how many milliliters of10.0m Hc(aq) are needed to prepare 890.0 ml of 1.00m Hcl(aq)
One thing you need to learn in chemistry:
A MUST. Locate the caps key on your computer and USE it.
m is not M. I assume you meant molarity (M) and not molality(m). If it is m you can't work the problem with the information provided. If it is M, then
mL1 x M1 = mL2 x M2
mL1 x 10.0M = 890.0 mL x 1.00M
Solve for mL 1 you need to use to make the 890.0 of 1.00M HCl.
To determine how many milliliters of 10.0 M HCl(aq) is needed to prepare 890.0 ml of 1.00 M HCl(aq), we can use the formula:
C₁V₁ = C₂V₂
Where:
C₁ = concentration of the stock solution (10.0 M)
V₁ = volume of the stock solution needed
C₂ = concentration of the desired solution (1.00 M)
V₂ = volume of the desired solution (890.0 ml)
Rearranging the formula, we get:
V₁ = (C₂ * V₂) / C₁
Now, let's substitute the values into the formula:
V₁ = (1.00 M * 890.0 ml) / 10.0 M
V₁ = 89.0 ml
Therefore, you need 89.0 milliliters of 10.0 M HCl(aq) to prepare 890.0 ml of 1.00 M HCl(aq).