A25.00-ml aliquot of an unstandardized hcl solution is titrated with the previously standardized naoh solution from #1 above. if 32.55 ml of naoh titrant is required to reach the endpoint, what is the exact molarity of the hcl solution?

The neutralization reaction would be:

HCl + NaOH --> NaCl + H₂O

Next, determine the equivalent moles of HCl from the given amount of titrant. However, there is a missing information. You should know the molarity of the NaOH to be able to solve this. Suppose, the concentration is 0.1 M NaOH.

0.1 M NaOH * 0.3255 L = 0.03255 mol NaOH

Since, the molar ratio is 1:1, then that is also equivalent to 0.03255 mol HCl. The molarity would then be:

Molarity = 0.03255 mol/0.25 = 0.1302 M HCl

Ametal with a large number of free flowing electrons most likely has good conductivity

H tricky one let me think and get back to u