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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 beA. 105.7B. 106.7C. 115.3D. 93.9 |
Answer» Correct Answer - A `M = (K_(f) xx 1000 xx w)/(W xx DeltaT_(f))` `= (1.86 xx 1000 xx 1.25)/(20 xx 1.1) = 105.68 ~~ 105.7` |
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