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300K पर जल का वाष्प दाब `12.3K Pa` है . इसमें bane अवाष्पशील विलेय के एक मोलल विलयन का वाष्प दाब ज्ञात कीजिए .

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प्रश्नानुसार `n=1` तथा `N=1000//18,` माना मोललता = मोललता
विलेय का मोल प्रभाज `=(1)/(1+55.5)=0.0177`
अतः `(p_(A)^(0)-P_(A))/(P_(A)^(0))=x_(B)`
अर्थात `(12.3-p_(A))/(12.3)=0.0177`
` p_(A)=12.08k pa.`

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