1. The problem below concern CACL2. a. 0.135 gram of cacl2 is dissolved to 250 m
ID: 595768 • Letter: 1
Question
1. The problem below concern CACL2.a. 0.135 gram of cacl2 is dissolved to 250 ml. What is the prepared molarity of the solution
b. what is the equilibrium molarity of chloride and the pcl value?
c. A sample of cacl2 is dissolved in 250ml. A 5.00ml aliquot of the solution is dissolved to 250ml. Then a 5.00ml of the second solution is dissolved in 500ml. The final solution is analyzed and the chloride concentration is 3.60*10^-5 M. If all the chloride came from cacl2 how many grams cacl2 were added to the first solution?
Explanation / Answer
a) Moles = 0.135 /111 =0.001216 Molarity = moles/litre = 0.001216/0.250 =0.004865 M b)Molarity of Chloride = moles of cholride / litre = 2 x Molratiy of soln = 0.00973 pCL = -log [CL]- = 2.0119 c)suppose initial moles = X molarity = x/0.25 next molarity = Xx0.005/0.25^2 moles of CaCl2 in last soln = 0.005^2 x X / 0.25^2 moles of Chlorine = 2x0.005^2 x X / 0.25^2 [Cl]- = 3.6x10^-5 M moles of [Cl]- = M x volume = 3.6x10^-5 x 0.505 = 2x0.005^2 x X / 0.25^2 X= 0.022725 gms of CaCl2 =111x moles = 2.52 gms
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