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What must be the normality of a solution of silver nitrate so that each ml shall be equivalent to 6 mg of NaCl?
Accepted Answer
To determine the normality of a silver nitrate solution that will be equivalent to 6 mg of NaCl per ml, we need to use the following steps:
1. Convert the mass of NaCl to moles: 6 mg NaCl * (1 g / 1000 mg) * (1 mol NaCl / 58.44 g NaCl) = 1.027 x 10^-4 mol NaCl
2. Since the reaction between silver nitrate and sodium chloride is a 1:1 mole ratio, the moles of silver nitrate required are also 1.027 x 10^-4 mol.
3. Calculate the volume of the silver nitrate solution needed: 1.027 x 10^-4 mol AgNO3 * (1000 mL / 1 L) / 1 mL = 0.1027 mL
4. Calculate the normality of the silver nitrate solution: Normality = (Moles of solute / Volume of solution in liters)
Normality = (1.027 x 10^-4 mol) / (0.1027 mL * (1 L / 1000 mL)) = 0.1 N
Therefore, the normality of the silver nitrate solution should be 0.1 N so that each ml of the solution is equivalent to 6 mg of NaCl.
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