The volume of distilled water that should be added to 10.0 ml of 6.00 m hcl(aq) in order to prepare a 0.500 m hcl(aq) solution is approximately

Volume can be defined as the capacity of a given container or object.

Dilution is the process of reducing the concentration of a given solution by adding more water to that solution.

The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution is approximately 110ml

The formula to solve  this question is given as;

C1V1 = C2V2

Where

C1 = Initial concentration of the solution

V1 = Initial volume of the solution

C2 = Fnal concentration of the solution

V2 =  Final volume of the solution.

We know from the question that

C1 = 6M of HCl(aq)

C2 = 0.5M of HCl(aq)

V1 = 10ml of HCl

V2= Volume of distilled water.

Applying the formula:

6 M x 10 ml = 0.5M x V2

Divide both sides by 0.5M

V2 = 6 M x 10 ml/0.5M

V2 = 120 ml

The Volume of water added = Final volume -Initial volume

Final volume = 120 ml

Initial volume = 10 ml

Hence,

120ml - 10ml = 110ml

The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution is approximately 110ml.

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Solution is defined as the mixture of two or more substance. In solution, the solute and solvent are present. The volume of solution in the given data is 110ml.

Volume can be defined as the capacity of the given container. The dilution is the process of reducing the concentration of the given solution by adding water.

Given that,

From the formula:

C1V1 = C2V2

where,

Applying the formula:

Dividing both sides by 0.5 M, we get:

Now, the volume required to added:

Therefore, the volume required to prepare 0.5 M HCl is 110 ml.

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43.452

step-by-step explanation:

284 times .153 = 43.452