`1.0xx10^(-3)` kg of urea when disso,ved in 0.0985 kg of a solvent, decreases freezing point of the solvent by `0.211K.1.6xx10^(-3)` kg of another non-electrolyte solute when disso,ved of 0.086 kg of the same solvent depresses the freezing point by 0.34 K. 0.35 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)

Explanation

Given : Mass of solute, Urea `=W_(2)`
`=1*0xx10^(-3)kg=1g`
Mass of solvent `=W_(2)`
`=0*0985kG=98*5g`
`:." "DeltaT=0*211k`
Molar mass of urea `(NH_(2)CONH_(2))`
`=M_(2)=60g//mol`
Mass of unknown sabstance
`=W_(2)^(')=1*6xx10^(-3)kg`
`=1*6g`
Mass of a solvent `=W_(1)^(')=0*086` kg
`:." "deltaT_(F)=0*34k`
Molar mass of unknown sabstance `=M_(2)^()=?`
This problem will be done in two parts :
(i) To calculate molal depression constant `'K_(f)'`,
For urea solution-
`DeltaT_(f)=K_(f)xx(W_(2)xx1000)/(W_(1)xxM_(2))`
`:." "K_(f)=(DeltaT_(f)xxW_(1)xxM_(2))/(W_(2)xx1000)`
`=(0*211xx98*5xx60)/(1xx1000)`
`=(1247*01)/(1000)=1*24701`
`:." "K_(f)=1*2g//mol`.
(ii) To calculate molar mass, `M_(2)`
For unknown solution
`M_(2)^(')=K_(f)xx(W_(2)^(')xx1000)/(W_(1)^(2)xx0*34)`
`:." "K_(f)=(1*2xx1*6xx1000)/(84xx0*34)`
`=(1920)/(29*24)`
`:." "M_(2)^(')=65*66g//mol`
Hence, the molar mass of abother solution is 65*66 g/mol.