Molarity = moles of solute/Liters of solution
Let's get moles AgNO3
0.125 M AgNO3 = moles AgNO3/0.250 liters
= 0.03125 moles AgNO3 (169.91 grams/1 mole AgNO3)
= 5.31 grams AgNO3 required
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Let's get moles AgNO3
0.125 M AgNO3 = moles AgNO3/0.250 liters
= 0.03125 moles AgNO3 (169.91 grams/1 mole AgNO3)
= 5.31 grams AgNO3 required
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