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The vapour pressure of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

It is given that:pA° = 450mm of HgpB0 = 700mm of Hgptotal = 600mm of Hgfrom Raoult's law , we have:pA = pA°xApB =pB0 xB  = pB°(1-xA)    therefore , total pressure, ptotal  = pA + pBptotal=   pA°xA + pB°(1-xA) ptotal =   pA°xA +  pB°-pB0 xA ptotal = ( pA° -  pB°)xA  +  pB°600= (450-750)xA + 700 -100 = -250xAxA = 0.4therefore ,xB  = 1-xA =1-0.4    =0.6Now, pA =   pA°xA               =450×0.4               =180 mm of HgpB  = pB0 xB      = 700×0.6      = 420mm of HgNow, fraction of liquid A  = pApA+pB          = 180180+420  = 180600  = 0.30

mole fraction of liquid b = 1-0.30 = 0.70

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Henry’s law constant for CO2 in water is 1.67 x 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.

Solution :
given that, 
pressure of CO2=2.5atm
1 atm = 1.01325x 105 pa  so that 
pressure of CO2 = 2.5x1.01325x105pa
   = 2.533125 x 10pa
KH = 1.67 x 10pa
ACcording to henry's law  p= KH*X
or X=P/KH 
     = 2.533125 x 105/1.67x 108
     = 1.52 x 10-3

But we have 500ML odf soda water so that 
Volume of water = 500mL
Density of water =1g/ml
mass = volume x density 
500 mL of water = 500g of water
molar mass f water (H2O) = 18g mol-1
number of moles =

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H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at S.T.P. is 0.195 m, calculate Henry's law constant.


It is given that the solubility of H2S in water at STP is 0.195m, i.e., 0.195 mol of H2S is dissolved in1000 g of water.
     
           =1000g18g mol-1moles of water =55.56 mol=moles of H2S moles of H2S+ moles of water =0.1950.195+55.56 = 0.0035

At STP pressure (P) = 0.987bar
According to henry's law p = kx

KH = px = 0.9870.0035bar = 282 bar

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Boiling point of water at 750 mm Hg 99.630C. How much sucrose is to be added to 500g of water such that it boils at 1000C.

Solution:
here, elevation of boiling point Tb= (100+273)-(99.63+273) = 0.37K
mass of water, w1 = 500g
Boiling point of water at 750 mm Hg = 99.63
0C
          MB = kb×1000×ωBωA×Tb
kb = 0.52 k kg mol-1,  ωB = 500 gTb = 100 - 99.63 = 0.37°C
Molar mass of sucrose, C12H22O11(Mg)
                        = 12×12+22×1+11×16= 144+22+176= 342 g mol-1

or         342 = 0.52×1000×WB500×0.37

or          WB = 342×500×0.370.52×1000=121.67 g

hence , 121.67 g of sucrose is to obtained.
               
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Vapour pressure of pure water at 298 K is 23.8 mm. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapoure pressure of pure water (solvent) at 298 K, p0 = 23.8 mm
Vapour pressure of solution, p = ?
Mass of solvent ,W = 850 g
Mass of solute,M = 50 g
Mol. mass of water (H2O), M = 18 g mol–1
Mol.mass of urea NH2 CO NH2
= 14 + 2 + 12 + 16 + 14 + 2
= 60 g mol–1
According to Raoult's law, p0-pp0=ωMWm

                     p=p0-w×Mm×W×p°

                   p=23.8-50×1860×850

                      =23.8-0.017=23.78
Hence, 23.78 mm Hg. Ans.


 
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