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Chemistry I

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Class 10 Class 12
Explain the following phenomena:
(i) Hemolysis
(ii) Crenation.

Answer:

(i) Hemolysis: The red blood cells are protected from the external environment by a semipermeable membrane. The red blood cells are placed in a hypotonic solution. Because the hypotonic solution is less concentrated than the interior of the cell, water moves into the cell. The cell swells and eventually burst, releasing hemoglobin and other molecules. This process is called hemolysis.

(ii) Crenation: When a bacterial cell is placed in a hypertonic (high concentration) sugar solution, the intracellular water tends to move out of the bacterial cell to be more concentrated solution by osmosis. This causes the cell to shrink and eventually, to stop functioning. This process is called crenation.

208 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 (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 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 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