A solution containing `Na_(2)CO_(3)` and NaOH requires 300 mL of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 N HCl is required. The amount of NaOH present in the original solution is :

Options

A. 0.5 g
B. 1 g
C. 2 g
D. 4 g

Answer

B

Explanation

300 mL HCl of 0.1 N neutralises entire amount of NaOH and `1//2 " of" Na_(2)CO_(3)`. Remaining `1//2 " of " Na_(2)CO_(3)` is neutralised by 25 mL 0.2 N HCl, i.e., 50 mL of 0.1 N HCl.
Thus, 250 mL of 0.1 N HCl is required to neutralise NaOH completely.
`N_(1)V_(1)(NaOH)=N_(2)V_(2)(HCl)`
`=0.1xx250`
=25
`W_(NaOH)=(ENV)/(1000)=(40xx25)/(1000)=1 g`