Calculate the concentration of NaOH solution in g/mL, which has the same normality as that of a solution of `HCl` of concentration 0.04 g/mL.

Explanation

`N_(HCl)=(w_Bxx1000)/(E_BxxV)=(0.04xx1000)/(36.5xx1)=1.095`
`N_(NaOH)=N_(HCl)`
`:.1.095=(w_Bxx1000)/(40xx1)`
`w_B=0.0438g//mL`