What is the molarity of $53%"w/w"$ $"HNO"_3$ ?

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Answer 1 · 2018-02-04T05:28:07

$M=0.53$

The concentration is the molarity, except that molarity is in decimal form.

Molarity, $c$ , is defined as $c=n/V$ , where $n$ is the moles of the solute and $V$ the volume of solution.

Here, the molarity will be $0.53$ .

Answer 2 · 2018-03-01T21:02:32

I'm getting a bit over $"11 M"$ .

You have

$"53 g HNO"_3/("100 g solution")$

And at $20^@ "C"$ , it has density of $"1.3278 g/mL"$ . Therefore, this has a volume of:

$100 cancel"g solution" xx "1 mL"/(1.3278 cancel"g soln") = "75.31 mL"$

$=$ $"0.07531 L"$

As a result, this has a molarity of:

$color(blue)(["HNO"_3]_(53%)) = (53 cancel"g" xx "1 mol"/(63.0119 cancel("g HNO"_3)))/("0.07531 L")$

$~~$ $color(blue)("11.17 M")$

What is the molarity of 53%"w/w"53%"w/w" "HNO"_3"HNO"_3?