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

Question

2134 views around the world

You can reuse this answer
Creative Commons License

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")#