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Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 3 K. (K for water = 1.86 K "mol"^(-1)). |
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Answer» Solution :KCI is a strong ELECTROLYTE and is completely DISSOCIATED in water as : `KCIto K^+ +Cl^-` Now, 1 mole of KCI will give 2 mole IONS. THEREFORE, the value of Van.t Hoff factor (i) will be equal to 2. `Deltat=i k_f m ` `K_f =1.86 (K kg "mol"^(-1))` `DeltaT = 3 K,i=2` `m= (Delta T )/(i K _f ) = ( 3 )/( 2 xx 1.86 ) = 0.81` mole Thus, 0.81 mole or `0.81 xx74.5 g = 60-34 g` must be added to 1 kg of water. |
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