Question

concentration of cl- ions in a mixture obtained by mixing 100 ml of 0.1 m nacl with 40 ml of 0.1 m agno3 is​

Answer 1

Answer:

0.285M

Using

no.of mol of Cl- = 100×0.1/1000 =0.01 mol

no.of mol of Ag+= 40×0.1/1000 = 0.004mol

Reaction is

Cl- + Ag+ = AgCl

AgNO3 is the limiting reactant.

so.no.of mol of AgCl formed

= 0.04 mol

so, 0.04mol of AgCl is formed which is present in

140ml of solution.

concentration of Cl- = [AgCl] = 0.04mol/0.140L

= 0.285M

Answer 2

Answer:

0.285M

Explanation:

Answer:

0.285M

Using

Cl- number of molecules = 100 0.1/1000 = 0.01 mol

Ag+ moles = 40 0.1 / 1000 = 0.004 moles.

Response is

Ag+ + Cl- = AgCl

The limiting reactant is AgNO3.

mol amount of AgCl generated

= 0.04 mol

Consequently, 0.04 mol of AgCl is created and is present in

a solution of 140 ml.

Cl- concentration is [AgCl] = 0.04 mol/0.140 litres.

= 0.285M