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Chemistry I

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
Benzene and toluene form iedal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole-fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Solution
Given that
PoBenzene = 50.51 mm Hg  
PoNaphthalene = 50.51 mm Hg
Mass of Benzene = 80 g
Mass of Toluene = 100 g
Molar mass of benzene(C6H6) = 6 × 12  +  6 × 1  = 78 g mol - 1
Molar mass of toluene(C6H5CH3) = 6 × 12 + 5 × 1 + 12 + 3 × 1  = 92 g mol – 1
Use the formula



Mole of benzene (nA)
= Mass of bezeneMolar mass of bezene (C6H6)= 8078 g mol-1 = 1.025 mol.

Mass of Toluene (nB)
= Mass of tolueneMolar mass of toulene(C7H8)=1= 10092 g mol-1=1.087 mol

Mole fraction of benzene (xA)
= 1.025 mol(1.025 mol + 1.087mol) = 1.0252.113 = 0.486

Similarly
Mole fraction of toluene, X Toluene =
 
1-XBenzene  = 1  - 0.486 = 0.514

Use the formula if Henry law
PA      = poA  ×  XA

Partial vapour pressure of benzene, PBenzene = poBenzene   ×  XBenzene     

PBenzene=0.487 × 50.71  = 24.645 mm Hg
Similarly

Partial pressure of Toluene, P toluene = 0.514 × 32.06 = 16.48 mmHg 

Use the formula of mole fraction using partial pressure

Mole fraction of benzene = PbenzenePbenzene + PToluene
  

Plug the values we get

Mole fraction of benzene = 24.645 /(24.645 +  16.48 )   
                                       
=  24.645/41.123  = 0.60


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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 . 3 Shares

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 . 11 Shares

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 . 6 Shares

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 . 3 Shares

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 . 10 Shares

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