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And only dispersion forces operate in carbon tetrachloride. We might assume that the latter is more volatile than methanol. But as physical scientists, we should consider the data, and the best metric of intermolecular attraction should be normal boiling point...

$"boiling point of methanol"=64.7$ $""^@C$

$"boiling point of carbon tetrachloride"=76.7$ $""^@C$

$"boiling point of ethanol"=78.4$ $""^@C$

So in fact carbon tet is LESS volatile than methanol. Mind you it is a much heavier molecule than methanol, with more electrons, and thus greater opportunity for dispersion forces. A better molecule for comparison might be chloromethane, or methylene chloride....

$"boiling point of chloromethane"=-24.2$ $""^@C$

$"boiling point of methylene chloride"=39.6$ $""^@C$

Are you happy with this rationale?

PS I do not even know whether methanol is soluble in carbon tet. Is it? This is the province of experiment. Certainly, methanol is INSOLUBLE in hexane, $n-C_6H_14$. Why should this be so?

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