Question

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?

Answer 1

Answer:

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%.