Determine the Weight f the Sample, if
it Used 24.9ml of 0.1Normal Sodium
Hydroxide of Titrant {33.29% of
Busulfan (C7H6O2)} with solution
Volume of solution = 24.9 mL, Normality of solution = 0.1 N, Molar mass of Bisulphan = C7H6O2 = 84+6+32 = 122 g/mole
Mole of NaOH = 0.0025 mole
Mole of Busulphan = 0.0025
Mass of Busulphan = 0.0025 "\\times122= 0.305"
Mass of titration = 14 2 g
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