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The vapour pressure of 2.1% solution of a non- electrolyte in water at `100^(@) C` is 755 mm Hg. Calculate the molar mass of the solute . |
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Answer» Correct Answer - Molar mass of non- electrolyte= 58. 69 g `mol^(-1)` At `100^(@)C` vapour pressure of water `=P_(o)`= 760 mm Hg Vapour pressure of the solutions = P= 755 mm Hg Since the solution is 2.1 by mass Mass of solute (non- electrolyte ) =`W_(2) =2.1 g` Mass of water `=W_(1) =100 -2.1 =97.9 g` Molar mass of water = `M_(1) =18 g mol^(-1)` Molar mass of non- electrolyte `=M_(2)= ?` `(P_(o) -P)/(P_(o)) = (W_(2) xx M_(1))/(W_(1) xx M_(2))` `:. M_(2) = ((P_(o))/(P_(o) - P)) xx (W_(2) xx M_(1))/(W_(1)) = (700xx 2.1 xx 18)/((760 - 755) xx 97.9) = 58.69 g mol^(-1)` |
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