Molarity word problem?

What molarity of acid would result if i added 200mL of distilled water to 25.0mL of 12 M HCl?

Answer:
M1 x V1 = M2 x V2

(12 M)(25.0 mL) = (M2)(225 mL)

[(300 mL-M) / (225 mL)] = M2
It's a dilution problem so you use

M1V1 = M2V2.

M1= Molarity of Solution 1
V1= Volume of Solution 1
M2= Molarity of Total Solution
V2= Volume of Total Solution

Plug in the numbers.

(12M HCl)(25.0mL) = (M2)(225 mL from 200+25)

You get 1.33 M
You start with 25.0mL of 12 M HCl

n = M * V = 12 mol/L * 0.025 L = 0.3 mol

This solution is diluted with 200 mL of distilled H2O

Total volume: 200 mL + 25 mL = 225 mL = 0.225 L

Molarity: M = n / V = 0.3 mol / 0.225 L = 1.333333 mol/L

End result: 1.333333 M HCL

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