Visha took two aqueous solutions - one containing 7.5 g of urea (Molar mass = 60 g/mol) and the other containing 42.75g of substance Z in 100 g of water, respectively. It was observed that both the solutions froze at the same temperature. Calculate the molar mass of Z.

Explanation

Apply the relationship for depression in freezing point.
`DeltaT_(f)=(K_(f)xx w_(2) xx 1000)/(M_(2)xx w_(1))`
Applying the relation for urea
`DeltaT_(f)"(urea)"=(K_(f) xx 7.5 xx 1000)/(60xx100)" "…(i)`
Applying the relation for the substance Z
`DeltaT_(f)`(substance Z) `=(K_(f) xx 42.75xx1000)/(M xx 100)," " ...(ii)`
where M is the molecular mass of the substance Z.
Both the solutions freeze at the same temperature i.e.,
`DeltaT_(f)`(urea)`=DeltaT_(f)`(substance Z)
Therefore we can equate the R.H.S of equations (i) and (ii), we have
`(K_(f) xx 42.75xx1000)/(M xx 100)=(K_(f)xx7.5xx1000)/(60xx100)`
`or M=(K_(f)xx42.75xx1000)/(K_(f)xx7.5xx1000)xx(60xx100)/(100)=(42.75)/(7.5)xx60=342`