of the same kinds and numbers of atoms. Hence, Dalton’s theory aligns well with the Law of Definite Proportions.

Key Concept: Chemistry and Economy

b) Through the development of semiconductor materials

[Solution Description] This solution focuses on understanding how chemistry aids technological advancement. Chemistry plays a critical role by enabling the development of semiconductor materials essential for manufacturing computer chips, thus enhancing computer technology.

Key Concept: Comparative Analysis

b) Definite Proportions deals with one compound; Multiple Proportions deals with different compounds from the same elements.

[Solution Description] The Law of Definite Proportions states that a given compound always contains exactly the same proportion of elements by weight. The Law of Multiple Proportions involves two elements forming more than one compound, where the masses of one element that combine with a fixed mass of another are in simple whole-number ratios.

Key Concept: Conceptual Understanding

b) It ensures unit consistency and accurate conversions.

[Solution Description] Dimensional analysis ensures that equations and measurements are consistent with units, allows for conversion between different measurement systems, helps verify calculations for accuracy, and aids in understanding the physical meaning of a quantity by analyzing its dimensions.

Key Concept: Subtraction with Significant Figures

a) $1.22$

[Solution Description]

When subtracting, the answer must have as many decimal places as the number with the least. Here, $7.56$ has two decimal places, while $6.342$ has three. Subtracting:

$7.56 – 6.342 = 1.218.$

Rounding to two decimal places, the result is $1.22$.

Key Concept: Critical Thinking, Conceptual Understanding

c) $C_4H_8$

[Solution Description] The empirical formula mass of $CH_2$ is $12.011\, \text{u} + 2 \times 1.008\, \text{u} = 14.027\, \text{u}$. To find the molecular formula, divide the molecular mass by the empirical formula mass: $\frac{56}{14.027} \approx 4$. The molecular formula is thus four times the empirical formula: $C_4H_8$.

Key Concept: Analytical Reasoning, Theoretical Implications

b) It ensures the same proportion of C and D across samples.

[Solution Description] By the Law of Definite Proportions, a pure compound will always exhibit the same fixed proportion of elements by mass. Given the atomic masses of C and D, we can compute the expected mass ratio from their atomic masses.

Calculate the expected mass ratio using the formula:

$\text{Mass ratio} = \frac{\text{atomic mass of C}}{\text{atomic mass of D}}$

By measuring the mass of C and D in various samples of the compound and comparing them to this calculated ratio, we can determine purity. If all measured ratios match the calculated mass ratio, then the samples are pure. Otherwise, they may contain impurities or be a mixture.

Key Concept: Measurement Devices

a) Analytical balance

[Solution Description] The most accurate device for measuring the mass of small objects in a laboratory is an analytical balance. It provides high precision and accuracy in measurements compared to other balances.

Key Concept: Theory Application

c) Atoms are rearranged without being created or destroyed.

[Solution Description] Dalton’s Atomic Theory states that chemical reactions involve the reorganization of atoms. This means atoms are neither created nor destroyed; they simply change partners or rearrange themselves.

Key Concept: Temperature Scale Relationship

a) 310.15 K

[Solution Description] To convert Celsius to Kelvin, add 273.15 to the Celsius temperature:

$T(K) = T(°C) + 273.15 = 37 + 273.15 = 310.15\,K$

Therefore, the temperature in Kelvin is approximately 310.15 K.

Key Concept: Volume Definition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that volume is the amount of space occupied by a substance, which is correct as per the definition. The reason states that volume is measured in units of $$(length)^3$$, which is also correct because volume is derived from three-dimensional space where each dimension is represented by length. Thus, both the assertion and reason are true, with the reason correctly explaining the assertion.

Key Concept: Real-World Application, Unified Mass Application

a) $1.992672 \times 10^{-23} \text{ g}$

[Solution Description] Convert the mass from unified atomic mass units (u) to grams by multiplying the atomic mass by the conversion factor:

$\text{Mass in grams} = 12 \text{ u} \times 1.66056 \times 10^{-24} \text{ g/u}$

Calculating this product:

$12 \times 1.66056 \times 10^{-24} = 1.992672 \times 10^{-23} \text{ g}$

Key Concept: Real-World Application, Interconversion of Mass, Volume, and Density

c) 1487.53 kg

[Solution Description] The balanced chemical equation for the combustion of benzene is:

$2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O$

First, convert volume to mass:

Mass of benzene $= 500 \text{ L} \times 1000 \text{ mL/L} \times 0.879 \text{ g/mL} = 439500 \text{ g}$.

Molar mass of benzene $C_6H_6 = (6 \times 12) + (6 \times 1) = 78 \text{ g/mol}$.

Moles of benzene $= \frac{439500 \text{ g}}{78 \text{ g/mol}} = 5634.62 \text{ moles}$.

Using stoichiometry, 2 moles of benzene produce 12 moles of $CO_2$:

Moles of $CO_2 = 5634.62 \times \frac{12}{2} = 33807.72 \text{ moles}$.

Molar mass of $CO_2 = 44 \text{ g/mol}$.

Mass of $CO_2 = 33807.72 \times 44 \text{ g/mol} = 1487529.68 \text{ g} = 1487.53 \text{ kg}$.

Key Concept: Properties Comparison

d) Reactivity with acids

[Solution Description] Chemical properties describe a substance’s potential to undergo chemical change. Reactivity with acids involves such a transformation where new substances are formed, indicating it is a chemical property.

Key Concept: Advanced Historical Development, Real-World Chemistry Applications

c) Arabs

[Solution Description] The Arabs were instrumental in developing early alchemical practices. They translated and preserved Greek texts, introduced systematic experimentation, and influenced European alchemy, paving the way for modern chemistry.

Key Concept: Stoichiometry, Limiting Reagent

b) 2 moles

[Solution Description] According to the balanced equation, 2 moles of $\text{H}_2$ react with 1 mole of $\text{O}_2$ to produce 2 moles of $\text{H}_2\text{O}$. Since we have 4 moles of $\text{H}_2$, it would require 2 moles of $\text{O}_2$ to completely react. However, only 1 mole of $\text{O}_2$ is available, making $\text{O}_2$ the limiting reagent. Therefore, only 2 moles of $\text{H}_2$ can react with 1 mole of $\text{O}_2$, producing 2 moles of $\text{H}_2\text{O}$.

Key Concept: Complex Molar Mass Calculation, Stoichiometry

c) 2.24 g

[Solution Description]

To solve this, we need to follow these steps:

1. Calculate the molar mass of $\text{CaCO}_3$:

$\text{Molar mass of CaCO}_3 = 40.08 + 12.01 + (16 \times 3) = 100.09 \text{ g/mol}$

2. Calculate the moles of $\text{CaCO}_3$ using its given mass:

$\text{Moles of CaCO}_3 = \frac{4.0 \text{ g}}{100.09 \text{ g/mol}} = 0.03997 \text{ mol}$

3. The balanced chemical equation for decomposition of calcium carbonate is:

$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$

From the equation, 1 mole of $$\text{CaCO}_3$$ produces 1 mole of $$\text{CaO}$$. Thus, 0.03997 mol of $$\text{CaCO}_3$$ will produce 0.03997 mol of $\text{CaO}$.

4. Calculate the molar mass of $\text{CaO}$:

$\text{Molar mass of CaO} = 40.08 + 16.00 = 56.08 \text{ g/mol}$

5. Calculate the mass of $\text{CaO}$ produced:

$\text{Mass of CaO} = 0.03997 \text{ mol} \times 56.08 \text{ g/mol} = 2.2428 \text{ g}$

Therefore, the mass of $

$\text{CaO}$ produced is approximately 2.24 g.

Key Concept: Chemical Properties

d) Reactivity with acids

[Solution Description] Reactivity with acids is categorized as a chemical property. It refers to a substance’s ability to undergo chemical changes when exposed to an acid, which can help identify the substance based on its reactions.

Key Concept: Complex Stoichiometry, Molecular Formula

a) $C_3H_6O_3$

[Solution Description] Vapor density is given by $\frac{\text{molar mass}}{2}$, so the molar mass of the compound is $2 \times 45 = 90 \text{ g/mol}$.

The empirical formula mass of $CH_2O$ is:

$12 + 2(1) + 16 = 30 \text{ g/mol}$

To find the molecular formula, divide the molar mass by the empirical formula mass:

$\text{Factor} = \frac{90}{30} = 3$

Thus, the molecular formula is $(CH_2O)_3$, or $C_3H_6O_3$.

Key Concept: Significant Figures Count

b) 3

[Solution Description]

The number $0.00780$ has three significant figures:

– The leading zeros $(0.00)$ are not considered as they only serve as placeholders.

– The digits $7$, $8$, and the trailing zero are significant since it appears after a non-zero digit and a decimal point.

Therefore, there are three significant figures.

Key Concept: Simple Calculation

b) 18 u

[Solution Description] The molecular mass of water can be calculated by adding the atomic masses of its constituent atoms. There are two hydrogen atoms and one oxygen atom in water.

Mass of 2 Hydrogen atoms $= 2 \times 1 \, \text{u} = 2 \, \text{u}$

Mass of 1 Oxygen atom $= 16 \, \text{u}$

Total molecular mass of water $= 2 \, \text{u} + 16 \, \text{u} = 18 \, \text{u}$

So, the molecular mass of water is 18 u.

Key Concept: Theoretical Implications

c) Atoms are rearranged but never lost or gained.

[Solution Description] Lavoisier’s experiments led to the understanding that chemical reactions involve the rearrangement of atoms, not their creation or destruction, supporting the concept of atoms being conserved in reactions.

Key Concept: Advanced Conceptual Understanding, Temperature Effect on Molarity

b), Molarity decreases due to expansion of solvent volume

[Solution Description] Molarity is defined as the number of moles of solute per litre of solution. As the temperature increases, the volume of the solution typically expands while the amount of solute stays the same. This causes the molarity, which depends inversely on volume, to decrease because:

$$M = \frac{\text{No. of moles of solute}}{\text{Volume of solution}}$$

Since the denominator increases with a rise in temperature, the value of molarity decreases.

Key Concept: Comparative Analysis

d) Law of Conservation of Mass

[Solution Description] This statement directly describes the Law of Conservation of Mass. Other laws like the Law of Definite Proportions or Avogadro’s Law pertain to different aspects of chemical reactions and compounds.

Key Concept: Complex Stoichiometry, Temperature Effect on Concentration

a) No change

[Solution Description]

First, find the moles of NaCl in the original solution:

$\text{Moles of NaCl} = \text{Molarity} \times \text{Volume (L)} = 3 \, \text{mol/L} \times 0.25 \, \text{L} = 0.75 \, \text{moles}$

When the solution is diluted to 1 L, its new molarity becomes:

$\text{New Molarity} = \frac{\text{Moles of NaCl}}{\text{New Volume (L)}} = \frac{0.75 \, \text{moles}}{1 \, \text{L}} = 0.75 \, \text{M}$

However, molality does not depend on volume, but on the mass of solvent. Here assume density of water is approximately $1 \, \text{kg/L}$.

Total mass of solvent initially at 250 mL (ignoring solute mass for simplicity):

$\text{Mass of solvent} = 0.25 \, \text{kg}$

New mass of solvent after dilution to 1 L:

$\text{Mass of solvent} = 1 \, \text{kg}$

Molality is calculated by:

$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.75 \, \text{moles}}{1 \, \text{kg}} = 0.75\, \text{mol/kg}$

Thus, there is no change in molality due to temperature since initial calculation was based on weight not volume.

Key Concept: Chemical Properties

d) Reactivity with acids

[Solution Description] Reactivity with other substances, such as oxygen, signifies a chemical property because it involves a change in the substance’s composition, unlike physical properties that don’t change the substance’s nature.

Key Concept: Advanced Dimensional Analysis, Real-World Application of Standards

c) 0.00591 inches

[Solution Description] To convert micrometers to inches, use dimensional analysis:

Given: 150 µm

Conversion factor: $1 \text{ inch} = 25400 \text{ µm}$

Length in inches = $\frac{150 \text{ µm}}{25400 \text{ µm/inch}}$

Performing the division:

$\frac{150}{25400} \approx 0.00590551181$

Thus, the length is approximately 0.00591 inches, rounded to five significant figures.

Key Concept: Advanced Scientific Notation, Dimensional Analysis

b) 31.25 kg

[Solution Description]

To find the mass, use the formula: $mass = density \times volume$.

Substitute the given values:

$mass = (1.25 \times 10^3 \, \text{kg/m}^3) \times (2.5 \times 10^{-2} \, \text{m}^3)$

Calculate as follows:

$mass = 1.25 \times 2.5 \times 10^3 \times 10^{-2}$

$mass = 3.125 \times 10^{1}$

$mass = 31.25 \, \text{kg}$

Key Concept: Real World Application, Multi-step Problem

a) $C_4H_{12}O$, $C_4H_{12}O$

[Solution Description]

Convert the mass percentages to grams:

$\text{Mass of H} = \frac{13.20}{100} \times 500 = 66\, \text{g}$,

$\text{Mass of C} = \frac{39.80}{100} \times 500 = 199\, \text{g}$,

$\text{Mass of O} = \frac{47.00}{100} \times 500 = 235\, \text{g}$.

Convert masses to moles:

$\text{Moles of H} = \frac{66}{1.008} \approx 65.48\, \text{mol}$,

$\text{Moles of C} = \frac{199}{12.011} \approx 16.57\, \text{mol}$,

$\text{Moles of O} = \frac{235}{16.00} \approx 14.69\, \text{mol}$.

Divide by the smallest number of moles (14.69):

$\text{H: } \frac{65.48}{14.69} \approx 4.46$,

$\text{C: } \frac{16.57}{14.69} \approx 1.13$,

$\text{O: } \frac{14.69}{14.69} = 1$.

Recognize an approximate 4:1:1 ratio, thus the empirical formula is $$C_4H_{12}O_1$$. Calculate the empirical formula mass:

$4(12.01) + 12(1.008) + 16.00 = 76.13\, \text{g/mol}.$

Determine the molecular formula factor:

$n = \frac{90}{76.13} \approx 1.18.$

Since n is approximately 1, the molecular formula matches the empirical formula: $C_4H_{12}O$.

Key Concept: Application Scenario

c) The compound obeys the Law of Definite Proportions.

[Solution Description] The consistent elemental composition across all samples aligns with the Law of Definite Proportions. This confirms that the compound’s identity is characterized by a fixed ratio of its constituent elements by mass.

Key Concept: Compound Formation

d) $H_2O$ and $H_2O_2$

[Solution Description] The Law of Multiple Proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in simple whole number ratios. $H_2O$ and $H_2O_2$ both consist of hydrogen and oxygen. In water, 2g of hydrogen combine with 16g of oxygen while in hydrogen peroxide, 2g of hydrogen combine with 32g of oxygen, demonstrating the ratio 1:2 for oxygen.

Key Concept: Advanced Stoichiometry, Mole and Avogadro’s Constant

b) $5.33 \text{ moles}$

[Solution Description]

According to the balanced chemical equation, 1 mole of $N_2$ reacts with 3 moles of $H_2$ to produce 2 moles of $NH_3$. We need to determine the limiting reagent first.

For 3 moles of $N_2$, the required moles of $H_2$ would be:

$= 3 \times 3 = 9 \text{ moles of } H_2$

However, only 8 moles of $H_2$ are available, making $H_2$ the limiting reagent.

Using the stoichiometry from the balanced equation for $H_2$:

$3 \text{ moles of } H_2 \rightarrow 2 \text{ moles of } NH_3$

Therefore, for 8 moles of $H_2$, the amount of $NH_3$ produced is:

$= \left(\dfrac{2}{3}\right) \times 8 = \dfrac{16}{3} \approx 5.33 \text{ moles of } NH_3$

Key Concept: Significant Figures Basics

c) 4

[Solution Description] Count all non-zero digits and any zeros between them or after the decimal point: In $0.004560$, the significant figures are 4 (4, 5, 6, and trailing zero).

Key Concept: Interconversion of States

c) It becomes a liquid

[Solution Description] When a gas is cooled sufficiently, it loses energy, which reduces the motion of its particles. As the temperature drops further, the particles come close enough to enter the liquid phase, a process known as condensation.

Key Concept: Complex Problem Solving, Density and Buoyancy

c) 5.2 $g/cm^3$

[Solution Description]

First, determine the volume of the metal using the water displacement method. The volume displaced by the metal is equal to the volume of the metal, which is given as 150 cm$^3$. Next, use the formula for density:

$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

Substituting the values,

$\text{Density} = \frac{780 \, \text{g}}{150 \, \text{cm}^3} = 5.2 \, \text{g/cm}^3$

Therefore, the density of the metal is 5.2 g/cm$^3$.

Key Concept: Isotope Abundance Impact

b) 16.0086 u

[Solution Description] To find the new average atomic mass of oxygen, we use the formula:

$\text{Average Atomic Mass} = (f_{16} \times m_{16}) + (f_{17} \times m_{17}) + (f_{18} \times m_{18})$

where $f_{16}, f_{17}, f_{18}$ are the fractional abundances, and $m_{16}, m_{17}, m_{18}$ are the atomic masses.

Given:

$f_{16} = \frac{99.76}{100} = 0.9976, \quad f_{17} = \frac{0.04}{100} = 0.0004, \quad f_{18} = \frac{0.25}{100} = 0.0025$

Substitute these values into the formula:

$\text{Average Atomic Mass} = (0.9976 \times 15.9949) + (0.0004 \times 16.9991) + (0.0025 \times 17.9992)$

Calculate each term:

$0.9976 \times 15.9949 = 15.9568$

$0.0004 \times 16.9991 = 0.0068$

$0.0025 \times 17.9992 = 0.0450$

Add these results together:

$\text{Average Atomic Mass} = 15.9568 + 0.0068 + 0.0450 = 16.0086 \, \text{u}$

The new average atomic mass of oxygen is approximately 16.0086 u.

Key Concept: Advanced Stoichiometry, Reaction Yield

a) 149.63 g

[Solution Description] Firstly, calculate the theoretical yield of $CO_2$ using stoichiometry.

Molar mass of $C_2H_5OH = 46.08 \, \text{g/mol}$, thus moles of $C_2H_5OH = \frac{92 \, \text{g}}{46.08 \, \text{g/mol}} = 2.00 \, \text{mol}$.

From the balanced equation, $1 \, \text{mol}$ of $C_2H_5OH$ yields $2 \, \text{mol}$ of $CO_2$, so $2.00 \, \text{mol} \, C_2H_5OH$ yields $4.00 \, \text{mol} \, CO_2$.

Molar mass of $CO_2 = 44.01 \, \text{g/mol}$, so theoretical mass of $CO_2$ is $4.00 \, \text{mol} \times 44.01 \, \text{g/mol} = 176.04 \, \text{g}$.

With 85% yield, actual mass of $CO_2$ produced is $0.85 \times 176.04 \, \text{g} = 149.634 \, \text{g}$.

Key Concept: Metallurgy

b) Alloying copper with tin and arsenic

[Solution Description] Ancient Indian metallurgy included techniques to improve the hardness and durability of copper by alloying it with tin and arsenic. This process created stronger and more durable alloys like bronze, which marked a significant development in metallurgical practices.

Key Concept: Mixture Separation

c) Separating funnel

[Solution Description] The most suitable method for separating a mixture of oil and water is using a separating funnel. This takes advantage of the immiscible nature of oil and water, which allows them to separate into layers based on their densities.

Key Concept: Rounding Off

b) 5.68

[Solution Description]

When rounding off a number to a specified number of significant figures, you look at the digit following the last significant figure you want to keep. If this digit is 5 or greater, round up. Otherwise, leave the last significant figure unchanged.

For $5.6789$, rounding to three significant figures:

The third significant figure is 7 and the next digit (8) is greater than 5, so we round up:

Result: $5.68$

Key Concept: State Change Process

d) Sublimation

[Solution Description] Sublimation is the process where a solid changes directly into a gas without going through the liquid phase. This happens at specific conditions of pressure and temperature.

Key Concept: Average Atomic Mass Calculation

b) 35.485 amu

[Solution Description] The average atomic mass is calculated by multiplying each isotope’s mass by its relative abundance and summing up these values. For chlorine, this can be calculated as follows:

$\text{Average atomic mass} = (35 \times 0.7577) + (37 \times 0.2423)$

Calculating each term:

$35 \times 0.7577 = 26.5195$

$37 \times 0.2423 = 8.9651$

Adding them together gives:

$26.5195 + 8.9651 = 35.4846 \, \text{amu}$

Thus, the average atomic mass of chlorine is approximately 35.485 amu.

Key Concept: Multiplication/Division Rules

a) $4 \times 10^3$

[Solution Description]

To divide these numbers in scientific notation, divide the coefficients and subtract the exponents:

$\frac{1.2}{3} = 0.4$

$10^{7-3} = 10^4$

Combine them:

$0.4 \times 10^4 = 4 \times 10^3$

The result is $4 \times 10^3$.

Key Concept: Philosophical Speculations

c) Acharya Kanda

[Solution Description] Acharya Kanda, also known as Kashyapa, was an ancient Indian philosopher who proposed the idea of “Anu,” which means atom or an indestructible particle. He laid down the foundation of Vaisheshika philosophy, which includes the concept of Anu being the smallest unit of matter.

Key Concept: Length Measurement Evolution

c) As the length between two marks on a platinum-iridium bar

[Solution Description] The metre was originally defined as the distance between two marks on a platinum-iridium bar kept at the International Bureau of Weights and Measures. This physical representation has since been replaced by more precise definitions based on fundamental constants.

Key Concept: Historical Figures

c) Nagarjuna

[Solution Description] Nagarjuna was a prominent ancient Indian scientist who contributed to various fields including alchemy and metallurgy. He is credited with pioneering work on the extraction of metals such as zinc.