At 25°C the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and `BaCl_(2)` are `248 xx 10^(-4) ,126 xx 10^(-4)` and `280xx10^(-4)Sm^(2)mol^(-1)` respectively, `lambda_(m)^(o) Ba(OH)_(2)` in ` Sm^(2)mol^(-1)`is

Options

A. `52.4 xx 10^(-4)`
B. `524 xx 10^(-4)`
C. `402 xx 10^(-4)`
D. `262 xx 10^(-4)`

Answer

B

Explanation

`BaCl_(2) + 2NaOH to Ba(OH)_(2) + 2NaCl`
From the reaction
` lambda_(m)^(infty) = lambda_(m)^(infty) BaCl_(2)+2 lambda_(m)^(infty) NaOH -2 lambda_(m)^(infty) NaCl`
`therefore lambda_(m)^(infty) Ba(OH)_(2)=(280+2xx248-2xx126)10^(-4)`
`524xx10^(-4) Sm^(2)mol^(-1)`