How many milliliters of 0.100 m naoh are required to neutralize 9.00 ml of 0.0500 m hcl?

V  ( NaOH ) = mL ?

M ( NaOH ) = 0.100 M

V ( HCl ) = 9.00 mL / 1000 => 0.009 L

M ( HCl ) = 0.0500 M

number of moles HCl:

n = M x V

n = 0.009 x 0.0500 => 0.00045 moles HCl

mole ratio:

HCl + NaOH = NaCl + H2O

 1 mole HCl 1 mole NaOH
 0.00045 moles HCl ??

0.00045 x 1 / 1 => 0.00045 moles of NaOH

M = n / V

0.100 = 0.00045 / V

V = 0.00045 / 0.100

V = 0.0045 L

1 L 1000 mL
0.0045 L ??

0.0045 x 1000 / 1 => 4.5 mL of NaOH

The answer is a) carbon-12