Class 9 Science Chapter 2 Is Matter Around Us Pure?

Key Concept: Advanced Calculation, Temperature-Solubility Relationship

c) 28.57%

[Solution Description] Initially, the mass of solute is 30 g in 100 mL of water. The initial solubility is therefore 30 g per 100 mL.

Upon increasing the temperature from 298 K to 348 K (a change of 50 K), the solubility increases by 10 g per 100 mL according to the given condition. Therefore, the new solubility becomes:

$30\, \text{g} + 10\, \text{g} = 40\, \text{g per 100 mL}$

Assuming that the additional solute can dissolve, the total mass of solute becomes 40 g in 100 mL of water.

The mass of the solvent initially was 100 g of water (assuming density of water is around 1 g/mL).

Now, calculate the new mass percentage:

$\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 = \left(\frac{40}{140}\right) \times 100 = 28.57\%$

Thus, the final mass percentage of the solution is approximately 28.57%.

Key Concept: Concentration Understanding

c) 50 g/L

[Solution Description] The concentration of a solution can be calculated using the formula: concentration = $\frac{\text{mass of solute}}{\text{volume of solution}}$.

Here, mass of solute = 10 g, and volume of solution = 200 mL or 0.2 L.

$\text{Concentration} = \frac{10 \text{ g}}{0.2 \text{ L}} = 50 \text{ g/L}$

Key Concept: Real-World Context

c) Add 12 g more solute to reach 20 g total in 200 mL

[Solution Description] To maintain the concentration of 10%, the required mass of solute should follow:

$\frac{x}{200} \times 100 = 10$

Solving for $x$: $x = \frac{10 \times 200}{100} = 20 \, \text{g}$

Therefore, 12 g more solute must be added to achieve 20 g of solute in 200 mL to keep the concentration at 10%.

Key Concept: Real-world Application, Comparative Concentration

d) Solution B: 23.81%, more concentrated.

[Solution Description]

For Solution A:

$\text{Mass of solution A} = 30\,g + 120\,g = 150\,g$

$\text{Mass by mass % of A} = \left(\frac{30}{150}\right) \times 100 = 20%$

For Solution B:

$\text{Mass of solution B} = 25\,g + 80\,g = 105\,g$

$\text{Mass by mass % of B} = \left(\frac{25}{105}\right) \times 100 \approx 23.81%$

Comparing both concentrations, Solution B has a higher mass by mass percentage, making it more concentrated.

Key Concept: Property Identification

b) Fixed melting point

[Solution Description] Compounds have properties that are different from those of their constituent elements due to the chemical bonding between them, unlike mixtures which show the properties of their components.

Key Concept: Separation Techniques

a) Decantation

[Solution Description] Given that oil and water form a heterogeneous mixture due to differences in density and immiscibility, they can be efficiently separated via decantation. This method involves pouring off the top layer (oil), leaving the bottom layer (water) behind. Hence, option a is the appropriate choice for this scenario.

Key Concept: Real-world Examples

b) Saline water

[Solution Description] In the pharmaceutical industry, solutions where active ingredients are dissolved in a solvent form homogeneous mixtures essential for consistent medication delivery. An example of such a solution is saline water, where salt is uniformly dissolved in water, making it widely used for medical purposes.

Key Concept: Composition Analysis

d) Mixtures have variable compositions

[Solution Description] Mixtures have variable compositions, meaning their constituent elements or compounds can be present in varying ratios without any fixed proportion. This contrasts with compounds, which have a fixed composition determined by chemical bonds between elements.

Key Concept: Light Scattering

d) The solute particles are too small (less than 1 nm) to scatter light

[Solution Description] In a solution, the solute particles are very small, typically less than 1 nm in size. Because of this small size, these particles cannot scatter light, allowing the beam to pass through without scattering.

Key Concept: Comparative Analysis, Advanced Solubility

c) Solution B is supersaturated while Solution A is just saturated

[Solution Description] To find out which statement is correct, we first determine whether each solution is saturated or unsaturated.

For Solution A:

– The given condition states it is already saturated with 40 g solute in 100 g of water at 30°C.

For Solution B:

– Convert the amount of solvent to its equivalent if there were 100 g of water:

$$\text{Converting to equivalent for 100 g of water} = \frac{20 \, \text{g solute}}{80 \, \text{g water}} \times 100 = 25 \, \text{g solute per 100 g of water}$$

The solubility at 25°C is 18 g/100 g of water. Since 25 g is greater than 18 g, Solution B is supersaturated.

Key Concept: Real-World Application

d) Freezing juice to make popsicles

[Solution Description] Freezing juice to make popsicles involves a physical change where the liquid state changes to solid due to lowering temperature. This change doesn’t affect the chemical composition of the juice.

Key Concept: Concentration Calculation

b) 10%

[Solution Description]

The mass by mass percentage is calculated using the formula:

$\text{Mass by Mass %} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100$

Given, mass of solute = 15g and mass of solvent = 135g. So, mass of solution = 15g + 135g = 150g.

$\text{Mass by Mass %} = \left( \frac{15}{150} \right) \times 100 = 10\%$

Key Concept: Example Identification

d) Melting ice cream

[Solution Description] Melting ice cream is a physical change because it involves a change in the state of matter from solid to liquid without altering the chemical composition.

Key Concept: Non-Metal Identification

c) Iodine

[Solution Description] Iodine is a non-metal that sublimates easily and forms a violet vapour when it changes from solid to gas without passing through the liquid phase.

Key Concept: Comparison with Physical Change

a) Formation of a new substance

[Solution Description] A chemical change involves a change in the chemical properties and results in the formation of new substances, whereas a physical change does not alter the chemical composition of a substance.

Key Concept: Compound vs Mixture Analysis, Chemical Change Impact

c) The new substance has different properties from both sodium and chlorine as it forms a compound.

[Solution Description] When sodium and chlorine react chemically in the presence of electricity, they form sodium chloride through a chemical change. Sodium chloride is a compound with properties distinct from both sodium and chlorine. This chemical reaction results in a fixed composition and uniform properties throughout.

Key Concept: Complex Reaction Analysis, Real-world Application

c) The combustion of charcoal into ash, carbon dioxide, and heat

[Solution Description] During barbecuing, various processes occur. Lighting the charcoal initiates the combustion of carbon, a chemical process resulting in ash, carbon dioxide, and heat. This differs from simply placing food on the grill or heating the metal, which are physical processes without changing chemical composition.

Key Concept: Application of Formula

b) 10%

[Solution Description]

To find the volume by volume percentage, we use the formula:

$\text{Volume by volume percentage} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100$

Here, the volume of solute (glycerol) is 25 mL and the total volume of the solution is 25 + 225 = 250 mL. Plugging these values into the formula, we get:

$\text{Volume by volume percentage} = \frac{25}{250} \times 100 = 10%$

Therefore, the volume by volume percentage of glycerol is 10%.

Key Concept: Advanced Separation, Chemical vs Physical Separation

b) Magnet $\rightarrow$ Filtration $\rightarrow$ Evaporation

[Solution Description] The ideal process involves using a magnet to remove the iron filings first due to their magnetic properties. Next, add water to dissolve the salt, then filter out the sand. Finally, evaporate the water to recover the dissolved salt. This approach utilizes magnetic separation, filtration, and evaporation.

Key Concept: Conceptual Comparison, Advanced Reasoning

b) Heterogeneous mixtures have larger visible particles that can be separated by simple physical methods, while solutions do not.

[Solution Description]

In heterogeneous mixtures, the particles are large enough to be seen and separated by simple mechanical means, unlike solutions where the solute is uniformly dispersed at the molecular level and cannot be separated by ordinary filtration.

Key Concept: Problem Solving, Concept Integration

b) Filtration to remove sand, then evaporation for the salt-water solution.

[Solution Description] Begin by using filtration to separate the sand, which is insoluble, from the saltwater solution. Then, employ evaporation to remove the water, leaving behind pure salt. This approach leverages both the solubility properties of sand and salt as well as the volatility of water.

Key Concept: Chemical Reaction Context, Elemental Behavior

a) Sulfur

[Solution Description] Non-metals tend to form oxides with acidic properties when they react with oxygen. Among the given options, sulfur readily forms sulfur dioxide and sulfur trioxide, which are known for being acidic oxides. Therefore, sulfur is the element most likely to combine with oxygen to form such compounds.

Key Concept: Advanced Application, Concept Integration

c) Colloid

[Solution Description] The Tyndall effect is a key characteristic of colloids due to their particle size, which scatters light. Since the particles do not settle upon standing and cannot be separated by simple filtration, it fits the definition of a colloidal solution. Therefore, the mixture is most accurately described as a colloid.

Key Concept: Element vs Compound

a) Nitrogen

[Solution Description] An element is a pure substance that cannot be broken down into simpler substances using chemical reactions. Among the given options, nitrogen is an element as it exists in nature as discrete atoms of the same type, unlike water which is a compound made up of hydrogen and oxygen.

Key Concept: Advanced Application, Chemical Reactions

b) $2K + 2HCl \rightarrow 2KCl + H_2$

[Solution Description] When potassium reacts with hydrochloric acid, hydrogen gas is released, and potassium chloride ($KCl$) is formed. The balanced chemical equation for this reaction is:

$2K + 2HCl \rightarrow 2KCl + H_2$

Key Concept: Real-world Application, Concentration Calculation

b) 30 g

[Solution Description] To find the mass of salt required, we use the formula for mass by mass percentage:

$\text{Mass by mass percentage} = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 100$

Given that the mass by mass percentage is 15% and the total mass of the solution is 200 g, we solve for the mass of salt (solute):

$15 = \left(\frac{\text{mass of salt}}{200}\right) \times 100$

Solving for the mass of salt,

$\text{mass of salt} = \frac{15 \times 200}{100} = 30 \, \text{g}$

Thus, the technician needs to add 30 g of salt.

Key Concept: Conceptual Application, Stability Analysis

d) Stirring provides kinetic energy to counteract sedimentation.

[Solution Description]

Continuous stirring prevents the clay particles from settling down due to gravity. The energy from stirring keeps the particles suspended by constantly agitating them and preventing aggregate formation. This process reduces the effect of gravitational force that normally causes the particles to settle.

Key Concept: Concentration Calculation, Solution Stability Analysis

a) 10%, due to strong solvent-solute interactions

[Solution Description] To calculate the volume/volume percentage:

Total volume of solution = 10 mL + 90 mL = 100 mL

Volume/volume percentage of iodine:

$\left( \frac{10 \text{ mL}}{100 \text{ mL}} \right) \times 100% = 10%$

Iodine remains stable because it forms a homogeneous solution with ethanol due to strong interactions between iodine molecules and ethanol, preventing separation or sedimentation.

Key Concept: State Change

b) Physical change

[Solution Description] Freezing occurs when water transitions from a liquid state to a solid state without altering its chemical composition. This process is an example of a physical change because it only affects the state of matter and not the chemical structure.

Key Concept: Compound Properties

d) Its properties differ from those of its constituents.

[Solution Description]

A compound consists of two or more elements chemically combined in a fixed proportion, giving it properties distinct from its constituent elements.

Key Concept: Mixture Complexity, Chemical Composition Analysis

c) It is a homogeneous mixture because sugar remains dissolved until water evaporates.

[Solution Description] A sugar-water solution is an example of a homogeneous mixture or solution. The fact that sugar can be recovered by evaporation when water leaves as vapor indicates it was a physical mixture. The sugar retains its properties after separation, demonstrating the non-chemical combination characteristic of mixtures rather than pure substances.

Key Concept: Critical Thinking, Mixture Classification

c) Milk

[Solution Description] The Tyndall effect occurs when light scatters in a colloid due to the size of its dispersed particles. Among the options, milk is known to scatter light because it’s a colloidal suspension where the particles are large enough to scatter light but not settle, confirming the presence of the Tyndall effect.

Key Concept: Advanced Application, Synthesis and Evaluation

d) Silicon

[Solution Description] Metalloids are known for their intermediate properties between metals and non-metals. For applications requiring thermal stability at high temperatures, it is crucial to select a metalloid that does not lose its structure or properties when exposed to heat. Among the commonly known metalloids, silicon has excellent thermal stability and is used extensively in environments with elevated temperatures. Silicon’s ability to maintain structural integrity and electrical performance under such conditions makes it ideal for creating temperature sensors.

Key Concept: Historical Context

d) Antoine Lavoisier

[Solution Description] The syllabus specifies contributions from Robert Boyle and Antoine Lavoisier regarding the concept of elements. Antoine Lavoisier is particularly noted for defining elements based on experimental evidence.

Key Concept: Compound Properties

c) It involves chemical or electrochemical reactions

[Solution Description] Compounds can only be separated into their individual elements through chemical reactions or electrochemical methods. Unlike mixtures, compounds cannot be separated by physical means because their composition is uniform throughout.

Key Concept: Element vs Compound

c) Gold

[Solution Description] An element is a pure substance consisting of only one type of atom, and it cannot be broken down into simpler substances by chemical reactions. Gold is an element, represented by the symbol Au, and consists solely of gold atoms.

Key Concept: Complex Classification, Advanced Tyndall Effect

c) A gel such as jelly

[Solution Description] The Tyndall effect is best observed in systems where there is a significant difference in the refractive indices of the dispersed phase and the dispersing medium, which enhances the scattering of light. Gels typically have larger particle sizes than aerosols or foams, making them more suitable for exhibiting the Tyndall effect. In emulsions, the particles are generally too small or similar to the surrounding medium to scatter light effectively.