The Amazing World of Solutes, Solvents, and Solutions

(i) False [F]
Correction: Oxygen gas is *less* soluble in hot water than in cold water. Generally, the solubility of gases *decreases* with an increase in temperature.

(ii) False [F]
Correction: A mixture of sand and water is *not* a solution. It is a heterogeneous (non-uniform) mixture because sand does not dissolve in water and the particles are visible and settle down.

(iii) False [F]
Correction: The amount of space occupied by any object is called its *volume*, not mass. Mass is the amount of matter present in an object.

(iv) False [F]
Correction: A *saturated* solution has the maximum amount of solute dissolved at a given temperature. An unsaturated solution has *less* solute dissolved than a saturated solution (i.e., more solute can still be dissolved in it).

(v) True [T]
The atmosphere is a uniform mixture of different gases such as nitrogen, oxygen, carbon dioxide, argon, etc., so it qualifies as a solution (gaseous solution).

(i) The volume of a solid can be measured by the method of displacement, where the solid is submerged (immersed) in water and the rise in water level is measured.

(ii) The maximum amount of solute dissolved in a fixed quantity of solvent (or solution) at a particular temperature is called solubility at that temperature.

(iii) Generally, the density decreases with increase in temperature.

(iv) The solution in which glucose has completely dissolved in water, and no more glucose can dissolve at a given temperature, is called a saturated solution of glucose.

Correct Answer: (ii) Water is denser than oil

Justification: When two immiscible liquids are mixed, the denser liquid sinks to the bottom and the less dense liquid floats on top. Since oil floats on water, it means oil is less dense than water, i.e., water is denser than oil. The density of water is approximately $1\ \text{g/cm}^3$ while the density of oil is approximately $0.8{-}0.9\ \text{g/cm}^3$.

Given:
- Mass of stone sculpture $= 225\ \text{g}$
- Volume of stone sculpture $= 90\ \text{cm}^3$

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

Calculation:
$$\text{Density} = \frac{225\ \text{g}}{90\ \text{cm}^3} = 2.5\ \text{g/cm}^3$$

Prediction:
The density of water is $1\ \text{g/cm}^3$. Since the density of the stone sculpture ($2.5\ \text{g/cm}^3$) is greater than the density of water ($1\ \text{g/cm}^3$), the stone sculpture will sink in water.

Answer: Density $= 2.5\ \text{g/cm}^3$; the sculpture will sink in water.

Most appropriate statement: (iii) No more solute can be dissolved into the saturated solution at that temperature.

Explanation:
A saturated solution is defined as the solution in which the maximum amount of solute has already been dissolved at a given temperature, and no more solute can be dissolved in it at that temperature. Statement (iii) correctly captures this definition.

Why the other statements are not appropriate:

- (i) is incorrect: A saturated solution has already reached its maximum capacity of dissolving solute. It *cannot* dissolve any more solute at the same temperature. This is the opposite of what a saturated solution means.

- (ii) is incorrect: It is the *saturated* solution (not the unsaturated solution) that has dissolved the maximum amount of solute possible at a given temperature. An unsaturated solution still has the capacity to dissolve more solute.

- (iv) is incorrect: A saturated solution can form at *any* temperature, not just at high temperatures. The amount of solute that saturates a solution may differ with temperature, but saturation is not restricted to high temperatures only.

Given:
- Total volume of the bottle $= 2\ \text{litres} = 2000\ \text{mL}$
- Volume of water already poured $= 500\ \text{mL}$

Calculation:
$$\text{Remaining capacity} = \text{Total volume} - \text{Volume already filled}$$
$$\text{Remaining capacity} = 2000\ \text{mL} - 500\ \text{mL} = 1500\ \text{mL}$$

Answer: The bottle can hold 1500 mL (or 1.5 litres) more water.

Given:
- Mass of the object $= 400\ \text{g}$
- Volume of the object $= 40\ \text{cm}^3$

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

Calculation:
$$\text{Density} = \frac{400\ \text{g}}{40\ \text{cm}^3} = 10\ \text{g/cm}^3$$

Answer: The density of the object is $\mathbf{10\ g/cm^3}$.

*(Note: The figures show an unpeeled orange floating in water and a peeled orange sinking in water.)*

Explanation:

An object floats in water if its density is less than the density of water ($1\ \text{g/cm}^3$), and sinks if its density is greater than $1\ \text{g/cm}^3$.

- Unpeeled orange (floats): The peel (skin) of an orange is thick and spongy. It contains a large number of tiny air pockets (pores). These air pockets increase the overall volume of the orange significantly without adding much mass. As a result, the average density of the unpeeled orange (orange flesh + peel with air pockets) becomes less than the density of water, so it floats.

- Peeled orange (sinks): When the peel is removed, the air pockets are gone. The remaining orange flesh is compact and denser. The average density of the peeled orange becomes greater than the density of water, so it sinks.

Conclusion: The peel with its air pockets lowers the overall density of the orange below that of water, causing it to float. Removing the peel removes the air pockets, increasing the average density above that of water, causing it to sink.

Given:
- Object A: Mass $= 200\ \text{g}$, Volume $= 40\ \text{cm}^3$
- Object B: Mass $= 240\ \text{g}$, Volume $= 60\ \text{cm}^3$

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

Density of Object A:
$$\text{Density}_A = \frac{200\ \text{g}}{40\ \text{cm}^3} = 5\ \text{g/cm}^3$$

Density of Object B:
$$\text{Density}_B = \frac{240\ \text{g}}{60\ \text{cm}^3} = 4\ \text{g/cm}^3$$

Comparison:
\text{Density}_A = 5\ \text{g/cm}^3 > \text{Density}_B = 4\ \text{g/cm}^3

Answer: Object A is denser than Object B.

Given:
- Mass of modeling clay $= 120\ \text{g}$ (remains constant throughout)
- Volume when moulded into a cube $= 60\ \text{cm}^3$

Density of the compact cube:
$$\text{Density} = \frac{120\ \text{g}}{60\ \text{cm}^3} = 2\ \text{g/cm}^3$$

When flattened into a thin sheet:
- The mass of the clay does not change — it is still $120\ \text{g}$.
- The volume of the clay also does not change — clay is a solid and reshaping it does not compress or expand the material itself, so the volume remains $60\ \text{cm}^3$.

Therefore:
$$\text{Density (thin sheet)} = \frac{120\ \text{g}}{60\ \text{cm}^3} = 2\ \text{g/cm}^3$$

Prediction: The density of the modeling clay will remain the same ($2\ \text{g/cm}^3$) even after flattening, because both mass and volume remain unchanged when the shape is changed. Density is an intrinsic property of a substance and does not depend on its shape.

Given:
- Mass of iron block $= 600\ \text{g}$
- Density of iron $= 7.9\ \text{g/cm}^3$

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

Rearranging for Volume:
$$\text{Volume} = \frac{\text{Mass}}{\text{Density}}$$

Calculation:
$$\text{Volume} = \frac{600\ \text{g}}{7.9\ \text{g/cm}^3} \approx 75.95\ \text{cm}^3$$

Answer: The volume of the iron block is approximately $\mathbf{75.95\ cm^3}$ (or $\approx 76\ \text{cm}^3$).

*(Note: The figures show a test tube with a glass tube inserted; when placed in hot water, the water level in the glass tube rises, indicating that the water inside the test tube expands.)*

Observation: When the test tube is placed in hot water (~70 °C), the water inside the test tube gets heated and its level rises in the glass tube. This indicates that the volume of water increases (water expands) on heating.

Effect on Density:

Using the formula:
$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

- On heating, the mass of the water remains the same (no water is added or removed).
- However, the volume increases because water expands on heating.
- Since density $= \dfrac{\text{Mass}}{\text{Volume}}$, and the volume increases while mass stays constant, the density decreases.

Conclusion: When water is heated, its volume increases (as seen by the rising water level in the glass tube), while its mass remains unchanged. Therefore, the density of water decreases with an increase in temperature. This is consistent with the general principle that density decreases with an increase in temperature.