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

0.0165 m The balanced equation for the reaction is AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2 So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights. Atomic weight silver = 107.8682 Atomic weight chlorine = 35.453 Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol Now how many moles were produced? 0.1183 g / 143.3212 g/mol = 0.000825419 mol So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division. 0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m Rounding to 3 significant figures gives 0.0165 m

Answer 2

Answer:

$0.00826\ molar.$

Explanation:

The balanced reaction is:

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

We know, number of moles = $\dfrac{Given\ mass}{Molar\ mass}.$

Therefore, moles of AgCl formed=$\dfrac{0.1183}{143}=0.000827\ moles.$        ( 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

$\dfrac{0.000827}{2}=0.0004135\ mol$

Now concentration of $MgCl_2=\dfrac{moles\ of\ MgCl_2}{Volume\ in\ Liters}=\dfrac{0.000413}{0.050}\ molar=0.00826\ molar.$

Hence, this is the required solution.