6 g of a substance is dissolved in 100 g of water depresses the freezing point by `0.93^(@)C`. The molecular mass of the substance will be: (`K_(f)` for water ` = 1.86^(@)C`/molal)

Explanation

`M_(B) = (K_(f) xx 1000 xx w_(B))/(w_(A) xx Delta T_(f))`
`K_(f) = 1.86 Km^(-1) , w_(B) = 9 g , w_(A) = 100g`
`Delta T_(f) = 0.93 K`
`therefore " " M_(B) = (1.86 xx 1000 xx 9)/(100 xx 0.93)`
` = 180 g mol^(-1)`