What volume of a 2.00 M HCL solution can be prepared by diluting 0.350L of 14.0 M HCl solution?
To determine the volume of a 2.00 M HCl solution that can be prepared by diluting 0.350 L of a 14.0 M HCl solution, we can use the equation for dilution:
M₁V₁ = M₂V₂
Where:
M₁ = initial concentration (14.0 M)
V₁ = initial volume (0.350 L)
M₂ = final concentration (2.00 M)
V₂ = final volume (unknown)
To solve for V₂, we rearrange the equation:
V₂ = (M₁V₁) / M₂
Substituting the given values:
V₂ = (14.0 M * 0.350 L) / 2.00 M
V₂ = 2.45 L
Therefore, a volume of 2.45 L of a 2.00 M HCl solution can be prepared by diluting 0.350 L of a 14.0 M HCl solution.