A0.2589 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.

Moles of  = 0.00026 moles

Explanation:

Mass of = 0.2589 g

Molar mass of = 100.0869 g/mol

The formula for the calculation of moles is shown below:

Thus,

Also,

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 of the sample = 0.0104 M

Considering:

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 :

Moles of  = 0.00026 moles

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explanation:

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