Question

When 50.0 ml of a 0.3000 m agno3 solution is added to 50.0 ml of a solution of mgcl2, an agcl precipitate forms immediately. the precipitate is then filtered from the solution, dried, and weighed. if the recovered agcl is found to have a mass of 0.1183 g, what as the concentration of magnesium ions in the original mgcl2 solution?

Answer 1

`0.00826\ molar.`

Step-by-step explanation:

The balanced reaction is:

`2AgNO_3+MgCl_2-->2AgCl+Mg(NO_3)_2`

We know, number of moles =
`(Given\ mass)/(Molar\ mass).`

Therefore, moles of AgCl formed=
`(0.1183)/(143)=0.000827\ moles.` ( Molar mass of AgCl is 143 gm)

From the balanced equation 1 mol of
`MgCl_2` forms 2 mol of AgCl.

Therefore, 0.000827 mol of AgCl was formed by

`(0.000827)/(2)=0.0004135\ mol`

Now concentration of
`MgCl_2=(moles\ of\ MgCl_2)/(Volume\ in\ Liters)=(0.000413)/(0.050)\ molar=0.00826\ molar.`

Hence, this is the required solution.