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Calculate the degree of ionisation of normal ZnSO4 solutionat 373 k if its vapour pressure is 750mm.

7 years ago


Answers : (1)


Recall that the total pressure of a solution is the sum of the partial pressures of the solvent and solute

psolution = psolvent + psolute = xsolvent P°solvent + xsolute P°solute
If the solute is non-volatile (no vapour pressure: P°solute = 0) then the total vapour pressure of solution is

psolution = psolvent = xsolvent P°solvent (where x= mol fractn)

now v.p of water at 373 k = 760 mm

750= x*760

=> x = 750/760

1-x = mole fractn of solute = mol of solute/total moles(nearly = moles of water)

= mol of solute/moles of water

ZnSO4 -> Zn(+2) +SO4(-2)
0.5-y y y (1N= 0.5M)

total moles of solute = 0.5+y

total moles of solvent(water)=55.55 (in 1 lt)

=> 1-x = 0.5 + y /55.55

=>1-750/760= 1/76= 0.5+y /55.55

=> y = 0.23
y = dissociation = 23%

7 years ago

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