Question

Manganese (II) chloride has a solubility of 5.75 M at 0 degrees C. If 125 mL of saturated MnCl2 is evaporated to dryness, what mass of MnCl2 will be left?

Answer 1

And so the solubility of `MnCl_2` is `5.75*mol*L^-1`..I am not sure if I believe that....I get a mass of approx. `90*g`...

Explanation:

We gots .................

`125*mLxx10^-3*L*mL^-1xx5.75*mol*L^-1=0.719*mol`

...the which represents a mass of .......`125.84*g*mol^-1xx0.719*mol=??*g`

The question proposes quite an unrealistic scenario...