48.70 mL of an HCl stock solution was transferred into a 500 mL volumetric flask and filled to the mark. 47.67 mL of the diluted solution was titrated with 0.179 M NaOH and found to have the concentration of 0.885 M. What is the concentration of the stock solution?
The concn of the titrated solution is 0.885M. It came from the 500 mL flask; therefore, it must be 0.885M also. How many moles HCl are in the 500 mL flask? That will be 0.885 x 0.500L = 0.4425 moles.
All of those moles came from the 48.70 mL stock solution; thus, the M of the stock soln is 0.4425 moles/L = about 9M but that's just an estimate.