Question

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 1

Correct Answer - -9
Moles of solute, `n_(1) = (w_(1))/(m_(1))`, Moles of solvent, `n_(2) = (w_(2))/(m_(2))`
`chi_(1)("solute") = 0.1` and `chi_(2)("solvent") = 0.9`
`:. (chi_(1))/(chi_(2)) = (n_(1))/(n_(2)) = (w_(1))/(m_(1)).(m_(2))/(w_(2)) = (1)/(9)`
Molarity `= ("Solute(moles)")/("Volume (L)") = (w_(1) xx 1000 xx 2)/(m_(1) (w_(1)+w_(2)))`
Note `"Volume" = ("Total mass of solution")/("Density") = ((w_(1)+w_(2))/(2))mL`
Molarity `= ("Solute(moles)")/("Solvent(kg)") = (w_(1) xx 1000)/(m_(1) xx w_(2))`
Given, molarity = molality
hence, `(2000 w_(1))/(m_(1) (w_(1)+w_(2))) = (1000 w_(1))/(m_(1) w_(2))`
`:. (w_(2))/(w_(1)+m_(2)) = (1)/(2) rArr w_(1) = w_(2) = 1`
`:. (w_(1) m_(2))/(m_(1) w_(2)) = (1)/(9) rArr (m_(1) ("solute"))/(m_(2)("solvent")) = 9`