nK = \(\dfrac{19,5}{39}\) = 0,5 mol
2K + 2H2O -> 2KOH + H2
0,5 ->0,5 ->0,25
=>C% = \(\dfrac{0,5.56}{19,5+261-0,25.2}\).100% = 10%
\(C\%=m_{ct}:m_{dd}.100\%=19,5:261.100\%=7,4\%\)
\(C\%_{dd}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{19,5}{280,5}.100\%\approx6,9\%\)
\(C\%_{dd}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{19,5}{19,5+261}.100\%\approx6,95\%\)