InterviewSolution
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`1.4 g` of acetone dissolved in `100 g` of benzene gave a solution which freezes at `277.12 K`. Pure benzene freezes at `278.4 K`.`2.8` of solid `(A)` dissolved in `100 g` of benzene gave a solution which froze at `277.76 K`. Calculate the molecular mass of `(A)`. |
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Answer» We know that `DeltaT = (K_(f)) xx (W_(2) xx 1000)/(Mw_(2) xx W_(1))` where`DeltaT` = Depression in freezing point `K_(f)` = Molal depression constant of benzene` `W_(2)` = Mass of solute `Mw_(2)` = Molecular mass of solute `W_(1)` = Mass of solvent Case i: `(278.4 - 277.12) = K_(f)xx(1.4 xx 1000)/(58 xx 100)` `1.28 = K_(f) xx (14)/(58)` Case ii: `(278.4 - 277.76) = K_(f)xx(2.8 xx 1000)/(Mw_(A) xx 100)` `0.64 = K_(f) xx (28)/(Mw_(A))` Dividing Eq. (i) by (ii), we get `Mw_(A) = 232` |
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