What volume in mL of your HCl solution did you use for each of the three acceptable trial samples for the standardization? Your volume of HCl should be approximately 5 mL. Report your volume to 4 significant figures.

The question is incomplete, here is the complete question:

What volume in mL of your HCl solution did you use for each of the three acceptable trial samples for the standardization?

Trials          Molarity of HCl        Volume of NaOH

  1                     0.1 M                      24.80 mL

 2                     0.2 M                     19.20 mL

 3                     0.01 M                    18.00 mL

The average molarity of NaOH is 0.0755 M

For Trial 1: The volume of HCl required is 18.72 mL

For Trial 2: The volume of HCl required is 7.25 mL

For Trial 3: The volume of HCl required is 135.9 mL

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is HCl

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in above equation, we get:

Hence, the volume of HCl required is 18.72 mL

We are given:

Putting values in above equation, we get:

Hence, the volume of HCl required is 7.25 mL

We are given:

Putting values in above equation, we get:

Hence, the volume of HCl required is 135.9 mL

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3