The freezing point of a solution prepared from `1.25 g` of non-electrolyte and `20 g` of water is `271.9 K`. If the molar depression constant is `1.86 K mol^(-1)`, then molar mass of the solute will be

Explanation

Given `=(T_(f))_(s) = 271.9 K`
`w = 1.25 g W = 20g K_(f) 1.86`
`DeltaT_(f) = T_(0) -(T_(f))_(s) = 273 - 271.9 = 1.1 K`
`DeltaT_(f) =` molality `xx K_(f) rArr DeltaT_(f) =(w)/(m xx W) xx 1000 xx K_(f)`
or `m = (w xx 1000 xx K_(f))/(DeltaT_(f) xx W) = (1.25 xx 1000 xx 1.86)/(1.1 xx 20) = 105.68 mol//kg`