Answer: In this case, you can estimate that the solubility of potassium nitrate in water at
60
∘
C
is equal to about
solubility
≈
110 g / 100 mL water
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
This tells you that a saturated solution of potassium nitrate will hold about
110 g
of dissolved salt, i.e. of dissociated salt, per
100 mL
of water at
60
∘
C
.
A highly corrosive acid should have a Ph balance between 0-6
I believe your answer is Favorable.
Hope this helps, and happy studying~!
~{Dunsforhands}
