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

Answer:

Moles of $CaCO_3$  = 0.00026 moles

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.2589\ g}{100.0869\ g/mol}$

$Moles\ of\ CaCO_3= 0.0026\ mol$

Also,

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

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,

$Molarity=\frac{0.0026}{0.250}$

Molarity of the sample = 0.0104 M

Considering:

$Moles =Molarity \times {Volume\ of\ the\ solution}$

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 $CaCO_3$ :

$Moles=0.0104 \times {25.0\times 10^{-3}}\ moles$

Moles of $CaCO_3$  = 0.00026 moles