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How would you determine molar mass from relative lowering of vapour pressure. |
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Answer» <P> Solution :(i) The measureement of relative lowering of vapour pressure can be used to determine the molar mass of a non-volatile solute.(ii) A known mass of the solute is dissolved in a known quantity of solvent. The relative lowering of vapour pressure is measured experimentally. (iii) According to Raoult.s law, the relative lowering of vapour pressure is `(P_("solvent")^(0) -P_("solution"))=x _(B)` `W _(A) = weight of solvent, `W_(B)=` weight of solute `M _(A)=` Molar mass of solvent,` M_(B)=` molar mass of solute `therefore x _(B) = (n_(B))/(n _(A)+ n _(B))` where `n _(A)=` NUMBER of moles of solvent, `n _(B)=` number of moles of solute. For dilute solution, `n _(A) gt gt n _(B), n _(A) + n _(B) ~~ n _(A).` Then,` x _(B) = (n_(B))/(n _(A))` Number of moles of solvent and solute are `n _(A) = (W_(A))/(M_(A)) , n _(B) = (W_(B))/( M _(B))` `therefore x _(B) = ((W _(B))/(M _(B)))/((W _(A))/(M _(A)))` Thus, relative lowering of vapour PRESURE ((W _(B))/(M _(B)))/((W _(A))/(M _(A)))` Relative lowering of vapour pressure ` = (P^(@)-P)/( P^(@))` `(P^(@) -P)/(P^(@)) = (W_(B) xxM _(A))/(W _(A) xx M _(B))` From the above equation, molar mass of the folute `M_(B)` can be calculated using the knonw values of `W_(A), W_(B), M_(A)` and the measured relative lowering of vapour pressure. |
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