A sample of freshly precipitated silver chloride weighs 704.7mg. Before the solution can be filtered and washed 10.0mg of AgCl is photodecomposed. After the precipitate is filtered and washed another 10.0 mg of AgCl is photodecomposed. Calculate the weight of the precipitate.
AgCl is converted to Ag metal in the photodecomposition so the 10 mg that decomposed weighs
10mg x (atomic mass Ag/molar mass AgCl) = ?
The other 10 mg does the same = ?
So the new weight of the AgCl ppt is
704.7 mg - ? - ? = x mg AgCl + Ag.
689.64 mg is the new ppt? am i right?
To calculate the weight of the precipitate, we need to subtract the weight of the decomposed AgCl from the initial weight of the sample.
Given:
Weight of freshly precipitated AgCl = 704.7 mg
Weight of AgCl photodecomposed before filtering and washing = 10.0 mg
Weight of AgCl photodecomposed after filtering and washing = 10.0 mg
Step 1: Calculate the total weight of AgCl photodecomposed.
Total weight of AgCl photodecomposed = Weight of AgCl photodecomposed before filtering and washing + Weight of AgCl photodecomposed after filtering and washing
Total weight of AgCl photodecomposed = 10.0 mg + 10.0 mg
Total weight of AgCl photodecomposed = 20.0 mg
Step 2: Calculate the weight of the precipitate.
Weight of the precipitate = Weight of freshly precipitated AgCl - Total weight of AgCl photodecomposed
Weight of the precipitate = 704.7 mg - 20.0 mg
Weight of the precipitate = 684.7 mg
Therefore, the weight of the precipitate is 684.7 mg.
To calculate the weight of the precipitate, we need to determine the remaining weight of silver chloride after the photodecomposition.
Let's break down the information provided step by step:
1. The initial weight of the freshly precipitated silver chloride sample is 704.7 mg.
2. Before the filtration and washing, 10.0 mg of AgCl is photodecomposed. Therefore, we deduct this from the initial weight.
Initial weight - First photodecomposition = Remaining weight
704.7 mg - 10.0 mg = 694.7 mg
3. After the precipitate is filtered and washed, an additional 10.0 mg of AgCl is photodecomposed. Now, we again deduct this weight from the previous remaining weight.
Remaining weight - Second photodecomposition = Final weight of the precipitate
694.7 mg - 10.0 mg = 684.7 mg
Therefore, the weight of the precipitate is 684.7 mg.