After a bottle of 14% potassium hydroxide solution was heated and evaporated off 100 g of water,80 mL of 28% potassium hydroxide solution was obtained at a concentration of () A.5 mol/L B.6 mol/L C.6.25mol /L D.6.5mol/L

After a bottle of 14% potassium hydroxide solution was heated and evaporated off 100 g of water,80 mL of 28% potassium hydroxide solution was obtained at a concentration of () A.5 mol/L B.6 mol/L C.6.25mol /L D.6.5mol/L

If the mass of the 14% potassium hydroxide solution is x and the mass of the solute before and after evaporation remains unchanged, then x×14%=(x-100g)×28%, x=200g,
The amount of the substance containing potassium hydroxide in the 28% potassium hydroxide solution was 100 g×28%
56G/mol=0.5mol,
The mass concentration of the resulting solution was 0.5 mol
0.08L=6.25mol/L,
Therefore, C.

After evaporating 100 g of water from 14% KOH solution,80 mL of 28% KOH solution was obtained, and the mass concentration of this solution was ______.

Assuming that the mass of solution before evaporation is m, it can be seen from the constant mass of substance before and after evaporation that:14%×m=28%×(m-100), the solution is: m=200g,
So n (KOH)=200g×14%
56G/mol=0.5mol,
Mass concentration of solution after evaporation C=n
V=0.5mol
0.08L=6.25mol/L,
Therefore, the answer is:6.25mol/L