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

Let total moles be 1
`n_("solute") = 0.1 , n_("solvent") = 0.9`
At 298 K, molality = molarity (say x)
‘x’ M `rArr` x mole solute in 1 L solution,
x mole solute in 1l solution (2 gm/cc) = 2000 gm solution (d = 2 gm/cc)
‘x’ molal `rArr` and x mole solute in 1000 g solvent
x mole solute in 1000 gm solvent.