2KOH+ H2SO4 ------> K2SO4+ 2H2O
0.2...........0.1...................0.1.........0.2
nH2SO4=(200.4.9%)/98=0.1 mol
a) VddKOH=\(\dfrac{0.2\cdot56}{1.12}\)=10 ml ( V=m/D)
b)mK2SO4=174*0.1=17.4 g
mddKOH=1.12*10=11.2 g (m=D*V)
c) mdd=200+11.2=211.2 g
=>C%K2SO4=(17.4*100)/211.2=8.24%