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Define the term solution. How many types of solutions are formed? Write briefly about each type with an example. 

Answer:

A solution is a homogenous mixture of two or more than two substances whose composition can change within a certain limits.  A solution of two substances is called binary solution.
In solution, the component that present in small amount is known as solute and the component present in larger amount known as solvent.

Nine kinds of solution are possible.

(i) Gas in gas. When one gas is mixed with another gas, it is called solution of gas in gas. Example : Air is a mixture of nitrogen and oxygen.

(ii) Liquid in gas. When liquid is mixed with large amount of gas, it is called liquid in gas solution. Example: Moisture (water in air).

(iii) Solid in gas. When small amount of solid particles are dispersed in gas, it is called solution of solid in gas. Example: Smoke.

(iv) Gas in liquid. When gas is dissolved in liquid, it is called gas in liquid solution. Examples: CO2 gas dissolved in water, oxygen dissolved in water.

(v) Liquid in liquid. When a liquid is miscible with another liquid, it forms solution of liquid in liquid. Examples. Ethanol dissolved in water, methanol dissolved in water.

(vi) Solid in liquid. When solid is dissolved in water, the solution is called solid in liquid. Examples: Cane sugar dissolved in water, sodium chloride dissolved in water.

(vii) Gas in solid. When gas is present, the solution is called gas in solid. Example: H2 gas in palladium.

(viii) Liquid in solid. When liquid is present in solid, the homogeneous mixture is called solution of liquid in solid. Example: Amalgam of mercury with sodium.

(ix) Solid in solid. When solid is dissolved in another solid, the homogeneous mixture is called solution of solid in solid. Examples: Alloys are solid in solid solution, copper dissolved in gold.

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Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5 0C. Kf= 3.9 k kg mol–1.

Solution:
Mass of acetic acid  W1= 75g 
Molar mass oC6H8O6

                = 6×12+8×1+6×16= 72+8+96 = 176 g mol-1
lowering of melting point , 

Tf = 1.5 kWe know that:Tf  =Kf×w2×1000M2×w1  =  w2 = Tf×M2×w1Kf×1000         =1.5×176×753.9×1000  = 5.08g approx.  

 

Hence, 5.08 g of ascorbic acid is needed to be dissolved. 

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Define the following terms:
Molality

Answer:

Molality. The moles of the solute dissolved in one kilogram of the solvent is called the molality of the solution.
    Moality = m  = Moles of BMass of solvent in kg

∵  Moles = Moles of BMolar mass of B = ωBMB    m = ωBMB×ωAwhere  ωA = Mass of solvent in kg.
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Define the following terms:
Mole fraction 

Answer: 
mole fraction can be defined as :

Mole fraction of a component =


Number of moles of the componentTotal number of moles of all the components 


 for example
Mole fraction. The ratio of the moles of a component A to the total moles of the solution is called the mole fraction of A. It is denoted by symbol x. For a binary solution of components A and B Mole fraction of A = x
A
= Moles of ATotal moles of solution  = nAnA+nB

Mole fraction of  B = xB
    =Moles of BTotal moles of solution= nBnA+nB
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Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL at 370C.


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