Solubility product of `PbC_(2)` at 298K is `1.0 xx 10^(-6)` At this temperature solubility of `PbCI_(2)` in moles per litre is

Solubility product of PbCl_(2) at 298 K is 1.0xx10^(-6) . At this temperature solubility of PbCl_(2) in moles per litre is :

Solubbility product of PbCl_2 at 298 K is 1.0 xx 10^-6. At this temperature solubility of PbCL_2 in mol per litre is :

The solubility product of PbCl_(2) at 298K is 1.7 xx 10^(-5) . Calculate the solubility of PbCl_(2) in g L^(-1) at 298K

The solubility product of PbCl_2 at 298 K is 7.1 xx 10 ^(-5) Calculate the solubility of PbCl_2 " in " g L^(-1) at 298 K.

The solubility product of Fe(OH)_3 at 298K is 1.1 xx 10^(-36) . Find its solubility in kgdm^-3 at the same temperature. (Given: At Wts. Fe = 56, O =16, H =1).

Solubility product of BaCl_(2) is 4xx10^(-9) . Its solubility in moles//litre would be

Solubility product of BaCl_(2) is 4xx10^(-9) . Its solubility in moles//litre would be