Answer:
Explanation: n=m/M(molar mass)
n=24.3 grams/(16+2x1.008)grams/moles(molar mass of H2O)
n=24.3grams/18.016grams/moles
n=1.35moles
This problem is providing information about the volume and concentration of HCl as 505 mL and 0.160 M respectively, and it is asking for the water that have to be added to prepare a 0.100-M solution.
In such a way, we work over the assumption of constant moles in dilution processes, so that we are able to write:

Which relates de volume and concentration at the beginning and end of the experiment. This means we can solve for the resulting volume of the solution as a first calculation:

Hence, the following amount of water must be added:

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Two things can apply here he can either broaden his scope based on the existing Data or Come up with a new way to test the question