Mole fraction of `A` in `H_(2)O` is `0.2` The molality of `A` in `H_(2)O` is :

Options

A. `13.8`
B. `15.5`
C. `14.5`
D. `16.8`

Answer

A

Explanation

`x_(Lambda)=0.2`
`x_(H_(2))O = 1-0.2=0.8`
`"wt of" H_(2)O = 0.8xx18 = 14.4 g`
Molality
`=("moles of solute")/("wt. of solvent" (H_(2)O)"in kg")`
`= (2xx1000)/(14.4)=13.8`