Question

To neutralize 8.34 ml of HCl 36.5% by mass and density =
1, 2 g/cm3, the required volume of 5 mol/L NaOH will be?

Answer 1

What is the molarity of the HCl? That's 1.2 g/cc x 1000cc x 0.65 x (1/36.5) = ?
Then mols HCl = M x L = ?
mols NaOH = mols HCl
Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and M NaOH, solve for L NaOH. Convert to mL if needed.

Answer 2

Awesome ! thanks!@ Drbob

Answer 3

What is the molarity of the HCl? That's 1.2 g/cc x 1000cc x 0.65 x (1/36.5) = ?

That line should read
1.2 g/cc x 1000 cc x 0.365 x (1/36.5) = ?
It's the .65 that is not right. It should be 0.365 and that comes from the 36.5%