Given that a 0.160 M HCl(aq) solution costs $39.95 for 500 mL , and that KCl costs $10/ton, which analysis procedure is more cost-effective?
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KCl
To determine which analysis procedure is more cost-effective, we need to compare the costs of using HCl solution versus KCl. Let's break down the steps to calculate the costs for each procedure.
1. HCl Solution:
The given concentration of HCl solution is 0.160 M, which means it contains 0.160 moles of HCl per liter. We have a 500 mL solution, so we need to calculate the number of moles in this volume:
0.160 moles/L * 0.5 L = 0.080 moles of HCl
The cost of the HCl solution is $39.95 for 500 mL or 0.5 L. So, the cost per mole of HCl is:
$39.95 / 0.080 moles = $499.375/mole
2. KCl:
The cost of KCl is given as $10/ton. To convert this to the cost per mole, we need to know the molar mass of KCl. The molar mass of KCl is:
K (39.10 g/mol) + Cl (35.45 g/mol) = 74.55 g/mol
Next, we need to convert the cost of KCl into the cost per mole:
$10/ton = $10 / (1000 kg/ton) * (1 ton/2000 lbs) * (2000 lbs/453.59 g) * (1 g/0.07455 moles) ≈ $0.877/mole
Now that we have the cost per mole for both HCl and KCl, we can compare them:
Cost of HCl (per mole) = $499.375/mole
Cost of KCl (per mole) = $0.877/mole
Clearly, the cost of using KCl is significantly lower than the cost of using HCl solution. Therefore, the analysis procedure involving KCl is more cost-effective.