<span>same no of atoms
b/c
no. of atoms = moles * avogadro's no (6.022 X 10^23)</span>
<span>1.67 x 10^-3 moles
Calculate the molar mass of C9H8O4
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
C9H8O4 = 9 * 12.0107 + 8 * 1.00794 + 4 * 15.999 = 180.1558
Now divide the mass you have by the molar mass
0.300 g / 180.1558 g/mole = 1.665225 x 10^-3
Round to 3 significant figures
1.67 x 10^-3</span>
Answer:
<u>20.25 mL.</u>
Explanation:
The volume of base required for the titration can be derived by removing the base titrant volume from the volume at endpoint.
i.e Final volume - initial volume
= (22.08 - 1.83)mL
<u>= 20.25 mL.</u>
<em>(Repeat and average volume results for accuracy.)</em>
