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Calculate the mass of a non-volatile solute (molar mass = 40 g/mol) which should be dissolvedin 114 g octane to reduce its vapour pressure to 80%. |
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Answer» SOLUTION :Let the vapour pressure of PURE octane = `p_0` Vapour pressure of the solution = `0.8p_0` Lowering of vapour pressure = `0.2p_0` Relative lowering `= (0.2p_0)/(p_0)` According to Raoult.s law, Relative lowering of vapour pressure of a solution is equal to the mole fraction of the solute. Let the mass of solute be = w g Its molecular mass is GIVEN as = 40 u No. of moles of the solute =`(w)/(40)` No. of moles of the solvent ` = 114/114 = 1`[ mol. mass of octane = 114] Mole fraction of the solute ` = (w//40)/(1 + w//40)` According to Raoult.s law, `(0.2p_0)/(p_0) = (w/40)/(1 + w//40) ` on simplification `0.2 = (w//40)/((40 + w)/(40)) ` `0.2 = (w)/(40 + w)` or 0.2(40 + w) = WOR 8 + 0.2w - w = 0 or - 0.8w = -8or w = 10 g |
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