The dissoctiation of water of `25^(@)C` is `1.9 xx 10^(-7)%` and the density of water is `1g//cm^(3)` the ionisation constant of water is

The dissociation of water at 25^@C is 1.9xx 10^(-7) percent and the density of water is 1 gm /cm_3 the ionisation constant of water is :

The degree of dissociation of water at 25^@C is 1.9xx10^-7% and density is 1g cm cube. The ionic constant for water is

The degree of dissociation of water is 1.8 xx 10^(-9) at 298 K. Calculate the ionisation constant and ionic product of water at 298 K

If degree of dissociation of pure water at 100^@ C is 1.8 xx 10^(-8) , then the dissociation constant of water will be: (Density of H_2O = 1 g/cc)

At 277 K, degree of dissociation water is 1xx10^(-7)% . The value of ionic product of water is

At 277 K, degree of dissociation water is 1xx10^(-7)% . The value of ionic product of water is

12.25g of CH_(3)CH_(2)CHClCOOH ia added to 250g of water to make a solution. If the dissociation constant of above acid is 1.44 xx 10^(-3) , the depression in freezing point of water in .^(@)C is : ( K_(f) for water is 1.86 "K kg mol"^(-1) )