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How would you prepare exactly `3.0 "litre"` of `1.0M NaOH` by mixing proportions of stock solutions of `2.50M NaOH` and `0.40M NaOH`? No water is to be used. |
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Answer» Let `V mL` of `2.50M NaOH` be mixed with `(3000-V) mL` of `0.40M NaOH` Meq. Of `2.50 M NaOH`+ Meq. Of `0.4 M NaOH`=Meq. Of `1.0 M NaOH` `2.50xxV+0.4(3000-v)=3xx1000xx1(M=N)` `:. V=857.14 mL` `:. 857.14 mL` of `2.50 M NaOH` and `2142.96 mL` of `0.4 M NaOH` are to be mixed. |
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