How to prepare 250.00 ml of approximately 1.0 m hcl solution from the 2.5 m hcl solutio?
To solve this we use the
equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the
volume of the stock solution, M2 is the concentration of the new solution and
V2 is its volume.
2.5 M x V1 = 1.0 M x .250 L
V1 = 0.10 L or 100 mL of the 2.5 M HCl solution is needed
Hope this helps.
Answer: Add 150 ml of water to 100 ml of 2.5 M stock solution.
Explanation: According to the neutralization law,
\(M_1V_1=M_2V_2\)
where,
\(M_1\) = molarity of stock solution = 2.5 M
\(V_1\) = volume of stock solution = ?
\(M_2\) = molarity of desired solution = 1.0 M
\(V_2\) = volume of desired solution = 250 ml
Now put all the given values in the above law, we get the volume of stock solution.
\((2.5M)\times V_1=(1.0M)\times (250ml)\)
\(V_1=100ml\)
Therefore, 100 ml of 2.5 M of stock solution is taken and 150 ml of water is added to make the volume to 250 ml.