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 ______.

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

After the 14% KOH solution is heated and evaporated off 100 g of water, it becomes 92 ml of 28% KOH solution. What is the mass concentration of this solution?

KOH solution Xg
After evaporation, the mass fraction of solution is doubled.
X-100=1/2 x
X =200
N (KOH)=0.14*200/56
C=n/v=0.14*200/[56*92*10^(-3)]=5.43mol/l