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

7 years ago

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```										Hi,

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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