Answer: There is 0.8 liters (L) of water required to be added to prepare 4800mL of 0.5M NaCl from a 3M NaCl stock solution.
Explanation:
Given: [tex]M_{1}[/tex] = 0.5 M
[tex]V_{1}[/tex] = 4800 mL
Convert mL into L as follows.
[tex]1 mL = 0.001 L\\4800 mL = 4800 mL \times \frac{0.001 L}{1 mL}\\= 4.8 L[/tex]
[tex]M_{2}[/tex] = 3 M
Formula used to calculate the volume of water required as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
Substitute the values into above formula.
[tex]M_{1}V_{1} = M_{2}V_{2}\\0.5 M \times 4.8 L = 3 M \times V_{2}\\V_{2} = \frac{0.5 M \times 4.8 L}{3 M}\\= 0.8 L[/tex]
Thus, we can conclude that 0.8 liters (L) of water need to be added to prepare 4800mL of 0.5M NaCl from a 3M NaCl stock solution.
