Answer:
1.0876 M
Explanation:
Initial Volume of NaOH (V1) = 1234.56 mL
Convert mL to L
1 mL = 0.001 L
1234.56 mL = 1234.56 x 0.001 = 1.2346 L
Initial Molarity of NaOH (M1) = 3.33 M
final Volume of NaOH (V2) = 3.78 L
Final Molarity of NaOH (M2) = ?
Solution:
As we diluting the concentrated NaOH
Note: Molarity = M= mol/L
So, Dilution formula will be used
M1 V1 = M2 V2
Put values in above equation
3.33 mol/L x 1.2346 L = M2 x 3.78 L
Rearrange the above equation
M2 = 3.33 mol/L x 1.2346 L / 3.78 L
M2 = 4.1112 mol / 3.78 L
M2 = 1.0876 M
So, the final molarity of the diluted Sodium hydroxide (NaOH) solution is 1.0876 M.
