27. Calculate the amount of Fe+2 ions in 500 ml Mohr’s salt solution if 10ml of this solution is titrated with 7.5ml 0.06N KMnO4 solution. (1ml 1N KMnO4= 0.05584 gm of F+2). Ans: 1.256 gm
Question
- Calculate the amount of Fe+2 ions in 500 ml Mohr’s salt solution if 10ml of this solution is titrated with 7.5ml 0.06N KMnO4 solution. (1ml 1N KMnO4= 0.05584 gm of F+2). Ans: 1.256 gm
Solution
Sure, here is the step by step solution:
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First, we need to calculate the amount of Fe+2 ions in 10 ml of Mohr’s salt solution. We know that 1 ml of 1N KMnO4 reacts with 0.05584 gm of Fe+2. Therefore, 7.5 ml of 0.06N KMnO4 will react with:
7.5 ml * 0.06N * 0.05584 gm/ml = 0.0252 gm of Fe+2
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Now, we know that 10 ml of Mohr’s salt solution contains 0.0252 gm of Fe+2. Therefore, 500 ml of the same solution will contain:
(0.0252 gm/10 ml) * 500 ml = 1.26 gm of Fe+2
So, the amount of Fe+2 ions in 500 ml Mohr’s salt solution is 1.26 gm.
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