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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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Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 250 C, assuming that it is completely dissociated. 

Answer:
When K2SO4 is dissolved in water , K+ and SO42- ion produced

K2SO4  2K+ + SO42-
total number of ion produced 3
therfore, i = 3
given that 

w = 25 mg = 0.025g
T = 250C + 273 = 298 K
Also  we know that 
R = 0.0821 L atm K-1 mol-1
M= (2 x39)+(1x 32)+(4x16)
Osmotic pressure, π=?
 
 V = 2L,  i = 3
π = i n2 RTV = i w RTmVπ = 3×25×10-3×0.0821×298174×2    = 5.27 × 10-3 atm.
202 Views

Determine the amount of CaCl(i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 270C.

Answer;
π = inRTVπ = iWRTMV     where n = w/mw = πMViRT

CaCl
2 → Ca2+ + 2Cl

π = 0.75 atm


T = 273 + 27 = 300 K,
M = 1 x 40 + 2x 35.5 = 111g mol-1
V = 2.5 L
i = 2.47
       osmatic pressure, π = i n2 RTV 

                  w= πMViRTtherefore w= 0.75×111×2.52.47×0.082×300  = 3.42 g
 

amount of calcium chloride dissolved is 3.42g
434 Views

Vapour pressures of pure acetone and chloroform at 328 k are 741.8 nm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot Ptotal , Pchlroform and Pacelone as a function of Xactone. The experimental date observed for different composition of mixture is:

100 x xacetone

0

11.8

23.8

36.0

50.8

58.2

64.5

72.1

Pacetone / mm Hg

 

0

54.9

110.1

202.4

327.7

405.9

454.1

521.1

Pchloroform/ mm Hg

632.8

548.1

469.4

359.7

257.7

193.6

161.2

120.7

Plot this data also on the same graph paper, indicate whether it has positive deviation or negative deviation from the ideal solution.


The plot is given below:

The plot is given below:It has negative deviation from the ideal solu
It has negative deviation from the ideal solution.
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At 300 K, 36 g of glucose (C6H12O6) present per litre in its aqueous solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of another solution of glucose is 1.52 bar at the same temperature, what would be its concentration? 


Answer:

π1 = C1RT                         ....................1π2 = C2RT                         .................... 2therfore on dividing the 1 by 2 we get π1π2 = C1C2

Let us calculate the concentration of the first solution with osmatic pressure 4.98bar

mass of glucose = 36g
molar mass of gulcose = 180g/ mol
therefore number of moles of gulcose =  36/180
                                                         = 0.2 moles

volume of the solution  = 1L
  molarity  = No. of moles of glucose / vol.of solution

molarity  =0.2/1L

C1 = 0.2 moles/L
C2 = π2C1π2C2 = 1.52×0.24.98 = 0.61M


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