Subject: Science
A mixture is defined as a mass obtained by mixing up two or more chemically non-reactive substances in any proportion by weight. There are two types of mixture. They are as follows:
A homogeneous mixture is defined as a mass in which the particles of the components of the mixture are equally distributed. Examples: Sugar solution, air, alcohol water, brass, soda water etc
A heterogeneous mixture is defined as a mass in which the particles of components of the mixture are not equally distributed. Examples: Smoke, Oily water, Sandy water, etc.
On the basis of the size the particles are divided into following types:
The solution is defined as the homogeneous mixture of two or more than two substances that can be formed either by mixing suitable solid in the liquid or by the liquid in liquid.
Solution = solute + solvent
Solute: It is the component of a solution, which gets dissolved into other substance to form a solution.
Solvent: It is the substance, which dissolves solute to form a solution.
Water: It is a universal solvent that has the capacity of dissolving many substances like common salts, copper sulphate, sugar, etc.
Alcohol: It found in the form of spirit and can dissolve resin and iodine.
Petrol and kerosene: It dissolves ghee, grease, oil, fat, etc and is used for removing clothes stain.
Ether: It is an organic solvent that can dissolve fat, oil, resin, etc.
Colloid is a homogeneous mixture in which the diameter of particles of components ranges in between that of particles of a solution and suspension, i.e. a particle’s size ranges from 10-7 cm to 10-5 cm. The colloidal solution of liquid in liquid is called emulsion. Some of the examples of colloids are blood, milk, gum, wax, etc.
The suspension is a heterogeneous mixture in which the diameter of the particle is 10-5 cm or larger. Their particles are visible under a simple microscopic as well as to the naked eyes. Some of the examples of suspensions are sand water, muddy water, smoke in air, etc.
Solute is defined as the substance which gets dissolved and is present in smaller proportion is called the solute. Example: In sugar solution, sugar is solute.
At 20\(^0\)C, 4.1 gram of sugar dissolves in 2.5 gram of water to form a saturated sugar solution. Find the solubility of sugar.
Here, Weight of solute(W1) = 4.1gm.
Weight of solvent(W2) = 2.5 gm.
Solubility at 20\(^0\)C(s) = ?
We know that,
Solubility (S) = \(\frac{Weight\;of\;solute (W1) }{Weight\;of\;solvent (W2)}\) \(\times\) 100
= \(\frac{4.1}{2.5}\) \(\times\) 100 = 164
Therefore , the solubility of sugar at 20\(^0\)C is 164 .
The solubility of common salt at 30\(^0\)C is 100. What amount of water is required to prepare a saturated solution of 1kg of salt ?
Here,
Weight of solute (W1) = 1KG = 1000gm
Solute of salt at 30\(^0\)C (S) = 100
Weight of solvent (W2) = ?
We know that,
Solubility(s) = \(\frac{Weight\;of\;solute}{Weight\;of\;solvent}\) \(\times\) 100
or, 100 = \(\frac{100}{W2}\) \(\times\) 100
or, W2 = \(\frac{1000 \times 100}{1}\)
= 10000g.
Therefore, the weight of water in the saturated solution = 10000g.
If 70 gram of a saturated solution of sodium nitrate in water at 30\(^0\)C , contains 15 gram of solute , find the solubility of sodium nitrate at that temperature.
Here,
Weight of saturated solution(w) = 70gm
Weight of solute(W1) = 15gm
Weight of solvent (w2) = Weight of saturated solution - Wt. of solute
= 70-15 = 55
Solubility(S) = ?
We know that,
Solubility(S) = \(\frac{weight\;of\;soluten}{weight\;of\;solvent}\) \(\times\) 100
= \(\frac{15}{55}\) \(\times\) 100 = 27.27
Therefore, the solubility of sodium nitrate at 30\(^0\)C is 27.27
When 15gm of a saturated solution of sodium nitrate at 30 \(^0\)C is cooled down at 10 \(^0\)C, then how much sodium nitrate will be precipitated if solubility of sodium nitrate at 30 \(^0\)C is 95 and 10 \(^0\)C is 30 respectively ?
According to the statement :
At 30 \(^0\)C , 95 f of NaNO3 form 195 g of saturated solution
At 10 \(^0\)C, 30 g of NaNO3, form 130 g of saturated solution
The difference in the weight of solution is : 195 - 130 = 65g
It shows that
\(\therefore\) 195 g of saturated solution is cooled from 30 \(^0\)C to 10 \(^0\)C
it seperates \(\frac{65}{195}\)g of NaNO3
\(\therefore\) 15 g of saturated solution is when cooled from 30 \(^0\)C to 10 \(^0\)C
it seperates \(\frac{65}{195}\) \(\times\) 15g of NaNO3 = 5g
Thus, 5g sodium nitrate is seperated by cooling the solution.
What is the saturated solution?
The solution in which more amount of solute can dissolve in given solvent at given temperature.
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