The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is `2.0 g cm^(-3)`. The ratio of the molecular weights of the solute and solvent, `((MW_("solite"))/(MW_("solvent")))` is .........

Answer

9

Explanation

1 mole solutions has `0.1` mole solute and `0.9` mole solvent
Let `M_(1) =` molar mass solute
`M_(2) =` Molar mass solvent
Molality, `m = (0.1)/(0.9 M_(2)) xx 1000` ...(1)
Molarity, `M = (0.1)/(0.1M_(1) + 0.9 M_(2)) xx 2 xx 1000` ...(2)
`:' m = M`
`rArr (0.1 xx 1000)/(0.9 M_(2)) = (200)/(0.1 M_(1) +0.9 M_(2))`
`rArr (M_(1))/(M_(2)) = 9`
Alternate solution:
`:' M = m`
`rArr` volume of solution = mass of solvent
`rArr (W_(solute)+W_(solvent))/(2) = W_(solvent)`
`w_(solute) = W_(solvent)`
`0.1 xx M_(solute) = 0.9 xx M_(solvent)`
`(M_(solute))/(M_(solvent)) = 9`