How many ml of 0.01-m naoh will be required to titrate 0.061-g of khp to it's endpoint?

The volume of NaOH needed for the titration is calculated by equating the number of moles of KHP and NAOH.

(1) number of moles of KHP:
         n = mass of KHP / molar mass of KHP
Substituting,
         n = 0.061 g / (204.22 g/mol) = 2.987 x 10⁻⁴ moles KHP

(2) number of moles of NaOH
         n = (volume NaOH in L)(molarity of NaOH)
          n = V x 0.01

Equating,
       0.01V = 2.987 x 10⁻⁴

The value of V from the equation is 0.0299L of NaOH. 

Converting the value to mL will give us the answer of 29.87 mL. 

It’s b. mass i searched it up

Density and pressure