The correct option is D 50
Let x and y are the mEq of NaOH and Na2CO3, respectively.
Phenolphthalein as indicator:
mEq of NaOH=12 mEq of Na2CO3= mEq of HCl
x+y2=12×1×20=10 ....(i)
Methyl orange as indicator:
mEq of Na2CO3(left)2= mEq of HCl
y2=12×8=4 ....(ii)
y=8,x=6
Weight of Na2CO3=8×10−3×1062=0.424g
% of Na2CO3=0.4240.848×100=50