Chapter 9: The Amazing World of Solutes, Solvents, and Solutions

The picture at the beginning of the chapter shows a person stirring a substance (like sugar or salt) into a glass of water. As stirring continues, the solid particles gradually seem to disappear — they cannot be seen by our eyes any more, but the liquid tastes sweet or salty.

This is the process of dissolution. The solid (the solute) breaks into extremely tiny particles and spreads uniformly throughout the water (the solvent). The uniform mixture that is formed is called a solution. Even though we cannot see the solute particles, they are still present in the liquid.

Every liquid can dissolve only a limited amount of a solid at a given temperature. When we keep adding sugar to a cup of tea, a stage comes when the tea cannot hold any more sugar; the extra sugar simply settles at the bottom of the cup and does not dissolve. At this stage the tea has become a saturated solution of sugar.

To make the extra sugar dissolve, we can do one of the following:

• Heat the tea. The solubility of sugar in water increases with temperature, so hot tea can hold more sugar than cold tea.
• Add more liquid (more tea or hot water). A larger amount of solvent can dissolve more solute.
• Stir the tea. Stirring speeds up dissolution, though it does not increase the total amount that can dissolve.

Whether a solute dissolves in a particular solvent depends on the nature of both. Sugar and common salt ($$\mathrm{NaCl}$$) are made of particles that are strongly attracted to water particles. Water is able to surround these particles and pull them apart from one another, so the solid breaks up and spreads uniformly through the water.

Oil, on the other hand, is made of particles that do not attract sugar or salt particles strongly. So the water-loving solids (sugar and salt) cannot be pulled apart by oil and remain undissolved.

Water is called a universal solvent or a good solvent because it can dissolve a very large variety of substances — many solids (sugar, salt, copper sulphate), many liquids (vinegar, alcohol), and many gases (oxygen, carbon dioxide). This is why water plays such an important role in living cells, cooking, and industries.

A tall, cylindrical bottle has several practical advantages over a spherical one:

• Easy to hold and carry. The straight sides of a cylinder fit comfortably in the hand, while a sphere would keep slipping out.
• Stable and does not roll. A cylinder placed upright stays where it is kept; a sphere would roll off any surface that is not perfectly flat.
• Fits neatly in bags, shelves and refrigerators. Cylinders can be lined up side by side and stacked, using space more efficiently than spheres, which leave large empty gaps between them.
• Easy to pour and to measure water. A tall, narrow container allows even a small change in the amount of water to be seen clearly as a change in level. In a spherical container, a small change in volume gives an almost invisible change in level near the widest part.
• Easy to manufacture and label. Cylindrical bottles are simpler to mould and their flat side surface is convenient to print or paste a label on.

Fig. 9.1 shows sugar being stirred into water. After enough stirring the sugar completely disappears from view and the composition of the liquid becomes the same everywhere — any spoonful taken from the top, middle or bottom tastes equally sweet. Such a mixture is a uniform (or homogeneous) mixture. So the sugar–water mixture is a uniform mixture, i.e. a solution.

When chalk powder is stirred into water, the tiny chalk particles do not dissolve. They remain suspended for some time, giving the water a milky look, and then slowly settle at the bottom. Different parts of this liquid do not have the same composition — the bottom is thick with chalk while the top is nearly clear. Therefore chalk powder in water is a non-uniform (heterogeneous) mixture and is not a solution.

A solution is any uniform (homogeneous) mixture — the same composition is found everywhere in the mixture. The definition does not restrict solutions to only solids dissolved in liquids.

Air consists mainly of nitrogen (about $$78\%$$), oxygen (about $$21\%$$) and small amounts of carbon dioxide, water vapour and other gases. These gases mix so thoroughly that any sample of clean air, taken from any place, has practically the same composition. So air is a uniform mixture of gases and is therefore a solution — a gaseous solution in which nitrogen acts like the solvent and the other gases are the solutes.

Initially each spoon of salt added to the water dissolves completely on stirring and disappears from view. As we keep adding more and more salt, the water dissolves less and less of each new spoonful. Finally a stage comes when no more salt will dissolve at that temperature — the extra salt simply collects as undissolved crystals at the bottom of the vessel.

At this stage the water has taken the maximum amount of salt it can hold at that temperature. Such a solution is called a saturated solution. If we want the leftover salt to dissolve, we must either heat the water (to raise its solubility) or add more water.

This is an activity-based question — the number depends on the actual size of the spoon and the amount of water used. For the setup described in the textbook (about $$100 \, \mathrm{mL}$$ of water at room temperature, roughly $$25\text{–}30^{\circ}\mathrm{C}$$, stirred with a small teaspoon), students usually observe that about $$6\text{–}7$$ level teaspoons of common salt dissolve; from the $$7^{\text{th}}$$ or $$8^{\text{th}}$$ spoon onwards the extra salt starts settling at the bottom.

This is consistent with the known solubility of salt in water at room temperature: about $$36 \, \mathrm{g}$$ of salt dissolves in $$100 \, \mathrm{g}$$ of water at $$25^{\circ}\mathrm{C}$$, which corresponds to roughly $$6\text{–}7$$ level teaspoons.

Since only a limited number of spoons of salt could be dissolved and the extra salt did not go into the solution, we learn that water cannot dissolve an unlimited amount of a solute at a given temperature. Every solvent can hold only a fixed maximum quantity of a particular solute at a particular temperature. Once this limit is reached the solution is called saturated and cannot accept any more solute.

The maximum amount of a solute that dissolves in a fixed amount of solvent at a particular temperature is called the solubility of that solute in that solvent at that temperature.

The concentration of a solution tells us how much solute is dissolved per unit amount of solvent. To compare the two solutions we work out the amount of salt in the same volume of water — say $$100 \, \mathrm{mL}$$.

Solution A: $$2$$ spoons of salt in $$100 \, \mathrm{mL}$$ of water $$\;\Rightarrow\;$$ $$2$$ spoons per $$100 \, \mathrm{mL}$$.

Solution B: $$4$$ spoons of salt in $$50 \, \mathrm{mL}$$ of water. If we had $$100 \, \mathrm{mL}$$ of water (double the amount), it would need double the salt to keep the same concentration, i.e. $$8$$ spoons per $$100 \, \mathrm{mL}$$.

So comparing per $$100 \, \mathrm{mL}$$ of water:

Solution B has $$4$$ times as much salt in the same volume of water, so Solution B ($$4$$ spoons of salt in $$50 \, \mathrm{mL}$$ of water) is much more concentrated.

Yes, temperature has a strong effect on how much of a given solute a solvent can dissolve.

• For most solids (sugar, salt, baking soda, copper sulphate, potassium nitrate, etc.) the solubility increases when the solvent is heated. Hot water can hold much more sugar than cold water. This is why we add sugar to hot milk more easily, and why crystals of copper sulphate come out of a solution as it cools.
• For gases dissolved in liquids the trend is the opposite — solubility decreases when the liquid is heated. That is why boiling water shows bubbles of dissolved air escaping.

So, temperature definitely affects solubility, though the direction of the effect depends on the state of the solute.

At $$20^{\circ}\mathrm{C}$$ the water in Activity 9.2 had already dissolved as much baking soda ($$\mathrm{NaHCO_3}$$) as it could hold, so the extra baking soda lay at the bottom of the beaker as an undissolved layer — the solution was saturated.

When this beaker is heated to $$50^{\circ}\mathrm{C}$$, the solubility of baking soda in water increases. Water at $$50^{\circ}\mathrm{C}$$ can now hold more baking soda than it could at $$20^{\circ}\mathrm{C}$$. As the water warms up, the undissolved baking soda at the bottom starts dissolving. On stirring at $$50^{\circ}\mathrm{C}$$, the whole of the earlier undissolved baking soda goes into solution and the bottom of the beaker becomes clear again.

Activity 9.2 compares the amount of baking soda that dissolves in the same amount of water at two temperatures — $$20^{\circ}\mathrm{C}$$ and $$50^{\circ}\mathrm{C}$$. At $$20^{\circ}\mathrm{C}$$ some baking soda remains undissolved even after stirring, but at $$50^{\circ}\mathrm{C}$$ all of it goes into solution.

We can draw two important inferences:

1. At any given temperature, water has a fixed (limited) capacity to dissolve baking soda; once that limit is reached, no more baking soda dissolves.
2. The solubility of baking soda (and, in general, of most solid solutes) in water increases as the temperature is raised. So hot water can dissolve more solute than cold water.

Asima Chatterjee (1917–2006) was a great Indian chemist who spent her life studying the chemistry of Indian medicinal plants. Her inspiration came from two sources:

• Her father, Dr. Indranarayan Mukherjee, was a medical doctor with a deep interest in traditional Indian herbal medicine. From childhood she watched him collect and study plants used in Ayurveda, and this sparked her curiosity about the healing power of plants.
• She also saw that India has an enormous variety of medicinal plants and a long Ayurvedic tradition, but very little scientific work had been done to isolate the actual chemical substances in those plants that make them useful as medicines. She wanted to bring the tools of modern chemistry to this rich but under-explored knowledge.

These two influences led her to develop new drugs from Indian plants — most famously anti-malarial and anti-epileptic medicines derived from plants such as Alstonia scholaris and Rauwolfia serpentina.

Yes. Water can dissolve gases just as it dissolves solids and liquids. Some familiar examples are:

• Oxygen dissolved in water: Fish, prawns and other aquatic animals breathe by taking in the oxygen dissolved in pond, river and sea water.
• Carbon dioxide in soft drinks: The fizz in cola, lemonade and soda water is $$\mathrm{CO_2}$$ that has been dissolved in the water under pressure. When we open the bottle the pressure falls and bubbles of $$\mathrm{CO_2}$$ escape.
• When freshly drawn tap water is kept in a glass, tiny bubbles collect on the inside of the glass — these are gases (mostly air) that were dissolved in the water and slowly come out.

When a gas such as oxygen or carbon dioxide dissolves in water, its tiny particles spread evenly throughout the water. Every drop of water taken from the top, middle or bottom of the container has the same amount of dissolved gas in it, and the water still looks perfectly clear — we cannot see the gas separately.

Because the composition is the same in every part of the mixture, the gas-in-water mixture is a uniform (homogeneous) mixture. It is therefore a solution, in which water is the solvent and the gas is the solute.

Yes. Temperature affects the solubility of gases too, but the effect is opposite to what we see with most solid solutes.

The solubility of a gas in a liquid decreases when the temperature of the liquid is raised. As we heat the liquid, the dissolved gas particles gain energy, move about faster and escape out of the liquid as bubbles. This is easy to observe:

• When water is heated on a stove, tiny bubbles of dissolved air begin to appear on the walls of the vessel much before the water actually boils.
• A cold bottle of a soft drink retains its fizz for a longer time; a warm bottle loses its fizz quickly because the dissolved $$\mathrm{CO_2}$$ escapes faster.
• Fish are more comfortable in cool water; warm pond or lake water holds less dissolved oxygen, which is one reason fish struggle in polluted, warm water.

Whether an insoluble substance floats or sinks in water depends on how heavy the substance is for its size — that is, on its density. Density is defined as

The density of water is about $$1 \, \mathrm{g/cm^3}$$. When a substance is placed in water:

• If its density is less than that of water, it floats.
• If its density is greater than that of water, it sinks.

Sawdust is made of wood, and wood has a lot of tiny air-filled cells inside it. So a piece of sawdust is very light for its size — its density (about $$0.4\text{–}0.7 \, \mathrm{g/cm^3}$$) is less than $$1 \, \mathrm{g/cm^3}$$. Therefore sawdust floats on water.

Sand, on the other hand, is made of small hard grains of rock (mostly silica). It is much heavier for its size; its density (about $$2.6 \, \mathrm{g/cm^3}$$) is greater than that of water. So sand sinks.

Scientists define the density of a substance as the mass of a unit volume of that substance. In other words,

Density tells us how much matter is packed into a given amount of space. Two objects of the same size can have very different densities: a small piece of iron is much heavier than a piece of wood of the same volume, so iron has a much higher density than wood.

The SI unit of density is kilogram per cubic metre ($$\mathrm{kg/m^3}$$); in the laboratory it is often measured in grams per cubic centimetre ($$\mathrm{g/cm^3}$$). For example, the density of pure water at $$4^{\circ}\mathrm{C}$$ is $$1 \, \mathrm{g/cm^3} \;(= 1000 \, \mathrm{kg/m^3})$$.

Density is mass per unit volume:

Density of the oil. The packet contains $$V = 1 \, \mathrm{L} = 1000 \, \mathrm{mL} = 1000 \, \mathrm{cm^3}$$ of oil weighing $$m = 910 \, \mathrm{g}$$. So

Density of water. $$1 \, \mathrm{L}$$ of water weighs about $$1000 \, \mathrm{g}$$, so

Since $$0.91 \, \mathrm{g/cm^3} < 1.00 \, \mathrm{g/cm^3}$$, the oil is less dense than water. This is why, when we pour cooking oil into a glass of water, the oil rises and floats on top of the water instead of mixing with it or sinking.

The maximum volume that a measuring cylinder can measure is simply the highest volume mark on its scale — the total capacity of the cylinder.

In Fig. 9.16 the scale runs from $$10 \, \mathrm{mL}$$ at the bottom up to $$100 \, \mathrm{mL}$$ at the very top, and the label on the cylinder is $$100 \, \mathrm{mL}$$. So the maximum volume this cylinder can measure is $$100 \, \mathrm{mL}$$.

The smallest volume that a measuring cylinder can read is the volume represented by one small division on its scale — this is its least count.

Between two big marks on the cylinder in Fig. 9.16 (say $$10 \, \mathrm{mL}$$ and $$20 \, \mathrm{mL}$$) there are $$10$$ equal small divisions. The gap between the two big marks corresponds to $$10 \, \mathrm{mL}$$, so each small division represents

Therefore the smallest volume this measuring cylinder can measure is $$1 \, \mathrm{mL}$$.

The bigger (numbered) marks on the cylinder in Fig. 9.16 are labelled $$10, 20, 30, \ldots, 100 \, \mathrm{mL}$$. Two consecutive bigger marks differ by

The same $$10 \, \mathrm{mL}$$ gap is seen between any two neighbouring bigger marks (for example, between $$40 \, \mathrm{mL}$$ and $$50 \, \mathrm{mL}$$).

If we look carefully at the space between any two neighbouring bigger marks on the measuring cylinder in Fig. 9.16 (for example, between $$10 \, \mathrm{mL}$$ and $$20 \, \mathrm{mL}$$), we can count $$10$$ equal small divisions.

The value of one small division is obtained by dividing the volume difference between two bigger marks by the number of small divisions between them.

From the previous two questions:

• Volume difference between two bigger marks $$= 10 \, \mathrm{mL}$$.
• Number of small divisions between them $$= 10$$.

Therefore

So each small division on this measuring cylinder represents a volume of $$1 \, \mathrm{mL}$$.

Measuring cylinders are designed narrow and tall so that even a small change in the amount of liquid produces a clearly visible change in the height of the liquid column. This lets us measure the volume accurately.

Suppose $$1 \, \mathrm{mL}$$ (i.e. $$1 \, \mathrm{cm^3}$$) of water is added to two containers with the same cross-sectional area formula, $$\text{Volume} = \text{cross-sectional area} \times \text{height}$$, so

• In a narrow tall measuring cylinder the cross-sectional area is small, so the rise in level for the same $$1 \, \mathrm{mL}$$ is large — easy to see and read off precisely.
• In a wide short beaker the cross-sectional area is large, so the rise in level for $$1 \, \mathrm{mL}$$ is very small — hard to see and hard to read accurately.

A narrow tall shape therefore stretches the same volume across a long scale, letting us mark small divisions (e.g. $$1 \, \mathrm{mL}$$) that are far enough apart to be read without confusion.

When a liquid is put in a narrow measuring cylinder, its upper surface is not flat — it curves near the walls of the cylinder because of the attraction between the liquid and the glass. This curved surface is called the meniscus.

For a colourless liquid such as water, the meniscus is concave (dips downwards in the middle). We read the mark that coincides with the bottom of the meniscus, with our eye at the same level as the meniscus.

For a coloured liquid the bottom of the meniscus is hidden under the dark colour of the liquid and cannot be seen clearly. Therefore, in the case of coloured liquids we read the mark on the cylinder that coincides with the top of the meniscus (the upper curved surface of the liquid). While doing so, the eye must again be kept exactly at the level of that surface to avoid parallax errors.

Whether an object floats or sinks in a liquid depends on the comparison of their densities:

• If the object is less dense than the liquid, it floats.
• If the object is more dense than the liquid, it sinks.

The egg sinks because the density of a raw whole egg (about $$1.03 \, \mathrm{g/cm^3}$$) is slightly greater than the density of tap water (about $$1.00 \, \mathrm{g/cm^3}$$). To make the egg float we must change the density of the water so that it becomes greater than the density of the egg. The easiest way to do this is to dissolve a large amount of common salt (or sugar) in the water and stir well.

The dissolved salt adds mass to the water without appreciably changing its volume, so the density of the water rises well above $$1 \, \mathrm{g/cm^3}$$. Once the salt solution is denser than the egg, the egg rises up and floats. (This is the same reason why it is much easier to float in sea water, or in the very salty Dead Sea, than in ordinary pond or river water.)

An object (or a liquid) floats on another liquid only if it is less dense than the liquid below.

When oil is poured into a glass containing water, it stays as a separate layer on top of the water. So the oil must be less dense than the water — equivalently, water must be denser than oil.

Also, oil and water do not mix; they form two separate layers. So (iv) is wrong. Their densities are clearly different, so (iii) is wrong. Since oil floats, (i) is wrong.

The correct choice is (ii) Water is denser than oil.

Given. Mass $$m = 225 \, \mathrm{g}$$, volume $$V = 90 \, \mathrm{cm^3}$$.

Formula. Density is mass per unit volume:

Calculation.

Comparison with water. The density of water is $$1 \, \mathrm{g/cm^3}$$. Since

the stone sculpture is denser than water, so it will sink in water.

Recall the definitions:

• Saturated solution — the solvent has dissolved the maximum possible amount of solute at that temperature; no more solute will dissolve at that temperature.
• Unsaturated solution — the solvent has dissolved less than the maximum possible amount of solute; more solute can still dissolve at that temperature.

Now let us check each option:

• (i) Wrong. A saturated solution has already reached the limit and cannot dissolve any more solute at the same temperature.
• (ii) Wrong. It is the saturated (not the unsaturated) solution that has the maximum amount of solute dissolved.
• (iii) Correct. This is exactly the definition of a saturated solution.
• (iv) Wrong. A solution can be saturated at any temperature — low, room, or high. What changes with temperature is the amount needed to saturate it. A saturated solution can also be prepared at, say, room temperature.

Hence the most appropriate statement is (iii).

Given. Total capacity of the bottle $$V_{\text{total}} = 2 \, \mathrm{L}$$ and water already poured $$V_{\text{poured}} = 500 \, \mathrm{mL}$$.

Convert to the same unit. Since $$1 \, \mathrm{L} = 1000 \, \mathrm{mL}$$,

Extra water the bottle can still hold.

That is, the bottle can hold $$1500 \, \mathrm{mL}$$ (i.e. $$1.5 \, \mathrm{L}$$) more water.

Given. Mass $$m = 400 \, \mathrm{g}$$, volume $$V = 40 \, \mathrm{cm^3}$$.

Formula.

Calculation.

So the density of the object is $$10 \, \mathrm{g/cm^3}$$. (Since this is much greater than the density of water, $$1 \, \mathrm{g/cm^3}$$, the object would sink in water.)

Whether an object floats or sinks depends on how its density compares with that of water. Water has a density of about $$1 \, \mathrm{g/cm^3}$$.

Unpeeled orange (Fig. 9.25a) — floats. The peel (rind) of an orange is spongy and full of tiny air-filled pockets. This air makes the peel very light for its size. When the orange is unpeeled, the trapped air greatly increases the total volume of the orange without adding much to its mass, so the overall density of the whole orange (fruit + peel + trapped air) becomes less than the density of water. Being less dense than water, the unpeeled orange floats.

Peeled orange (Fig. 9.25b) — sinks. When the peel is removed, the trapped air is also removed along with it. The remaining fruit pulp is heavier for its size — its density is a little greater than that of water. So the peeled orange sinks.

Thus the difference in behaviour is entirely explained by the air trapped in the peel, which lowers the density of the whole fruit when the peel is on.

To compare denseness, compute the density of each object using $$d = m/V$$.

Object A.

Object B.

Comparison. $$d_A = 5 \, \mathrm{g/cm^3} > d_B = 4 \, \mathrm{g/cm^3}$$.

So Object A is denser than Object B, even though Object B is heavier — it has enough extra volume to lower its density.

Density depends only on the mass and volume of the material:

Density of the compact cube. $$m = 120 \, \mathrm{g}$$, $$V = 60 \, \mathrm{cm^3}$$, so

What changes when the clay is flattened? Flattening the clay only changes its shape. The amount of clay (its mass) does not change — nothing is added or removed. Since the clay is not compressed or stretched into a different material either, the total volume it occupies also stays the same ($$60 \, \mathrm{cm^3}$$).

Because both mass and volume remain unchanged, the density remains the same:

Thus the density of the modelling clay does not change on flattening. Density is a property of the material itself, not of its shape.

Given. Mass $$m = 600 \, \mathrm{g}$$, density $$d = 7.9 \, \mathrm{g/cm^3}$$.

Formula. Since $$d = m/V$$, rearranging for volume gives

Calculation.

Rounded to a suitable number of significant figures, the volume of the iron block is about $$76 \, \mathrm{cm^3}$$ (more precisely $$75.95 \, \mathrm{cm^3}$$).

In Fig. 9.26 a test tube is completely filled with water and closed with a stopper through which a narrow glass tube passes. Some water stands in the glass tube at a certain level. When this test tube is placed in a beaker of hot water at about $$70^{\circ}\mathrm{C}$$, the water inside the test tube gets heated.

What happens to the water? On heating, the particles of water move about faster and spread apart. So the water inside the test tube expands. Since the test tube is closed, the extra volume has nowhere to go except up the narrow glass tube. That is why the level of water in the glass tube rises.

Effect on density. The amount of water in the closed system does not change on heating — its mass $$m$$ stays the same. But its volume $$V$$ has now increased. Applying

if $$m$$ is constant and $$V$$ increases, then $$d$$ must decrease.

So heating the water causes it to expand, and the density of the water decreases. This experiment shows, in general, that the density of a liquid decreases when its temperature is raised.