To prepare 500 ml of 0.0750 M AgNo3 from the solid reagent, we need to first calculate the amount of AgNo3 required.
The molar mass of AgNo3 is:
Ag = 107.87 g/mol
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of AgNo3 = 107.87 + 14.01 + (3 x 16.00) = 169.87 g/mol
To prepare 500 ml of 0.0750 M solution, we need:
Moles of AgNo3 = Molarity x Volume
Moles = 0.0750 mol/L x 0.500 L = 0.0375 moles
Now, we can calculate the mass of AgNo3 required:
Mass = Moles x Molar mass
Mass = 0.0375 moles x 169.87 g/mol = 6.37 g
Therefore, to prepare 500 ml of 0.0750 M AgNo3 solution, we need to weigh out 6.37 grams of AgNo3 and dissolve it in sufficient water to make a total volume of 500 ml.
Molar Analytical Concentration: Write the complete solution.
1. Describe the preparation of 500 ml of 0.0750 M AgNo3 from the solid reagent.
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