The conductivity of a saturated solution of AgBr at `25^@C` is `8.5xx10^-7 S cm^-1`. If the limiting molar ionic conductance of `Ag^+` and `Br^-` ions are 62 and `78 S cm^2 mol^-1`, then calculate the solubility product of AgBr.

The specific conductance of a saturated solution of silver bromide is kScm^(-1) . The limiting ionic conductivity of Ag^+ and Br^- ions are x and y respectively. The solubility of silver4 vromide in gL^(-1) is : (molar mass of AgBr=188)

If limiting molar conductivity of Ca^(2+) and Cl^(-) are 119.0 and 76.3 S cm^2 mol^(-1) , then the value of limiting molar conductivity of CaCl_2 will be :

The conductivity of 0.20 MKCl solution at 298 K is 0.0248 S cm^-1 . Calculate its molar conductivity.

Solubility of Ag_(2)CrO_(4) is 8 xx 10^(-5) mol L^(-1) . Calculate the solubility product.

The conductivity of 0.35M NaCl solution at 298K is 0.025 S/cm. Calculate its molar conductivity .

The molar conductivity at infinite dilution of Al_2(SO_4)_3 is 858S cm^2 mol^-1 . Caclulate the molar ionic conductivity of Al^(3+) ion given that lambda^@(SO_4^(2-))=160 S cm^2 mol^-1 .

The solubility of silver chloride (AgCl) in water at 25^(@)C ia 1.06 xx 10^(-5) mol L^(-1) . Calculate the solubility product of AgCl at this temperature.

The specific conductance of a saturated solution of AgCl at 298K is found to be 1.386xx10^-6S cm^-1 . Calculate its solubility (lambda^@_(Ag^+)=62.0 S cm^2 mol^-1 and lambda^@_(Cl^-)=76.3 S cm^2 mol^-1 ).

The limiting equivalent conductivity of NaCl, KCl and KBr are 126.5,150.0 and 151.5 S cm^(2) eq^(-1) , respectively. The limiting equivalent ionic conductance for Br^(-) is 78 S cm^(2) eq^(-1) . The limiting equivalent ionic conductance for Na^(+) ions would be :