The density of seawater is 1.025 g/ml. a. The mass percentage of chloride ion in
ID: 886730 • Letter: T
Question
The density of seawater is 1.025 g/ml.
a. The mass percentage of chloride ion in a 25.00-ml sample of seawater was determined by titrating the sample with silver nitrate,precipitating silver chloride. It took 42.58 mol of 0.2997 M AgN0 3 solution to
react completely with the chloride present. What is the mass percentage of chloride ion in seawater?
b. The average concentration of bromide ion in seawater is 65.0 mg per kg of seawater (i.e.,65 ppm).
What is the molarity of the bromide ion in seawater? Write a set of (bullet-point) directions for preparing 100.0 mol f a solution with this same molarity,beginning with a 0.00350 M stock solution of NaBr.
c. Br2 can be extracted from seawater by using Cl2(g) to oxidize 8( (after adjusting the seawater pH from
than the theoretical amount,and H 2S04 is used to adjust the pH. What masses of H 2S04 and Ch would be needed to extract the available bromine from 1.00 x 103 L of seawater that has the density and
bromide content described earlier in this problem?
Explanation / Answer
There are multiple questions here. . i am allowed to answer only 1 at a time. I will answer question 1 for you.Please ask other as different question.
I think something is wrong in wording of question.
. It took 42.58 mol of 0.2997 M AgN03.It should have been 42.58 mL
AgNO3 + CL- ---->AgCl + NO3-
1 mol of AgNO3 requires 1 mol of CL- to precipitate.
mol of AgNO3 = M*V = 0.04258*0.2997 = 0.01276 mol
Since it took 42.58 mol of AgNO3, Number of moles of Cl- should be 0.01276 mol
Molar mass of Cl= 35.5 gm
mass of Cl-= 0.01276 *35.5 =0.453 gm
Volume = 25mL
total mass= volume * density = 1.025*25 = 25.625 gm
mass % =mass of Cl - *100 /total mass
= 0.453*100 / 25.625
=1.77 %