1.16 g CH_(3)(CH_(2))_(n)COOH was burnt in excess air and the resultant gases (CO_(2) and H_(2)O) were passed through excess NaOH solution. The resulting solution was divided in two equal parts. One part requires 50 mL of 1 N HCl for neutralization using phenolphthalein as indicator. Another part required 80 mL of 1 N HCl for neutralization using methyl orange as indicator. Amount of excess NaOH solution taken initially

1.16 g CH_(3)(CH_(2))_(n)COOH was burnt in excess air and the resultan

1.	1.16 g CH_(3)(CH_(2))_(n)COOH was burnt in excess air and the resultant gases (CO_(2) and H_(2)O) were passed through excess NaOH solution. The resulting solution was divided in two equal parts. One part requires 50 mL of 1 N HCl for neutralization using phenolphthalein as indicator. Another part required 80 mL of 1 N HCl for neutralization using methyl orange as indicator. Amount of excess NaOH solution taken initially
Answer» 3.2 gm 6.4 gm 1.2 gm None of these ANSWER :B