How many mL of 6.0 M HCl solution will provide 164 g of HCl?

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How many mL of 6.0 M HCl solution will provide 164 g of HCl?

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Answer 1 · 2015-11-18T01:58:58

$0.75 L$ $"0.75 L"$

Explanation:

The idea here is that you need to determine how many moles you get in $164 g$ $"164 g"$ of hydrochloric acid, then use the given solution's molarity to figure out the volume that would contain that many moles.

So, hydrochloric acid has a molar mass of $36.46 g/mol$ $"36.46 g/mol"$ , which means that the $164-g$ $"164-g"$ sample will contain

$164 g \cdot \frac{1 mole HCl}{36.46 g} = 4.498 moles HCl$ $164color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "4.498 moles HCl"$

As you know, molarity is defined as moles of solute, which in this case is hydrochloric acid, divided by liters of solution

$molarity = \frac{moles of solute}{liters of solution}$ $color(blue)("molarity" = "moles of solute"/"liters of solution")$

A $6.0-M$ $"6.0-M"$ hydrochloric acid solution will contain $6$ $6$ moles of hydrochloric acid for every liter of solution. This means that $4.498$ $4.498$ moles will come with a volume of

$c = \frac{n}{V} \Rightarrow V = \frac{n}{c}$ $c = n/V implies V = n/c$

$V = \frac{4.498 moles}{6.0 \frac{moles}{L}} = 0.7497 L$ $V = (4.498color(red)(cancel(color(black)("moles"))))/(6.0color(red)(cancel(color(black)("moles")))/"L") = "0.7497 L"$

Rounded to two sig figs, the number of sig figs you have for the molarity of the solution, the answer will be

$V = 0.75 L$ $V = color(green)("0.75 L")$

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How many mL of 6.0 M HCl solution will provide 164 g of HCl?

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