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?

1 Answer
May 1, 2018

And so the solubility of MnCl2 is 5.75molL1..I am not sure if I believe that....I get a mass of approx. 90g...

Explanation:

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

125mL×103LmL1×5.75molL1=0.719mol

...the which represents a mass of .......125.84gmol1×0.719mol=??g

The question proposes quite an unrealistic scenario...