Chemistry - Liquid Solution Question with Solution | TestHub

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 . $(2.50x)+(0.40)(3.00-x)=3.00$ or $2.10x+1.2=3$ Or $2.10x=1.8$ or $x=0.857\mathrm{L}$ of 2.50 M NaOH $\therefore (3-0.857)=2.14$ L of 0.40 M NaOH