Explanation:

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in equals space 38 over 365 space mol space equals space 1.04 space mol therefore space Molarity space of space HCl space solution space equals space fraction numerator 1.04 space mol over denominator 84.1000 space straight L end fraction space equals space bold 12 bold. bold 4 bold space bold mol bold space bold L to the power of bold minus bold 1 end exponent left parenthesis straight b RIGHT parenthesis space Molarity space of space conc. space HCl space sample space equals space 12.38 space mol DIVIDED by straight L Molarity space of space HCl space solution space to space be space prepared space equals space 0.10 space mol divided by straight L Volume space of space HCl space solution space to space be space prepared space equals space 1.00 space straight L space equals space 1000 space mL By space using space molarity space equation comma straight M 1 straight V 1 space equals space straight M 2 straight V 2 VHCl space equals space fraction numerator 0.10 space cross times space 1000 over denominator 12.38 end fraction space mL space equals space bold 8 bold. bold 07 bold space bold mL Thus comma space to space obtain space 1.0 straight L space of space 0.10 space straight M space HCl comma space one space should space dissolve space 8.08 space mL space of space CONCENTRATED space HCl space to space make space the space volume space to space 1.0 straight L. end style