Question

A 0.2589 g sample of CaCO3 is dissolved in 6 M HCl and the resulting solution is diluted to 250.0 mL in a volumetric flask. Titration of a 25.00 mL sample of the solution requires 29.55 mL of EDTA to reach the Eriochrome Black T end point. How many moles of CaCO3 (CaCl2) are in the 25.00 mL aliquot? Select all that apply.

Answer 1

Moles of
`CaCO_3` = 0.00026 moles

Step-by-step explanation:

Mass of
`CaCO_3` = 0.2589 g

Molar mass of
`CaCO_3` = 100.0869 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (0.2589\ g)/(100.0869\ g/mol)`

`Moles\ of\ CaCO_3= 0.0026\ mol`

Also,

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Volume = 250 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 250×10⁻³ L = 0.250 L

So,

`Molarity=(0.0026)/(0.250)`

Molarity of the sample = 0.0104 M

Considering:

`Moles =Molarity * {Volume\ of\ the\ solution}`

Volume = 25.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.00×10⁻³ L

Thus, moles of
`CaCO_3` :

`Moles=0.0104 * {25.0* 10^(-3)}\ moles`

Moles of
`CaCO_3` = 0.00026 moles