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.

Explanation:

Given: $M_{1}$ = 0.5 M        

$V_{1}$ = 4800 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\4800 mL = 4800 mL \times \frac{0.001 L}{1 mL}\\= 4.8 L$

$M_{2}$ = 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.

$M_{1}V_{1} = M_{2}V_{2}\\0.5 M \times 4.8 L = 3 M \times V_{2}\\V_{2} = \frac{0.5 M \times 4.8 L}{3 M}\\= 0.8 L$

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.