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Calculate the depression in the freezing point of water when 10 g of CH_3 CH_2 CHCICOOHis added to 250 g of water. K_a = 1.4 xx 10^(-3) ,K_f = 1.86 K kg "mol"^(-1) |
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Answer» Solution : Molar mass of `CH_3 CH_2 CHCICOOH = 15 + 14 + 13 + 35.5 + 45 = 122.5 g "mol"^(-1)` Number of MOLES in 10 g of `CH_3CH_2CHCICOOH = (10)/(122.5) `MOLE ` = 8.16 XX 10^(-2)` mole MOLALITY of the solution `(m) = (8.16 xx 10^(-2))/(250) xx 1000 = 0.3264` ![]() `i=(1 + alpha)/(1) = 1 + alpha = 1+ 0.065 = 1.065` ` Delta T_f = iK_f m =(1.065)(1.86)(0.3264) = 0.65^@` |
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