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Chemistry I

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Class 10 Class 12
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.

184 Views

Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Mass % of benzene
                      = mass of benzenemass of solution ×100= 2222+122×100= 22144×100 = 15.28%
Mass% of carbon tetrachloride = 100 - 15.28
                          = 84.72%
1703 Views

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.


Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,



897 Views

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.


(a) 20% (mass/mass) means that 20 g of KI is present in 80 g of water.

Therefore, Moles of KI in solution
 



moles of KI = 20/166 =0.12mol
moles of water =80/18 =4.44mol
therefore, mole fraction of KI
 
=moles of  KImoles of KI + moles of water

=0.120.12+4.44= 0.0263            
1010 Views

Calculate the molarity of each of the following solution (a) 30 g of Co(NO3)2.6H2O in 4.3 L solution (b) 30 mL of 0.5 MH2SO4 diluted to 500 mL.

solution;

Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.


(a) Mol. mass of Co(NO3). 6H2O

               =58.9+(14+3×16)2+6(18)=58.9+(14+48)×2+108=58.9+124+108 = 290.9

Moles of Co(NO)3.6H2O
                                       =30290.9=0.103 mol.
Volume of solution = 4.3 L
Molarity, 
          M=Moles of soluteVolume of solution in litre    = 1034.3 = 0.024 M

(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=0.5×301000mol =0.015mol
therefore molarity =0.015/0.5L 

thus molarity is 0.03M

                 
  

844 Views

Calculate the mass of urea (NH2CONH2) required in making 2.5 kg 0.25 of molal aqueous solution.

Solution:

Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:

= Moles of soluteMass of solvent in kg
 
Mol. mass of urea NH2CONH2
                 = 14 + 2 + 12 + 16 + 14 + 2
                 = 60 g mol-1

Molality (m) = Moles of soluteMass of solvent in kg

25 = Moles of solute2.5

or Moles of solute
                = 0.25 x 0.25 =  0.625

  Mass of urea
                   = Moles of solute x Molar mass

                   = 0.625 x 60 = 37.5 g

1475 Views