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
anor277
May 1, 2018

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...

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