Answer:
1.451 %
Explanation:
Step 1: Write the balanced equation for the precipitation reaction
AgNO₃(aq) + Cl⁻(aq) = AgCl(s) + NO₃⁻(aq)
Step 2: Calculate the reacting moles of AgNO₃
42.00 mL of 0.2998 M AgNO₃ reacted.
0.04200 L × 0.2998 mol/L = 0.01259 mol
Step 3: Calculate the moles of Cl⁻ that reacted with 0.01259 moles of AgNO₃
The molar ratio of AgNO₃ to Cl⁻ is 1:1. The moles of Cl⁻ that reacted are 1/1 × 0.01259 mol = 0.01259 mol.
Step 4: Calculate the mass corresponding to 0.01259 moles of Cl⁻
The molar mass of Cl⁻ is 35.45 g/mol.
0.01259 mol × 35.45 g/mol = 0.4463 g
Step 5: Calculate the mass of the sample of seawater
We have 30.00 mL of a sample of seawater with a density of 1.025 g/mL.
30.00 mL × 1.025 g/mL = 30.75 g
Step 6: Calculate the mass percentage of Cl⁻in the sample of seawater
We will use the following expression.
%Cl⁻ = mass Cl⁻/ mass seawater × 100%
%Cl⁻ = 0.4463 g/30.75 g × 100% = 1.451 %
