The cryoscopic constant of water is 1.86 K `"mol"^(-1)` kg. An aqueous solution of cane sugar freezes at `-0.372^(@)C`. Calculate the molality of the solution.

Explanation

`K_(f)=1.86k g mol^(-1)`
Freezing point of solution `T_(f)=-0.327^(@)C`
`=[273+(-0.327)]`
`272.673K`
Freezing point of pure water,ds `T_(f)^(@)=0^(@)C=273K`
Depression in freezing point `DeltaT_(f)=T_(f)=T_(f)`
`=(273.272.673)=K=0.327K`
`DeltaT_(f)=i xx K_(f)xxm`
Since, cane sugar is a non-electrolyte, thus i (van.t Hoff factor) is equal to 1.
`0.327xx1.186xxm`
`m=0.327//1.86`
`=0.17` molal `~~`0.2 molal