FBISE Class 10 Chemistry Pre-board Exam: Test Paper and Guess Paper for Exam Preparation 2024
Pre-Board Exam - Chemistry SSC-II
Section-A (Marks-12)
Time Allowed: 20mins
Q1.Circle the correct option A/B/C/D each part carry equal marks
i-The Law of mass action was proposed by
a)Charles b)P. Waage c)M Gulberg d)Both b and c
b) P. Waage
ii-Which of the following is an lead ore
a)Gelna b)silica c)bauxite d)haematite
d) Haematite
iii-which fraction of petroleum is used in chemical feedstock
a)naphtha b)kerosene c)bitumen d)lubricating oil
a) Naphtha
iv- is used in drips to recover dehydration
a)maltose b)dextrose c)raffinose d)starch
b) Dextrose
v- In which of the following compounds, oxygen is attached to two alkyl carbon
atoms?
a)Alcohol b)Phenol c)Ether d)Ester
a) Alcohol
vi- Lewis acid base has following characteristics except
a) formation of adduct
b) formation of coordinate covalent bond
c) donation and acceptance of electron pair
d) donation and acceptance of a proton
c) Donation and acceptance of electron pair
vii- which of the following process is reduction in nature
a)roasting b)smelting c)flotation d)grinding
a) Roasting
viii-Percentage of sodium chloride in sea water is
a)3.4 b)0.02 c)97 d)2
c) 97
ix-percentage of nitrogen in urea is
a)75 b)46.4 c)80 d)21
b) 46.4
x-which is a reddish brown gas
a)chlorine b)nitrogen c)fluorine d)bromine
a) Chlorine
xi-milk of magnesia Mg(OH)2 is used as antacid it neutralizes excess stomach
acid.which salt is formed in this reaction
a)MgSO4 b)MgO c)MgCO3 d) MgCl2
c) MgCO3
xii-for which of the reaction kc has unit mol.dm-3
a)N2+3H2=2NH3
b)PCl5=PCl3+PCl2
c)2ICl=I2+Cl2
d)2NO2=N2O4
a) N2 + 3H2 = 2NH3
SECTION-B(Marks 33)
Time Allowed:2hrs 40min
Q2.Answer any eleven questions. Each question carry equal marks (11x3=33)
i)identify the branch of chemistry that deals with following examples.
a)sulphur dioxide is the major sources of acid rain.
i) a) Environmental Chemistry - Sulphur dioxide is a pollutant that reacts with water vapor in the atmosphere to form sulfuric acid, a major component of acid rain.
b)silver article tarnishes in air
c)Gasoline(hydrocarbon)is ignited in automobile
ii)What are defects in rutherford atomic model?
The Rutherford atomic model, while a groundbreaking discovery, had some key limitations:
Stability of Electrons: The model proposed electrons orbiting the nucleus like planets around the sun. However, according to classical physics, a charged particle like an electron undergoing acceleration would lose energy and eventually spiral into the nucleus. This contradicts the observed stability of atoms.
Emission Spectrum: The model couldn't explain the discrete atomic emission lines observed in experiments. Atoms emit light only at specific wavelengths, which the Rutherford model couldn't account for.
iii)Define the term scheilding affect and arrange the atom in order of increasing
having greater shielding affect from the list given below Li,Be and Mg
Shielding Effect and Atomic Order
Shielding Effect:
The shielding effect refers to the phenomenon where the inner core electrons (electrons closer to the nucleus) in an atom partially repel the electrostatic attraction between the nucleus and the outer valence electrons (electrons farther from the nucleus). This repulsion weakens the attraction experienced by the valence electrons, making them slightly less attracted to the nucleus.
Arranging Li, Be, and Mg by Increasing Shielding Effect:
- Li (Lithium): Lithium has only one electron (1s). This electron experiences no shielding effect from other electrons.
- Be (Beryllium): Beryllium has four electrons (2s²). The two inner electrons (1s²) provide some shielding to the outer two valence electrons (2s²).
- Mg (Magnesium): Magnesium has eight electrons (1s²2s²2p⁶). The inner six electrons (1s²2s²2p⁶) provide a stronger shielding effect for the outermost valence electrons compared to beryllium.
Therefore, the order of increasing shielding effect is: Li < Be < Mg
iv) Difference between amorphous and crystalline solids
Crystalline vs. Amorphous Solids
The key difference between crystalline and amorphous solids lies in the arrangement of their constituent particles (atoms, molecules, or ions).
Crystalline Solids:
- Have a highly ordered, regular arrangement of particles in a repeating three-dimensional pattern called a crystal lattice.
- This ordered structure leads to well-defined shapes and flat faces for crystals.
- They exhibit sharp melting points, where they abruptly transition from solid to liquid at a specific temperature.
- Examples: table salt (NaCl), diamond (C), quartz (SiO2).
Amorphous Solids:
- Lack a regular arrangement of particles. Their structure is more random and disorganized.
- They have irregular shapes and no flat faces.
- They exhibit a softening range rather than a sharp melting point. As temperature increases, they gradually soften and become more liquid-like.
- Examples: glass, plastic, rubber.
Here's a table summarizing the key differences:
v) Why does group 1 elements prefer to combine with group 17 elements.
Group 1 elements (alkali metals) and Group 17 elements (halogens) prefer to combine with each other due to their electron configurations and the desire to achieve a stable electron arrangement. Here's a breakdown:
Group 1 Elements (Alkali Metals):
- Have one valence electron (ns¹ configuration).
- They are highly reactive and readily lose this electron to achieve a stable noble gas electron configuration (ns²).
- Losing an electron results in a positive ion (cation) with a +1 charge.
Group 17 Elements (Halogens):
- Have seven valence electrons (ns²np⁵ configuration).
- They are also highly reactive and need to gain one electron to achieve a stable noble gas electron configuration (ns²np⁶).
- Gaining an electron results in a negative ion (anion) with a -1 charge.
The Attraction:
The attraction between these two groups arises from:
- Electrostatic Attraction: Oppositely charged ions (cations from Group 1 and anions from Group 17) attract each other due to Coulomb's law.
- Achieving Stable Electron Configuration: Both groups achieve a stable electron configuration by the transfer of electrons.
Ionic Bonding:
The transfer of the electron from the Group 1 element to the Group 17 element leads to the formation of an ionic bond. The metal loses its valence electron, becoming a positively charged cation. The non-metal gains the electron, becoming a negatively charged anion. The electrostatic attraction between these oppositely charged ions holds the compound together.
Example:
- Sodium (Na) from Group 1 readily loses its one valence electron and becomes a sodium ion (Na⁺).
- Chlorine (Cl) from Group 17 readily gains one electron to become a chloride ion (Cl⁻).
- The electrostatic attraction between Na⁺ and Cl⁻ forms sodium chloride (NaCl), table salt.
vi) briefly explain the formation of ionic bond between magnesium and Chlorine.
Ionic Bond Formation between Magnesium and Chlorine
Magnesium (Mg) and Chlorine (Cl) form an ionic bond through the following steps:
Electron Configuration:
- Magnesium has an electron configuration of [Ne]3s². It has two valence electrons in its outermost 3s orbital.
- Chlorine has an electron configuration of [Ne]3s²3p⁵. It needs one electron to fill its outermost 3p orbital and achieve a stable noble gas configuration.
Electron Transfer:
- Magnesium loses its two valence electrons due to its low ionization energy (energy required to remove an electron). This results in a positively charged magnesium ion (Mg²⁺).
- Chlorine gains the two electrons from magnesium to fill its outer shell. This results in a negatively charged chloride ion (Cl⁻).
Electrostatic Attraction:
- The positively charged magnesium ion (Mg²⁺) and the negatively charged chloride ion (Cl⁻) attract each other due to opposite charges (Coulomb's Law).
Ionic Compound Formation:
- The electrostatic attraction between the Mg²⁺ and Cl⁻ ions holds them together in a crystal lattice, forming the ionic compound magnesium chloride (MgCl₂).
Key Points:
- Magnesium loses electrons to achieve a stable configuration like a noble gas.
- Chlorine gains electrons to achieve a stable configuration like a noble gas.
- The oppositely charged ions attract to form a stable ionic compound.
vii) Write the valence shell electronic configuration for the following groups
a)Noble gases b)halogens c)alkali metals
The valence shell electronic configurations for the groups you requested are:
a) Noble Gases: ns²np⁶ (where n is the principal energy level of the valence shell) * Noble gases have a full valence shell, leading to their stability and unreactive nature.
b) Halogens: ns²np⁵ (where n is the principal energy level of the valence shell) * Halogens are one electron short of a full valence shell, making them highly reactive and seeking to gain one electron.
c) Alkali Metals: ns¹ (where n is the principal energy level of the valence shell) * Alkali metals have only one electron in their valence shell, making them very reactive and wanting to lose that electron.
viii) Draw bohr model for Chlorine atom.
Description of the Bohr model for a Chlorine atom:
Chlorine Atom (Cl):
- Atomic Number: 17
- Number of Protons: 17 (positively charged particles in the nucleus)
- Number of Electrons: 17 (negatively charged particles orbiting the nucleus)
Bohr Model Representation:
- Draw a small circle in the center to represent the nucleus.
- Label the nucleus with "17p" to indicate 17 protons.
- Draw three circles around the nucleus to represent the electron shells (energy levels).
- The first shell (closest to the nucleus) can hold a maximum of 2 electrons.
- The second shell can hold a maximum of 8 electrons.
- The third shell (outermost) can hold a maximum of 8 electrons.
- Distribute the 17 electrons in the shells following the Aufbau principle (filling lower energy levels first):
- Place 2 electrons in the first shell.
- Place 8 electrons in the second shell.
- Place the remaining 7 electrons in the third shell.
Important Note:
While the Bohr model provides a good visualization of the basic structure of the atom, it has limitations. Modern atomic models use orbitals to describe the electron probability distribution around the nucleus, which is more accurate for complex atoms.
ix) Define analytical chemistry and industrial chemistry.
Analytical Chemistry vs. Industrial Chemistry
Both analytical and industrial chemistry are important branches of chemistry, but they serve different purposes:
Analytical Chemistry:
- Focus: Studies the composition and structure of matter.
- Goal: Identifies the elements, compounds, and their quantities present in a sample.
- Techniques: Employs a wide range of techniques like spectroscopy, chromatography, titrations, and gravimetry.
- Applications: Used in various fields like environmental monitoring, forensics, quality control in industries, drug development, and medical diagnostics.
Industrial Chemistry:
- Focus: Large-scale production of chemicals and materials.
- Goal: Develops and implements efficient processes to manufacture chemicals, fuels, pharmaceuticals, polymers, and other products.
- Techniques: Employs various methods like chemical reactions, separations, purifications, and optimizations for large-scale production.
- Applications: Forms the backbone of various industries like pharmaceuticals, chemicals, plastics, fertilizers, energy production, and food processing.
Here's an analogy to understand the difference:
- Analytical Chemistry: Like a detective investigating the scene of a crime, analyzing the evidence (sample) to identify the culprit (components) and their involvement (quantity).
- Industrial Chemistry: Like a chef in a large kitchen, following recipes and techniques to prepare dishes (chemicals) in large quantities to meet customer demands.
x) Draw a label diagram of dry cell.
Labelled Diagram of a Dry Cell
Here's a description of a labelled diagram of a dry cell:
Cylindrical Dry Cell:
- Zinc Container: The outer shell of the dry cell is made of zinc metal. It acts as the anode (negative electrode) in the electrochemical reaction.
- Zinc Amalgam: A layer of zinc amalgam (a paste of zinc and mercury) lines the zinc container. This layer minimizes the reaction between zinc and the electrolyte, preventing self-discharge of the cell.
- Electrolyte Paste: A moist paste containing ammonium chloride (NH₄Cl) and other salts is present in the center of the cell. This paste acts as the electrolyte, allowing the flow of ions between the electrodes.
- Separator: A porous material like cardboard or paper separates the anode (zinc container) from the cathode (carbon rod). It allows the movement of ions but prevents physical contact between the electrodes.
- Carbon Rod: A central carbon rod acts as the cathode (positive electrode) in the electrochemical reaction.
- MnO₂ Mixture: Surrounding the carbon rod is a mixture of manganese dioxide (MnO₂) and other materials like graphite. This mixture acts as the depolarizer, facilitating the reduction reaction at the cathode and preventing the buildup of hydrogen gas, which would hinder the cell's performance.
- Wax Seal: The top of the cell is sealed with wax to prevent leakage of the electrolyte and moisture loss.
Labelled Diagram:
Note: Mercury is no longer commonly used in dry cells due to environmental concerns. Modern dry cells may use alternative materials in the zinc amalgam or different depolarizer mixtures.
xi) Differentiate diffusion and effusion
Diffusion and effusion are both processes involving the movement of particles from one region to another, but they occur under different conditions and involve distinct mechanisms. Here's how they differ:
Definition:
- Diffusion: Diffusion is the process by which particles move from an area of higher concentration to an area of lower concentration until equilibrium is reached.
- Effusion: Effusion is the process by which gas particles pass through a small opening or pinhole into a vacuum or a region of lower pressure.
Mechanism:
- Diffusion: In diffusion, particles move randomly due to their kinetic energy. There's no specific direction of movement, but on average, particles tend to move from regions of higher concentration to regions of lower concentration.
- Effusion: Effusion involves the movement of gas particles through a small opening. This occurs because gas particles possess kinetic energy, and some of them have sufficient energy to overcome the barrier presented by the opening and escape into the region of lower pressure.
Conditions:
- Diffusion: Diffusion occurs in all states of matter (solid, liquid, gas) and is influenced by factors such as temperature, concentration gradient, and the nature of the medium through which particles are diffusing.
- Effusion: Effusion typically occurs in gases and requires a pressure difference across the opening. It's more pronounced in gases because they have greater freedom of movement compared to liquids and solids.
Rate of Movement:
- Diffusion: The rate of diffusion depends on factors such as the concentration gradient, temperature, and the size and nature of the particles involved.
- Effusion: The rate of effusion is influenced by the size and mass of the gas particles. Lighter gas particles effuse more quickly than heavier ones, according to Graham's law of effusion.
In summary, while diffusion involves the movement of particles from areas of high concentration to low concentration, effusion specifically refers to the passage of gas particles through a small opening into a region of lower pressure.
xii) Explain zinc plating with help of chemical equation.
Zinc Plating with Chemical Equation
Zinc plating is an electroplating process where a thin layer of zinc is deposited onto the surface of another metal object. This process provides protection against corrosion for the underlying metal.
The process involves the following steps:
- Electrolyte Preparation: An aqueous solution containing a zinc salt, typically zinc sulfate (ZnSO₄), is used as the electrolyte.
- Electrodes: The object to be plated (cathode) is immersed in the electrolyte along with an anode made of pure zinc metal.
- Electrical Connection: An external electrical power source is connected to the electrodes, making the anode positive and the cathode negative.
Chemical Reactions:
- Anode (Oxidation): At the anode (positive electrode), zinc metal atoms lose two electrons and dissolve into the electrolyte as zinc ions (Zn²⁺).
Zn(s) → Zn²⁺(aq) + 2e⁻
- Cathode (Reduction): At the cathode (negative electrode), zinc ions from the electrolyte gain the electrons from the external power source and deposit as a thin layer of zinc metal on the object's surface.
Zn²⁺(aq) + 2e⁻ → Zn(s)
Overall Reaction:
The net reaction combines the anode and cathode reactions, showing the transfer of zinc metal from the anode to the cathode as a plating layer.
Zn(s) → Zn²⁺(aq) + 2e⁻
2e⁻ + Zn²⁺(aq) → Zn(s)
-----------------------
2Zn(s) → 2Zn²⁺(aq) + 2e⁻ (Not a single reaction, but represents the net transfer)
Benefits of Zinc Plating:
- Protects the underlying metal from corrosion by acting as a sacrificial anode.
- Provides a smooth, conductive surface.
- Improves the appearance of the object.
Applications:
Zinc plating is used in various applications like:
- Protecting steel from rust on car parts, screws, and nails.
- Providing a conductive surface for electrical components.
- Enhancing the appearance of jewelry and other decorative items.
xiii) Write any three uses of isotopes in medical field.
Isotopes have numerous applications in the medical field, here are three key uses:
Medical Diagnosis:
- Radioisotopes as tracers: Certain isotopes emit detectable radiation. Doctors can introduce small amounts of these radioisotopes (radiopharmaceuticals) into a patient's body. The radioisotope then accumulates in specific organs or tissues based on their biological function. Detectors can then track the radiation to create images of these organs or functions. This technique, known as nuclear medicine imaging, helps diagnose various conditions like:
- Bone scans to detect bone tumors or fractures.
- Thyroid scans to assess thyroid function and identify abnormalities.
- PET scans (Positron Emission Tomography) to study metabolic activity and diagnose cancers, heart disease, and brain disorders.
- Radioisotopes as tracers: Certain isotopes emit detectable radiation. Doctors can introduce small amounts of these radioisotopes (radiopharmaceuticals) into a patient's body. The radioisotope then accumulates in specific organs or tissues based on their biological function. Detectors can then track the radiation to create images of these organs or functions. This technique, known as nuclear medicine imaging, helps diagnose various conditions like:
Cancer Treatment:
- Radiation therapy: Certain isotopes emit high-energy radiation that can damage or destroy cancer cells. These isotopes can be used in two ways for cancer treatment:
- External beam radiation therapy: A focused beam of radiation from a machine containing a radioisotope is directed towards the tumor.
- Internal radiation therapy (brachytherapy): Radioisotopes are sealed in small capsules or seeds and implanted directly into or near the tumor site. This delivers a high dose of radiation to the cancer cells while minimizing damage to surrounding healthy tissues.
- Radiation therapy: Certain isotopes emit high-energy radiation that can damage or destroy cancer cells. These isotopes can be used in two ways for cancer treatment:
Sterilization of Medical Equipment:
- Gamma rays emitted by Cobalt-60 (⁶⁰Co) are a powerful form of radiation that can effectively kill bacteria and other microorganisms. This property makes it ideal for sterilizing medical equipment like syringes, bandages, and surgical instruments without damaging the materials. This ensures sterility and reduces the risk of infections during medical procedures.
xiv) Atomic mass of copper is 63amu . Calculate the mass of 3.5 moles of copper.
We can calculate the mass of 3.5 moles of copper using the concept of molar mass.
Molar Mass of Copper: You've correctly identified the atomic mass of copper as 63 amu (atomic mass units). However, for mass calculations, we typically use the molar mass in grams per mole (g/mol).
Moles of Copper: We are given that the quantity of copper is 3.5 moles.
Mass Calculation:
We can use the following formula to convert moles (n) of a substance to its mass (m) in grams, given its molar mass (M):
m = n * M
Here,
- n = 3.5 moles (quantity of copper)
- M = 63 g/mol (molar mass of copper)
Calculation:
m = 3.5 moles * 63 g/mol = 220.5 grams
Therefore, 3.5 moles of copper have a mass of approximately 220.5 grams.
xv) Compare solution and suspension in three points each.
Solution and suspension are both types of mixtures, but they have distinct characteristics. Here's a comparison of solution and suspension in three points each:
Solution:
- Homogeneity:
- Solutions are homogeneous mixtures where the solute is evenly distributed and thoroughly mixed with the solvent at the molecular level. This means that solutions have a uniform composition throughout, with no visible boundaries between the solute and solvent.
- Particle Size:
- The particles in a solution are typically very small, usually less than 1 nanometer in diameter. These particles are molecular or ionic in nature and are individually dispersed throughout the solvent. They do not settle out or exhibit the Tyndall effect (scattering of light).
- Stability:
- Solutions are generally stable over time, with the solute remaining dissolved in the solvent without settling out. As long as the solution is undisturbed, the solute particles remain uniformly distributed. However, certain conditions such as changes in temperature or pressure can affect solubility.
Suspension:
- Homogeneity:
- Suspensions are heterogeneous mixtures where the solute particles are not fully dissolved but instead are suspended in the solvent. As a result, suspensions often appear cloudy or opaque, with visible particles that settle out upon standing.
- Particle Size:
- The particles in a suspension are larger than those in a solution, typically ranging from a few nanometers to several micrometers in diameter. These particles may be visible to the naked eye and can settle out over time due to gravity.
- Stability:
- Suspensions are less stable than solutions, as the suspended particles tend to settle out over time, resulting in sedimentation. To maintain a suspension, constant agitation or stirring is often required. Additionally, suspensions may exhibit the Tyndall effect, where light is scattered by the suspended particles, causing the mixture to appear cloudy or milky when illuminated.
In summary, solutions are homogeneous mixtures with very small solute particles that are uniformly dispersed throughout the solvent, whereas suspensions are heterogeneous mixtures with larger solute particles that are suspended but not fully dissolved in the solvent, leading to visible settling and instability.
SECTION-C (Marks 20)
Attempt any two questions. All questions carry equal marks (2x10=20)
Q3.a) Define solubility.write a comprehensive note on affect of temperature on soluility of different solutes.also draw a graph to support your answer.[7]
Solubility
Definition: Solubility refers to the maximum amount of a solute (the substance being dissolved) that can dissolve in a specific solvent (the dissolving medium) at a particular temperature to form a stable solution.
Factors Affecting Solubility:
Several factors influence the solubility of a solute in a solvent. Here, we'll focus on the effect of temperature:
Effect of Temperature on Solubility:
The effect of temperature on solubility can be categorized into two main types:
Solid Solutes in Liquids:
- Generally, the solubility of most solid solutes in liquids increases with increasing temperature. As the temperature rises, the kinetic energy of both solute and solvent molecules increases. This increased energy makes it easier for the solvent molecules to overcome the attractive forces holding the solute particles together, leading to greater dissolution.
- Example: Sugar dissolves more readily in hot coffee than in cold coffee.
Gases in Liquids:
- The solubility of gases in liquids generally decreases with increasing temperature. This phenomenon seems counterintuitive, but it can be explained by the behavior of gas molecules. At higher temperatures, gas molecules have more kinetic energy and tend to escape from the liquid phase back into the gas phase, reducing their concentration in the solution.
- Example: Carbonated beverages tend to lose their fizz (dissolved CO₂) faster when warmed.
Graph:
The relationship between temperature and solubility can be depicted graphically:
The X-axis represents temperature, and the Y-axis represents the amount of solute dissolved (concentration).
- Solid Solutes: The graph for most solid solutes will show an upward slope, indicating increasing solubility with temperature.
- Gases: The graph for gases will show a downward slope, indicating decreasing solubility with temperature.
Important Note:
There are some exceptions to these general trends. The solubility of some solutes may exhibit a more complex relationship with temperature, or even decrease slightly with increasing temperature in a specific range.
b) State octet and duplet rule with one example each[3]
Octet Rule and Duplet Rule
These rules are used to predict the stable electron configurations of atoms in molecules.
1. Octet Rule:
The octet rule states that atoms tend to gain, lose, or share electrons in order to have eight electrons in their outermost valence shell, achieving a stable configuration similar to the noble gases.
Example:
- Chlorine (Cl): Chlorine has 17 electrons. Its electron configuration is [Ne]3s²3p⁵. In order to achieve an octet, it gains one electron, resulting in a stable Cl⁻ ion with the configuration [Ne]3s²3p⁶ (eight electrons in the valence shell).
2. Duplet Rule:
The duet rule applies specifically to the first electron shell of an atom. It states that the first shell can only accommodate a maximum of two electrons to achieve stability. This is especially true for the elements in the first row (Group 1 and Group 2) of the periodic table.
Example:
- Hydrogen (H): Hydrogen has one electron (1s¹). It can either lose this electron to become a positively charged H⁺ ion (empty valence shell) or share its electron to form a covalent bond, achieving the stable duet configuration (2 electrons in the first shell).
Q4.a)Explain electrolytic refining of copper with labelled diagram.[6]
Electrolytic Refining of Copper:
Electrolytic refining is a process used to purify metals, particularly copper, by the use of an electrolytic cell. In the case of copper, impure copper from smelting or ores is purified through this process to obtain high-purity copper suitable for various industrial applications.
Process Overview:
Preparation of Electrolyte: The electrolyte used in electrolytic refining of copper is typically an acidified copper sulfate solution.
Setup of Electrolytic Cell: The setup consists of a cathode (pure copper sheet), an anode (impure copper), and an electrolyte solution.
Electrolysis: When an electric current is passed through the electrolyte solution, copper ions (Cu²⁺) migrate towards the cathode (negative electrode), where they gain electrons and are deposited as pure copper atoms.
Oxidation of Impurities: The impurities in the anode (such as lead, arsenic, zinc, and other metals) are oxidized and form soluble compounds, which either dissolve in the electrolyte or settle as sludge at the bottom of the cell.
Formation of Copper Cathode: Pure copper gradually plates out onto the cathode, forming a high-purity copper cathode over time. The impurities settle or dissolve away from the anode.
Labelled Diagram:
Anode (Impure Copper) Electrolyte Solution +--------------------------------------+---------------------+ | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | +--------------------------------------+---------------------+ Cathode (Pure Copper)
Anode (Impure Copper): This is where the impure copper is placed. During electrolysis, the impurities in the anode oxidize and form soluble compounds or settle as sludge.
Cathode (Pure Copper): This is where the pure copper is deposited. Copper ions from the electrolyte are reduced at the cathode, forming a layer of pure copper over time.
Electrolyte Solution: Typically, an acidified copper sulfate solution is used as the electrolyte. It provides the necessary copper ions for the deposition of pure copper at the cathode.
This process allows for the purification of copper, producing high-purity copper cathodes suitable for various industrial applications such as electrical wiring, electronics, and construction.
b)Calculate the number of moles of solute present in 1.25cm3 of 0.5M H3PO4 [4]
Absolutely, we can calculate the number of moles of solute (H3PO4) present in 1.25 cm³ of 0.5 M H3PO4 solution using the following steps:
Concentration (Molarity): We are given the concentration of the solution as 0.5 M. Molarity (M) represents moles of solute per liter of solution.
Volume Conversion: The volume of the solution is provided in cubic centimeters (cm³). However, molarity is typically used with liters (L). Therefore, we need to convert the volume from cm³ to L.
Conversion factor: 1 L = 1000 cm³
Volume (L) = 1.25 cm³ * (1 L / 1000 cm³) = 0.00125 L
- Moles Calculation: Now, we can use the following formula to calculate the number of moles (n) of solute:
n = M * V
where:
- n = moles of solute (H3PO4)
- M = concentration (molarity) = 0.5 M
- V = volume of solution (in liters) = 0.00125 L
Calculation:
n = 0.5 M * 0.00125 L = 0.000625 moles
Therefore, there are 0.000625 moles of H3PO4 present in 1.25 cm³ of the 0.5 M solution.
Q5.a) How many ions of Na and Cl will be present in 30 g of NaCl? [3]
We can determine the number of Na and Cl ions present in 30 g of NaCl using the following steps:
Calculate Moles of NaCl:
- Molar mass of NaCl (Na = 23 g/mol, Cl = 35.5 g/mol) = 23 g/mol + 35.5 g/mol = 58.5 g/mol
- Moles of NaCl (n) = mass (m) / molar mass (M)
- n = 30 g / 58.5 g/mol = 0.513 mol (approximately)
Avogadro's Constant:
- Avogadro's constant (N_A) represents the number of particles (atoms, molecules, or ions) in one mole of a substance. It is approximately equal to 6.022 x 10^23 particles/mol.
Number of Ions:
- Since NaCl dissociates completely in water, one mole of NaCl produces one mole of Na⁺ ions and one mole of Cl⁻ ions.
- Therefore, 0.513 mol of NaCl will produce:
- Na⁺ ions: 0.513 mol NaCl * 1 mol Na⁺ ions / 1 mol NaCl = 0.513 mol Na⁺ ions
- Cl⁻ ions: 0.513 mol NaCl * 1 mol Cl⁻ ions / 1 mol NaCl = 0.513 mol Cl⁻ ions
Number of Individual Ions:
- To find the total number of individual ions, we multiply the moles of ions by Avogadro's constant:
- Number of Na⁺ ions = 0.513 mol * 6.022 x 10^23 particles/mol ≈ 3.09 x 10^23 ions
- Number of Cl⁻ ions = 0.513 mol * 6.022 x 10^23 particles/mol ≈ 3.09 x 10^23 ions
- To find the total number of individual ions, we multiply the moles of ions by Avogadro's constant:
Therefore:
- There are approximately 3.09 x 10^23 sodium (Na⁺) ions present in 30 g of NaCl.
- There are approximately 3.09 x 10^23 chloride (Cl⁻) ions present in 30 g of NaCl.
Important Note:
In reality, due to limitations in measurement and inherent statistical fluctuations, the actual number of ions will not be an exact whole number but will be very close to the calculated value.
b) Define boyles law. volume of ethane gas is 2.5dm3 at 2.10atm pressure. when its pressure is reduced to 1.05atm its volume become 5.0dm3. Expalin this change using boyles law [4]
Boyle's Law and Ethane Gas Volume Change
Boyle's Law:
Boyle's law states that at constant temperature, the volume (V) of a gas is inversely proportional to the pressure (P) exerted on the gas. Mathematically, it can be expressed as:
V ∝ 1 / P
This means that as the pressure on a gas increases, its volume decreases, and vice versa (assuming temperature remains constant).
Explanation for Ethane Gas:
You've provided the following information:
- Initial volume (V₁): 2.5 dm³
- Initial pressure (P₁): 2.10 atm
- Final volume (V₂): 5.0 dm³
- Final pressure (P₂): 1.05 atm
According to Boyle's law, the product of initial volume and pressure should be equal to the product of final volume and pressure, assuming constant temperature.
Verifying the Change:
Let's calculate the product of initial volume and pressure (V₁P₁) and compare it to the product of final volume and pressure (V₂P₂):
- V₁P₁ = 2.5 dm³ * 2.10 atm = 5.25 dm³ atm
- V₂P₂ = 5.0 dm³ * 1.05 atm = 5.25 dm³ atm
We can see that V₁P₁ is equal to V₂P₂, confirming that the change in volume follows Boyle's law.
Reasoning:
When the pressure on the ethane gas is reduced from 2.10 atm to 1.05 atm (almost half), the gas expands to occupy a larger volume (5.0 dm³), which is double the initial volume (2.5 dm³). This expansion occurs because the gas molecules have more space to move around due to the decrease in pressure. The total number of gas molecules remains constant, but they spread out to fill the available volume according to Boyle's law.
c) Write chemical formula for a)aluminium sulphate b)potassium phosphate c)sodium nitrate.[3]
The chemical formulas for the compounds you requested are:
a) Aluminium sulfate: Al₂(SO₄)₃ b) Potassium phosphate: K₃PO₄ (there are several potassium phosphates with different ratios of K to PO₄, but this is the most common one) c) Sodium nitrate: NaNO₃
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