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How much 0.1n AgNO3 could be prepared from a 5g pure silver?
Accepted Answer
The amount of 0.1N AgNO3 that can be prepared from 5g of pure silver can be calculated using the following steps:
1. Convert the mass of silver to moles using its molar mass (107.87 g/mol):
Moles of Ag = 5 g / 107.87 g/mol = 0.046 mol
2. Calculate the number of equivalents of silver present:
Equivalents of Ag = Moles of Ag x Valence of Ag
Equivalents of Ag = 0.046 mol x 1 = 0.046 equivalents
3. Use the definition of normality to determine the volume of 0.1N AgNO3 that contains 0.046 equivalents of Ag:
Normality = Equivalents of Solute / Volume of Solution (in liters)
0.1N = 0.046 equivalents / Volume of Solution
Volume of Solution = 0.046 equivalents / 0.1N = 0.46 liters = 460 mL
Therefore, 5g of pure silver can be used to prepare 460 mL of 0.1N AgNO3 solution.
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