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

##### 2 Answers

Feb 4, 2018

Mar 1, 2018

I'm getting a bit over

You have

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

And at

#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")#