How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?

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How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?

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Answer 1 · 2016-09-24T19:16:38

You must add 3 L of the 90 % acid.

Explanation:

For this problem, we can use the relation

$color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = "amount of solute" color(white)(a/a)|)))" "$

Let the $"volume of 90 % acid" = xcolor(white)(l) "L"$

Then, after mixing, we have $(6 + x") L of 40 % acid"$ .

This is made up of $"6 L of 15 % acid"$ and $xcolor(white)(l) "L of 90 % acid"$ .

$"Moles before = moles after"$

$c_1V_1 = c_2V_2 + c_3V_3$

$40 color(red)(cancel(color(black)(%))) × (6 + x) color(red)(cancel(color(black)("L"))) = 15 color(red)(cancel(color(black)(%))) × 6 color(red)(cancel(color(black)("L"))) + 90 color(red)(cancel(color(black)(%))) × x color(red)(cancel(color(black)("L")))$

$240 + 40x = 90 + 90x$

$50x = 150$

$x = 150/50 = 3$

So, we add $"3 L of 90 % acid"$ to $"6 L of 15 % acid"$ and get $"9 L of 40 % acid"$ .

Check:

$3 × 90 + 6 × 15 = 9 × 40$

$270 + 90 = 360$

$360 = 360$

It checks!

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How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?

1 Answer

Explanation:

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