Answer :
The volume of water that needs to be added to 100.0 mL of 10.0 M HCl(aq) to make 2.00 L of 2.00 M HCl(aq) is 1.98 L.
To calculate the volume of water that needs to be added, we can use the formula for dilution of solutions, which states that:
M₁V₁ = M₂V₂
where M₁ and V₁ are the initial molarity and volume of the concentrated solution, respectively, M₂ and V₂ are the final molarity and volume of the diluted solution, respectively.In this case, we know:
M₁ = 10.0 M (the molarity of the concentrated HCl solution)
V₁ = 100.0 mL (the volume of the concentrated HCl solution)
M₂ = 2.00 M (the desired molarity of the diluted HCl solution)
V₂ = 2.00 L (the desired volume of the diluted HCl solution)
We want to find the volume of water, which we can calculate as:
V_water = V₂ - V₁
where V_water is the volume of water to be added.
First, we need to convert the initial volume to liters:
V₁ = 100.0 mL = 0.100 L
Now we can substitute all the values into the formula and solve for V_water:
M₁V₁ = M₂V₂
10.0 M x 0.100 L = 2.00 M x 2.00 L + V_water
1.00 L = 4.00 L + V_water
V_water = 1.00 L - 4.00 L
V_water = -3.00 L
The result is negative, which means that we cannot obtain the desired concentration by adding water to the initial solution. Instead, we would need to dilute the initial solution by adding more water than the initial volume. Specifically, we would need to add:
V_water = |(M₂V₂ - M₁V₁) / M₂| - V₁
V_water = |(2.00 M x 2.00 L - 10.0 M x 0.100 L) / 2.00 M| - 0.100 L
V_water = 1.90 L
Therefore, we need to add 1.90 L of water to 100.0 mL of 10.0 M HCl(aq) to obtain 2.00 L of 2.00 M HCl(aq).
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