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:

Options

A. `1:3`
B. `1:4`
C. `2:3`
D. `1:2`

Answer

1

Explanation

Molality `=(x_(A)xx1000)/((1-x_(A)m_(B)))`
First solution's molality `=(0.25xx1000)/(0.75xx18)`
Second solution's molality `=(0.5xx1000)/(0.5xx18):. " " ` Molality ratio `=(1)/(3)`