Question

How many liters (L) of water need to be added to prepare 4800mL of 0.5M NaCl from a 3M NaCl stock solution? (Use the formula M1V1=M2V2)

Answer 1

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.

Step-by-step explanation:

Given:
`M_(1)` = 0.5 M

`V_(1)` = 4800 mL

Convert mL into L as follows.

`1 mL = 0.001 L\\4800 mL = 4800 mL * (0.001 L)/(1 mL)\\= 4.8 L`

`M_(2)` = 3 M

Formula used to calculate the volume of water required as follows.

`M_(1)V_(1) = M_(2)V_(2)`

Substitute the values into above formula.

`M_(1)V_(1) = M_(2)V_(2)\\0.5 M * 4.8 L = 3 M * V_(2)\\V_(2) = (0.5 M * 4.8 L)/(3 M)\\= 0.8 L`

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.