1.

The mole fraction of a solute in bezene solvent is 0.2. Themolality of the solution will be:

Answer»

14
3.2
1.4
2

Solution :`X_(B)=n_(B)/(n_(B)+n_(A))`
`0.2=n_(B)/(n_(b)+1000//78)`
`1/0.2=(n_(B)+12.82)/n_(B)`
`5=1+(12.82)/n_(B)n_(B)=(12.82)/4=3.2` (Mass of solvent is 1000 G)


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