Answer to Question #313555 in Physical Chemistry for Nelly

Question

25cm^3 of magnesium Chloride compleatly reacted with 12.5cm^3 of 0.2M silver nitrate to precipitate silver Chloride .Haw many grams of magnesium Chloride are contained in 500cm^3 of the solution

Answer 1

 

Moles of silver nitrate:

V = 12.5 cm³ = 12.5 mL = 0.0125 L

0.0125 L x 0.2 M = 0.0025 mol

MgCl₂ + 2AgNO₃ = Mg(NO₃)₂ + 2AgCl

The ratio of the reactants is 1 : 2

So, moles of MgCl₂: 0.0025 x 1/2 = 0.00125 mol – in 25 cm³ of the solution

In 500 cm³: 0.00125 x 500 / 25 = 0.025 mol

Molar mass of MgCl₂: 95.2 g/mol

Mass = 0.025 x 95.2 = 2.38 g