How many milliliters of 675 m hcl(aq) are needed to prepare 11.5 ml of 1.00 m hcl(aq)?

I'm going to make the assumption that it is meant to be 0.675 M HCl since the highest molarity of HCl is 12M. I have attached a conversion. The 11.5 ml was converted to L in my head. If you have any more questions or if my assumption was wrong, just tell me/ask

Hello! data: p (pressure) = 1.25 atmv (volume) = ? (in l)t (temperature) = 298 kn (number of mols) = 0.900 molesr (gas constant) = 0.082 (atm*l/mol*k)apply the data to the clapeyron equation (ideal gas equation), see: i hope this . =)