After evaporation of mg water from the KOH solution with a mass fraction of W, the mass fraction became exactly 3 W and the volume was VL (no crystal precipitation in the solution) The mass concentration of the concentrated solution is? Use a comprehensive expression to express it, not to count. After evaporation of mg water from the KOH solution with a mass fraction of W, the mass fraction becomes exactly 3 W and the volume is VL (no crystal precipitation in the solution) The mass concentration of the concentrated solution is? Use a comprehensive expression to express it, not to count.

After evaporation of mg water from the KOH solution with a mass fraction of W, the mass fraction became exactly 3 W and the volume was VL (no crystal precipitation in the solution) The mass concentration of the concentrated solution is? Use a comprehensive expression to express it, not to count. After evaporation of mg water from the KOH solution with a mass fraction of W, the mass fraction becomes exactly 3 W and the volume is VL (no crystal precipitation in the solution) The mass concentration of the concentrated solution is? Use a comprehensive expression to express it, not to count.

C=3mw/112Vmol/L

If a certain amount of KOH solution with a mass fraction of 14% becomes 28% by mass and 125 mL by volume after evaporation of 100 g of water, and no crystal precipitates during evaporation, the mass concentration of the concentrated KOH solution is () A.2.2 mol • L-1 B.4 mol•L-1 C.5 mol•L-1 D.6.25 mol • L-1

If the mass of 14% potassium hydroxide solution is m and the mass of solute before and after evaporation is unchanged, then m×14%=(m-100g)×28%, and m=200g,
The amount of substance containing potassium hydroxide in 28% potassium hydroxide solution is (200−100) g×28%
56G/mol=0.5mol,
The mass concentration of the resulting solution was 0.5 mol
0.125 L =4 mol/L,
Selected from: B.