Question

The volume (in mL) of 0.1 M AgNO₃ required for the complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H₂O)₅Cl]Cl₂, as silver chloride is close to:

A. 6 mL
B. 7 mL
C. 8 mL
D. 9 mL

Answer 1

The closest option to this is A. 6 mL. The volume of AgNO₃ required is 3 mL.

Step-by-step explanation:

To find the volume of 0.1 M AgNO₃ required for the complete precipitation of chloride ions, we need to determine the number of moles of chloride ions present in the solution of [Cr(H₂O)₅Cl]Cl₂. This can be done using the concentration and volume of the [Cr(H₂O)₅Cl]Cl₂ solution:

Number of moles of chloride ions = concentration of [Cr(H₂O)₅Cl]Cl₂ x volume of [Cr(H₂O)₅Cl]Cl₂ solution

= 0.01 mol/L x 0.03 L = 0.0003 mol

Since the reaction between chloride ions and silver nitrate is 1:1, the number of moles of AgNO₃ required will be the same as the number of moles of chloride ions. Therefore, we need 0.0003 mol of AgNO₃.

The volume of AgNO₃ can be calculated using its concentration:

Volume of AgNO₃ = number of moles of AgNO₃ / concentration of AgNO₃

= 0.0003 mol / 0.1 mol/L = 0.003 L = 3 mL

Therefore, the volume of 0.1 M AgNO₃ required for the complete precipitation of chloride ions is 3 mL. The closest option to this is A. 6 mL.