7) A 5000ppm solution of Fe3+ must be prepared. This involves the use of 6M HNO3

Question

7) A 5000ppm solution of Fe3+ must be prepared. This involves the use of 6M HNO3 and will need to be brought to 500mL total of solution. How many grams of Fe(NO3)3•9H2O [Iron Nitrate Nonahydrate] must be used to make this solution? FW of Iron Nitrate Nonahydrate: 403.925g/mol Show calculation steps: 7) A 5000ppm solution of Fe3+ must be prepared. This involves the use of 6M HNO3 and will need to be brought to 500mL total of solution. How many grams of Fe(NO3)3•9H2O [Iron Nitrate Nonahydrate] must be used to make this solution? FW of Iron Nitrate Nonahydrate: 403.925g/mol Show calculation steps: How many grams of Fe(NO3)3•9H2O [Iron Nitrate Nonahydrate] must be used to make this solution? FW of Iron Nitrate Nonahydrate: 403.925g/mol Show calculation steps:

Explanation / Answer

5000 ppm = 5000 mg/L

That means 1L=1000mL should contains 5000 mg of iron nitrate nanohydrate

500 mL should contain 5000/2=2500 mg of iron nitrate nanohydrate.

Therefore the mass of iron nitrate nanohydrate needed is 2500 mg=2500*10^-3 g= 2.50 g

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