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Calculate the concentration of NaOH solution g/mL, which has the same normality as that of a solution of `HCl` of concentration 0.04 g/mL. |
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Answer» `N_(HCl)=(w_(B)xx1000)/(E_(B)xxV)=(0.04xx1000)/(36.5xx1)=1.095` `N_(NaOH)-=N_(HCl)` `:. 1.095=(w_(B)xx1000)/(40xx1)` `w_(B)=0.0438g//mL`. |
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