The question is incomplete, the complete question is;
2. How many moles of NaOH are there in a 20.0mL sample of an 0.101M solution?
3. How many grams of benzoic acid would be required to react with the NaOH
sample mention in Question 2?
Answer:
0.247 g
Step-by-step explanation:
Recall that;
n =CV
n = number of moles
C = concentration
V = volume (in litres)
So;
n = 0.101 * 20/1000
n = 2.02 * 10^-3 moles
The equation of the reaction is;
C6H5COOH(s)+NaOH(aq) ---------> C6H5COO-Na+ (aq)+H2O(l)
Given that the reaction is in 1:1 molar ratio
number of moles of benzoic acid required= 2.02 * 10^-3 moles * 122.12 g/mol = 0.247 g