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.

Explanation

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.