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Find the molality of a solution containing a non-volatile solute if the vapour pressure is `2%` below the vapour pressure or pure water. |
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Answer» Given, `P_(S) = (98)/(100)P^(@)` Now `(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W)` `= (w)/(m xx W) xx 1000 xx (M)/(1000)` `{(P^(@) - (98)/(100).P^(@))/((98)/(100)P^(@))} = "Molality xx (18)/(1000)` `:. Molatity={(2P^(@))/(100xx98/100P^(@))}xx1000/18=1.1334` |
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