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What must be the normality of a silver nitrate solution so that each ml shall be equivalent to 6 mg sodium chloride
Accepted Answer
To determine the normality of the silver nitrate solution, we need to consider the chemical reaction between silver nitrate (AgNOₓ) and sodium chloride (NaCl):
2AgNOₓ + 2NaCl → 2AgCl + 2NaNOₓ
From the balanced equation, we can see that 2 moles of AgNOₓ react with 2 moles of NaCl. Therefore, the mole ratio of AgNOₓ to NaCl is 1:1.
Given that 1 mL of the silver nitrate solution should be equivalent to 6 mg of sodium chloride, we can calculate the number of moles of NaCl in 1 mL:
(6 mg NaCl) × (1 g NaCl / 1000 mg NaCl) × (1 mol NaCl / 58.44 g/mol NaCl) = 1.03 × 10⁻-4ₔ mol NaCl
Since 1 mL of the silver nitrate solution is equivalent to 1.03 × 10⁻-4ₔ mol NaCl, and the mole ratio of AgNOₓ to NaCl is 1:1, the number of moles of AgNOₓ in 1 mL of the solution is also 1.03 × 10⁻-4ₔ mol AgNOₓ.
Therefore, the normality of the silver nitrate solution is:
Normality = (Number of moles of AgNOₓ per liter of solution) / 1
Normality = (1.03 × 10⁻-4ₔ mol AgNOₓ / 1 mL) × (1000 mL / 1 L)
Normality = 0.103 N
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