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The vapour pressure of water is 12.3 kPa...

The vapour pressure of water is 12.3 kPa at 300 K. calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Text Solution

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Vapour pressure of pure water, `p^(0)=12.3kP_(a)`
Molecular mass of water = 18
Molarity of solution=1 M= 1 `"mol L"^(-1)`
Weight of 1L water = 1000 g
Number of moles of water `=(1000)/(18)=55.5` mol
Moles of solute = 1
Total number of moles in one litre
1M solution `= 55.5+1 = 56.5`
Mole fraction of solute `=(1)/(56.5)=0.0177`
Now `(p^(@)-p)/(p^(@))=x_(B)=0.0177`
`:.(12.3-p)/(12.3)=0.0177`
On solving, vapour pressure of solution,
`p_(A)=12.3 -0.0177 xx 12.3=12.3-0.2177=12.08kP_(a)`
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