A bottle of 14% KOH solution was heated and evaporated off 100 g of water to 80 mL of 28% KOH solution with an amount-concentration of () A.5 mol•L-1 B.6 mol•L-1 C.6.25 mol • L-1 D.6.75 mol • L-1 A bottle of 14% KOH solution was heated and evaporated off 100 g of water to 80 mL of 28% KOH solution in an amount of () A.5 mol•L-1 B.6 mol•L-1 C.6.25 mol • L-1 D.6.75 mol • L-1

A bottle of 14% KOH solution was heated and evaporated off 100 g of water to 80 mL of 28% KOH solution with an amount-concentration of () A.5 mol•L-1 B.6 mol•L-1 C.6.25 mol • L-1 D.6.75 mol • L-1 A bottle of 14% KOH solution was heated and evaporated off 100 g of water to 80 mL of 28% KOH solution in an amount of () A.5 mol•L-1 B.6 mol•L-1 C.6.25 mol • L-1 D.6.75 mol • L-1

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%, and x=200g is obtained.
Therefore, the mass of the solution after evaporation is 200 g-100 g=100 g,
Therefore, the amount of the substance containing potassium hydroxide in the 28% potassium hydroxide solution is 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 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.