Pure solvent A has freezing point 16.5^(@)C. On dissolving 0.4 g of B in 200 g of A, the solution freezing at 16.4^(@)C and on dissolving 2.24 g of C in 100 g of A, the solution has freezing point of 16.0^(@)C. If the molar mass of Bis "74 g mol"^(-1), what is the molar mass of C?

Pure solvent A has freezing point 16.5^(@)C. On dissolving 0.4 g of B

1.	Pure solvent A has freezing point 16.5^(@)C. On dissolving 0.4 g of B in 200 g of A, the solution freezing at 16.4^(@)C and on dissolving 2.24 g of C in 100 g of A, the solution has freezing point of 16.0^(@)C. If the molar mass of Bis "74 g mol"^(-1), what is the molar mass of C?
Answer» Solution :For solute B, dissolved in solvent A, `DELTA T_(F)=(1000K_(f)w_(2))/(w_(1)M_(2)),"i.e., 0.1"=(1000xxK_(f)xx0.4)/(200xx74)or K_(f)="3.7 K kg MOL"^(-1)` For solute C dissolved in the same solvent A, `M_(2)=(1000K_(f)w_(2))/(w_(1)DeltaT_(f))=(1000xx3.7xx2.24)/(100xx0.5)=165.8` Alternatively,`""((DeltaT_(f))_(B))/((DeltaT_(f))_(C))=(w_(B))/(w_(A)M_(B))xx(w_(A)'xxM_(C))/(w_(C))`