Correct Answer - B
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`