Question

A sample of 8.4 grams of NaOH is dissolved into 620 mL of aqueous 0.250 M NaOH (assume no volume change). This solution is then poured into 1.65 gallons of water. (You may assume that the two volumes can be added.) What is the concentration of NaOH in the final solution

Answer 1

Answer:

The concentration of NaOH in the final solution 0.0584 mole/lit

Explanation:

• $moles of NaOH =$ $\frac{Weight}{Molecular weight}$ $=\frac{8.4}{40}=0.21 mole$

• 620 ml of aqueous 0.25 M NaOH $=0.25X 620 = 0.155mole$

• Total moles = 0.21 + 0.155 = 0.365 moles
• 1.65 gallons = 3.785 lit x 1.65 = 6.245929 lit  (∵ 1 gallon = 3.785 lit)

• $Molarity =\frac{No. of moles}{Volume(lit)}$

So Molarity of the solution after mixing $=\frac{0.365}{6.245929} = 0.0584 (M)$

∴ The concentration of NaOH in the final solution 0.0584 mole/lit