Answer:
2.78 mL of solution will be needed to make 4.5 M solution with 0.50 grams of NaOH.
Explanation:
Moles of sodium hydroxide = n
[tex]n=\frac{0.50 g}{40 g/mol}=0.0125 mol[/tex]
[tex]Moles (n)=Molarity(M)\times Volume (L)[/tex]
Volume of sodium hydroxide solution =V
Molarity of the sodium nitrate solution = 4.5 M
[tex]V=\frac{0.0125 mol}{4.5 M}=0.00278 L[/tex]
V = 0.00278 L = 2.78 mL
2.78 mL of solution will be needed to make 4.5 M solution with 0.50 grams of NaOH.
