Question

Given that a 0.130 M HCl(aq) solution costs $39.95 for 500 mL, and that KCl costs $10/ton, which analysis procedure is more cost-effective

Answer 1

KCl is cost effective

Step-by-step explanation:

In order to know this, we need to see how much it cost 1 g of each reactant. Let's begin with HCl

HCl:

In this case, let's calculate the moles of HCl in a 0.130 M solution and then, the mass of HCl using the molecular weight of 36.5 g/mol, to get the cost the HCl at the end using the given price:

nHCl = 0.130 moles/L * 0.5 L = 0.065 moles

mHCl = 0.065 moles * 36.5 g/mol = 2.3725 g

Cost HCl = 39.95 $ / 2.3725 g = 16.84 $/g

Conclusion, 1 g of HCl costs 16.84 $

KCl:

In this case, it's pretty obvious that 1 ton of KCl cost 10$, so, there is no need to do further calculations because 1 ton (or more than 1000 kg of the salt) it's just 10$. This is less expensive than the 16.84$ for just 1 g of HCl, so, final conclusion, KCl is more cost-effective.

Hope this helps