An impure sample of table salt that weighed 0.8421 g, when dissolved in water and treated with excess AgNO3, formed 2.044 g of AgCl. what is the percentage of NaCl in the impure sample?

Question

Chemistry

Leonardo Huber

1 October, 21:51

An impure sample of table salt that weighed 0.8421 g, when dissolved in water and treated with excess AgNO3, formed 2.044 g of AgCl. what is the percentage of NaCl in the impure sample?

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Shania · Answer 1

99.24%.

Explanation:

NaCl reacted with AgNO₃ as in the balanced equation:

NaCl + AgNO₃ → AgCl (↓) + NaNO₃,

1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.

We need to calculate the no. of moles of AgCl produced:

no. of moles of AgCl = mass/molar mass = (2.044 g) / (143.32 g/mol) = 0.0143 mol.

Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:

using cross multiplication:

1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.

∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.

Now, we can get the mass of puree NaCl in the sample:

mass of pure NaCl = (no. of moles of pure NaCl) (molar mass of NaCl) = (0.0143 mol) (58.44 g/mol) = 0.8357 g.

∴ The percentage of NaCl in the impure sample = [ (mass of pure NaCl) / (mass of the impure sample) ] x 100 = [ (0.8357 g) / (0.8421 g) ] x 100 = 99.24%.