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

Options

A. 105.7
B. 106.7
C. 115.3
D. 93.9

Answer

A

Explanation

`Delta T_(f)=(1000K_(f) xx W_(2))/(M_(2) xx W_(1)), Delta T_(f)=273 - 271.9 =1.1`
`:. 1.1 = (1000 xx 1.86 xx 1.25)/(M_(2) xx 20) :. M_(2) = 105.68`