The solubility of `Li_3Na_3(AlF_6)_2` is 0.0744 g per 100 ml at 298 K. Calculate the solubility product of the salt. (Atomic masses: Li = 7, Na = 23, Al = 27, F = 19)
| A. |
`2.56 xx 10^(-3)` |
| B. |
`2 xx 10^(-3)` |
| C. |
`7.46 xx 10^(-19)` |
 |
| D. |
`3.46 xx 10^(-12)` |