Question

Question 11 Complete 0.00 points out of 3.00 P Flag question If 1.71 mL of a 0.118 M NaOH solution were required to completely neutralize an acid solution, what was the total number of moles of base dispensed? Make sure you have the correct number of significant figures. Do not include the units in your answer. Answer: 0.000201 ancante involved in this question from the link found

Answer 1

Q.8) We know that, Molarity = No. of moles of solute / volume of solution in L

$\therefore$ No. of moles of solute = Molarity $\times$ volume of solution in L

$\therefore$ No. of moles of NaOH added to acid solution = [ NaOH ]  $\times$ volume of solution in L

We have, [ NaOH ] = 0.118 mol / L & Volume of solution = 1.71 ml = 0.00171 L

$\therefore$ No. of moles of NaOH added to acid solution = 0.118 mol / L $\times$ 0.00171 L = 0.000202 mol

ANSWER : 0.000202

Q.9 ) Volume of titrant added = Final volume - Initial volume

Volume of titrant added = 1.54 ml - 0.40 ml = 1.14 ml

ANSWER : 1.14