1.

If 10.0 g of a non-electrolyte dissolved in 100 g of water lowers the freezing point of water by 1.86^(@)C, the molar mass of the non-electrolyte is ( K_(f)=1.86 Km^(-1))

Answer»

`10.0`
100
1000
186

Solution :`M_(B)=(K_(F)xx1000xx w_(B))/(w_(A) XX Delta T_(f))`
`=(1.86xx1000xx10.0)/(100xx1.86)=100`


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