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. Produced mole of the CO_(2)

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. Produced mole of the CO_(2)
Answer» 0.1 0.01 0.06 None of these ANSWER :C