Q.

If mole fraction of a solute in 1 kg benzene is 0.2 then molality of solute is

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3.2

3.6

answer is A.

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Detailed Solution

Let number of moles of solute in solution = x Moles of benzene in solution =1000g78gmol−1=12.82 moles Mole fraction of solute =xx+12.82⇒0.2=xx+12.82On solving x = 2. Molality (m)= Number of moles of solute Mass of solvent (in kg)=3.21=3.2

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