If the mole fraction of a solute is char...

If the mole fraction of a solute is charged from `(1)/(4)` to `(1)/(2)` by adding some solute in the `800g` of `H_(2)O` solvent, then the ratio of molality of two solutions will be `:`

`1:3`

`3:1`

`4:3`

`1:2`

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Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m. Molality(m) =("Number of moles of solute")/("Number of kilograms of the solvent") let w_(A) grams of the solute of molecular mass m_(A) be present in w_(B) grams of the solvent, then: Molality(m) = (w_(A))/(m_(A)xxw_(B))xx1000 Relation between mole fraction and molality: X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n) (X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A)) (X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m If the mole fraction of a solute is changed from (1)/(4) "to" (1)/(2) in the 800 g of solvent then the ratio tof molality will be:

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