What volume of 12.1M HCl(aq) is required to make 435 mL of 0.100M HCl?
would i just follow the equation
V1xM1=V2M2
V1x12.1=.435Lx.100M = 0.00360 L
Yes, you are correct. To solve this problem, you can use the equation V1 * M1 = V2 * M2, also known as the dilution equation. Let's go through the steps to find the volume of 12.1M HCl(aq) required to make 435 mL of 0.100M HCl.
First, let's assign the values to the variables in the equation:
V1 = volume of 12.1M HCl(aq) (unknown)
M1 = molarity of 12.1M HCl(aq) = 12.1 M
V2 = volume of 0.100M HCl = 435 mL = 0.435 L
M2 = molarity of 0.100M HCl = 0.100 M
Now, let's substitute these values into the equation and solve for V1:
V1 * M1 = V2 * M2
V1 * 12.1 = 0.435 * 0.100
V1 * 12.1 = 0.0435
To solve for V1, divide both sides of the equation by 12.1:
V1 = 0.0435 / 12.1
V1 ≈ 0.00360 L
So, the volume of 12.1M HCl(aq) required to make 435 mL of 0.100M HCl is approximately 0.00360 L.