1 mL of H3PO4 was diluted to 250 mL. 25 mL this solution required 40.0 mL of 0.10 N NaOH for neutralisation using phenolphthalein as indicator. The specific gravity of acid is

mEq H3PO4 =mEq of NaOH25 mL ×N = 40 mL × 0.1 NN of H3PO4 = 0.16N=W2×1000Ew2×volsolMw H3PO4=98,Ew=983(n factor =3)0.16=W2×100098/3×250,Wt H3PO4mL−1=1.32gmL−1dH3PO4=W2vol=1.321 mL=1.32gmL−1