Gravimetric Analysis (1 Viewer)

[Smithereens]

Got a question, but having difficulties solving it.

The percentage by weight of chloride in a sample of rock salt was determined by graviemtric analysis. 2.0 g of the rock salt was weighed, and dissolved in 50mL of distilled water. Silver nitrate was added in excess in order to precipitate the chloride ions as white silver chloride. The mixture was filtered, and the precipitate was washed and dried till constant weight was achieved.

The mass of the silver chloride precipitate was 4.653 g. Calculate the percentage weight of chloride in the rock salt.

 

[Forbidden.]

1M of Silver Chloride = 180.5g

You got 4.653g so that's 2.578 x 10^-2M

x = (4.653/180.5) = 2.578 x 10^-2M ... That's the molarity of your salt.
In percentage that's 2.578% of 1M.
Now Ag⁺ + Cl^- = AgCl ... all coefficents are 1 so they're all of equal concentration. No need to change

Now on the periodic table, Cloride is 35.5 in atomic weight, so 1M is 35.5g.
Multiply 35.5 by 0.2578 = 0.915132964....
(If you do Maths as well you will lose marks if you don't write full calculator display, in Chemistry I doubt that matters.)


Therefore there is 0.92g of Chloride in 4.653g of Silver Chloride salt.

 

[Aerlinn]

The percentage by weight of chloride in a sample of rock salt was determined by graviemtric analysis. 2.0 g of the rock salt was weighed, and dissolved in 50mL of distilled water. Silver nitrate was added in excess in order to precipitate the chloride ions as white silver chloride. The mixture was filtered, and the precipitate was washed and dried till constant weight was achieved.

The mass of the silver chloride precipitate was 4.653 g. Calculate the percentage weight of chloride in the rock salt.

Here's how I would do it:
-Using 4.653g, find the number of mole using n=m/M
-This = the no. mole of Choride ions.
-Convert this no. mole to a mass to find the mass of chloride ions.
-Divide this mass by 2.0g, then times by 100 to get the % you want for your answer...

 

[Kujah]

1M of Silver Chloride = 180.5g

You got 4.653g so that's 2.578 x 10^-2M

x = (4.653/180.5) = 2.578 x 10^-2M ... That's the molarity of your salt.
In percentage that's 2.578% of 1M.
Now Ag⁺ + Cl^- = AgCl ... all coefficents are 1 so they're all of equal concentration. No need to change

Now on the periodic table, Cloride is 35.5 in atomic weight, so 1M is 35.5g.
Multiply 35.5 by 0.2578 = 0.915132964....
(If you do Maths as well you will lose marks if you don't write full calculator display, in Chemistry I doubt that matters.)


Therefore there is 0.92g of Chloride in 4.653g of Silver Chloride salt.

Are you sure 1M of AgCl = 180.5?

 

[Flaming Ninja]

I think the real answer is that you should drop Chemistry. It is the worst subject ever.