The mole fraction of a solute in a solutions is `0.1`. At `298K` 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_("solute"))/(MW_("solvent"))` is

Answer

9

Explanation

Mole fraction of solute 0.1 means that 1 mol of solution has 0.1 mol of solute and 0.9 mol of solvent .
If `M_(1)` is molar mass of solute and `M_(2)` is molar mass of solvent , then
Mass of solvent ` = 0.9 xx M_(2)`
Molality `= (0.1)/(0.9M_(2)) xx 100`
Volume of solution ` = ("Mass of solute + Mass of solvent ")/("Density")`
` = (0.1 M_(1) + 0.9 M_(2))/(2)`
Molarity `= (0.1)/((0.1 M_(1) + 0.9 M_(2)) xx 2 xx 1000`
Since molality = molarity .
`(0.1)/(0.9 M_(2)) xx 1000 = (0.1 xx 2 xx 1000)/(0.1 M_(1) + 0.9 M_(2))`
` (100)/(0.9M_(2)) = (200)/(0.1 M_(1) + 0.9 M_(2))`
`10 M_(1) + 90 M_(2) = 180 M_(2)`
` 10 M_(1) = 90 M_(2) " " therefore " " (M_(1))/(M_(2)) = 9`