n = V/Vm = 0,56/22,4 = 0,025 moli NH3
NH3 + H2O => NH4OH
1 mol NH3..............18g H2O.........35g NH4OH
0,025 moli.............x.........................yg
x = 0,45g H2O
y = 0,875g NH4OH
mdf = 0,875g
msf = 0,875+600-0,45= 600,425g
Cf =mdf/msf×100 = 0,875/600,425×100 = 0,145%