For preparing 3.00 L of 1 M NaOH by mixing portions of two stock...

Moles needed $=(3.00 \mathrm{~L})(1.00 \mathrm{M})=3 \mathrm{~mol}$. Suppose that $x \mathrm{~L}$ of 2.50 M NaOH added, then $(3-x) \mathrm{L}$ of 0.40 M NaOH added. The number of moles of solute from the more concentrated solution is 2.50 , that become less concentrated solution is $(0.40)(3.00-x)$. The total number of moles is 3.00 . $(2.50 x)+(0.40)(3.00-x)=3.00$ or $2.10 x+1.2=3$ Or $2.10 x=1.8$ or $x=0.857 \mathrm{~L}$ of 2.50 M NaOH $\therefore \quad(3-0.857)=2.14$ L of 0.40 M NaOH